हल

$\sin 2 x$ का एंटी डेरिवेटिव एक ऐसे फलन है जिसका अवकलज $\sin 2 x$ होता है।

ज्ञात है कि,

$\dfrac{d}{d x}(\cos 2 x)=-2 \sin 2 x$

$\Rightarrow \sin 2 x=-\dfrac{1}{2} \dfrac{d}{d x}(\cos 2 x)$

$\therefore \sin 2 x=\dfrac{d}{d x}(-\dfrac{1}{2} \cos 2 x)$

इसलिए, $\sin 2 x$ का एंटी डेरिवेटिव $-\dfrac{1}{2} \cos 2 x$ होता है।

हल

$\cos 3 x$ का एंटी डेरिवेटिव एक ऐसे फलन है जिसका अवकलज $\cos 3 x$ होता है।

ज्ञात है कि,

$\dfrac{d}{d x}(\sin 3 x)=3 \cos 3 x$

$\Rightarrow \cos 3 x=\dfrac{1}{3} \dfrac{d}{d x}(\sin 3 x)$

$\therefore \cos 3 x=\dfrac{d}{d x}(\dfrac{1}{3} \sin 3 x)$

इसलिए, $\cos 3 x$ का एंटी डेरिवेटिव $\dfrac{1}{3} \sin 3 x$ होता है।

हल

$e^{2 x}$ का एंटी डेरिवेटिव एक ऐसे फलन है जिसका अवकलज $e^{2 x}$ होता है।

ज्ञात है कि, $\dfrac{d}{d x}(e^{2 x})=2 e^{2 x}$

$\Rightarrow e^{2 x}=\dfrac{1}{2} \dfrac{d}{d x}(e^{2 x})$

$\therefore e^{2 x}=\dfrac{d}{d x}(\dfrac{1}{2} e^{2 x})$

इसलिए, $e^{2 x}$ का एंटी डेरिवेटिव $\dfrac{1}{2} e^{2 x}$ होता है।

हल

$(a x+b)^{2}$ का एंटी डेरिवेटिव एक ऐसे फलन है जिसका अवकलज $(a x+b)^{2}$ होता है।

ज्ञात है कि,

$\dfrac{d}{d x}(a x+b)^{3}=3 a(a x+b)^{2}$

$\Rightarrow(a x+b)^{2}=\dfrac{1}{3 a} \dfrac{d}{d x}(a x+b)^{3}$

$\therefore(a x+b)^{2}=\dfrac{d}{d x}(\dfrac{1}{3 a}(a x+b)^{3})$

इसलिए, $(a x+b)^{2}$ का एंटी डेरिवेटिव $\dfrac{1}{3 a}(a x+b)^{3}$ होता है।

हल

अंतगत फलन $(\sin 2 x-4 e^{3 x})$ वह $x$ के फलन है जिसका अवकलज $(\sin 2 x-4 e^{3 x})$ है।

यह ज्ञात है कि, $\dfrac{d}{d x}(-\dfrac{1}{2} \cos 2 x-\dfrac{4}{3} e^{3 x})=\sin 2 x-4 e^{3 x}$

इसलिए, $(\sin 2 x-4 e^{3 x})$ का अंतगत फलन $(-\dfrac{1}{2} \cos 2 x-\dfrac{4}{3} e^{3 x})$ है।

हल

$ \begin{aligned} & \int(4 e^{3 x}+1) d x \\ & =4 \int e^{3 x} d x+\int 1 d x \\ & =4(\dfrac{e^{3 x}}{3})+x+C \\ & =\dfrac{4}{3} e^{3 x}+x+C \end{aligned} $

हल

$\int x^{2}(1-\dfrac{1}{x^{2}}) d x$

$=\int(x^{2}-1) d x$

$=\int x^{2} d x-\int 1 d x$

$=\dfrac{x^{3}}{3}-x+C$

हल

$\int(a x^{2}+b x+c) d x$

$=a \int x^{2} d x+b \int x d x+c \int 1 \cdot d x$

$=a(\dfrac{x^{3}}{3})+b(\dfrac{x^{2}}{2})+c x+C$

$=\dfrac{a x^{3}}{3}+\dfrac{b x^{2}}{2}+c x+C$

हल

$\int(2 x^{2}+e^{x}) d x$

$=2 \int x^{2} d x+\int e^{x} d x$

$=2(\dfrac{x^{3}}{3})+e^{x}+C$

$=\dfrac{2}{3} x^{3}+e^{x}+C$

हल

$ \begin{aligned} & \int(\sqrt{x}-\dfrac{1}{\sqrt{x}})^{2} d x \\ & =\int(x+\dfrac{1}{x}-2) d x \\ & =\int x d x+\int \dfrac{1}{x} d x-2 \int 1 \cdot d x \\ & =\dfrac{x^{2}}{2}+\log |x|-2 x+C \end{aligned} $

हल

$ \begin{aligned} & \int \dfrac{x^{3}+5 x^{2}-4}{x^{2}} d x \\ & =\int(x+5-4 x^{-2}) d x \\ & =\int x d x+5 \int 1 \cdot d x-4 \int x^{-2} d x \\ & =\dfrac{x^{2}}{2}+5 x-4(\dfrac{x^{-1}}{-1})+C \\

& =\dfrac{x^{2}}{2}+5 x+\dfrac{4}{x}+C \end{aligned} $

Solution

$ \begin{aligned} & \int \dfrac{x^{3}+3 x+4}{\sqrt{x}} d x \\ & =\int(x^{\dfrac{5}{2}}+3 x^{\dfrac{1}{2}}+4 x^{-\dfrac{1}{2}}) d x \\ & =\dfrac{x^{\dfrac{7}{2}}}{\dfrac{7}{2}}+\dfrac{3(x^{\dfrac{3}{2}})}{\dfrac{3}{2}}+\dfrac{4(x^{\dfrac{1}{2}})}{\dfrac{1}{2}}+C \\ & =\dfrac{2}{7} x^{\dfrac{7}{2}}+2 x^{\dfrac{3}{2}}+8 x^{\dfrac{1}{2}}+C \\ & =\dfrac{2}{7} x^{\dfrac{7}{2}}+2 x^{\dfrac{3}{2}}+8 \sqrt{x}+C \end{aligned} $

Solution

$\int \dfrac{x^{3}-x^{2}+x-1}{x-1} d x$

अंशांतर करने पर, हम प्राप्त करते हैं

$=\int(x^{2}+1) d x$

$=\int x^{2} d x+\int 1 d x$

$=\dfrac{x^{3}}{3}+x+C$

Solution

$\int(1-x) \sqrt{x} d x$

$=\int(\sqrt{x}-x^{\dfrac{3}{2}}) d x$

$=\int x^{\dfrac{1}{2}} d x-\int x^{\dfrac{3}{2}} d x$

$=\dfrac{x^{\dfrac{3}{2}}}{\dfrac{3}{2}}-\dfrac{x^{\dfrac{5}{2}}}{\dfrac{5}{2}}+C$

$=\dfrac{2}{3} x^{\dfrac{3}{2}}-\dfrac{2}{5} x^{\dfrac{5}{2}}+C$

Solution

$ \int \sqrt{x}(3 x^{2}+2 x+3) d x $

$=\int(3 x^{\dfrac{5}{2}}+2 x^{\dfrac{3}{2}}+3 x^{\dfrac{1}{2}}) d x$

$=3 \int x^{\dfrac{5}{2}} d x+2 \int x^{\dfrac{3}{2}} d x+3 \int x^{\dfrac{1}{2}} d x$

$=3(\dfrac{x^{\dfrac{7}{2}}}{\dfrac{7}{2}})+2(\dfrac{x^{\dfrac{5}{2}}}{\dfrac{5}{2}})+3 \dfrac{(x^{\dfrac{3}{2}})}{\dfrac{3}{2}}+C$

$=\dfrac{6}{7} x^{\dfrac{7}{2}}+\dfrac{4}{5} x^{\dfrac{5}{2}}+2 x^{\dfrac{3}{2}}+C$

Solution

$\int(2 x-3 \cos x+e^{x}) d x$

$=2 \int x d x-3 \int \cos x d x+\int e^{x} d x$

$=\dfrac{2 x^{2}}{2}-3(\sin x)+e^{x}+C$

$=x^{2}-3 \sin x+e^{x}+C$

Solution

$\int(2 x^{2}-3 \sin x+5 \sqrt{x}) d x$ $=2 \int x^{2} d x-3 \int \sin x d x+5 \int x^{\dfrac{1}{2}} d x$

$=\dfrac{2 x^{3}}{3}-3(-\cos x)+5(\dfrac{x^{\dfrac{3}{2}}}{\dfrac{3}{2}})+C$

$=\dfrac{2}{3} x^{3}+3 \cos x+\dfrac{10}{3} x^{\dfrac{3}{2}}+C$

Solution

$\int \sec x(\sec x+\tan x) d x$

$=\int(\sec ^{2} x+\sec x \tan x) d x$

$=\int \sec ^{2} x d x+\int \sec x \tan x d x$

$=\tan x+\sec x+C$

Solution

$\int \dfrac{\sec ^{2} x}{cosec^{2}x } d x$

$ \begin{aligned} & =\int \dfrac{\dfrac{1}{\cos ^{2} x}}{\dfrac{1}{\sin ^{2} x}} d x \\ & =\int \dfrac{\sin ^{2} x}{\cos ^{2} x} d x \\ & =\int \tan ^{2} x d x \\ & =\int(\sec ^{2} x-1) d x \\ & =\int \sec ^{2} x d x-\int 1 d x \\ & =\tan x-x+C \end{aligned} $

Solution

$\int \dfrac{2-3 \sin x}{\cos ^{2} x} d x$

$=\int(\dfrac{2}{\cos ^{2} x}-\dfrac{3 \sin x}{\cos ^{2} x}) d x$

$=\int 2 \sec ^{2} x d x-3 \int \tan x \sec x d x$

$=2 \tan x-3 \sec x+C$

Solution

$ (\sqrt{x}+\dfrac{1}{\sqrt{x}}) d x $

$=\int x^{\dfrac{1}{2}} d x+\int x^{-\dfrac{1}{2}} d x$

$=\dfrac{x^{\dfrac{3}{2}}}{\dfrac{3}{2}}+\dfrac{x^{\dfrac{1}{2}}}{\dfrac{1}{2}}+C$

$=\dfrac{2}{3} x^{\dfrac{3}{2}}+2 x^{\dfrac{1}{2}}+C$

इसलिए, सही उत्तर C है।

हल

दिया गया है कि,

$\dfrac{d}{d x} [f(x)]=4 x^{3}-\dfrac{3}{x^{4}}$

$\therefore$ $4 x^{3}-\dfrac{3}{x^{4}}$ का विपरीत समाकलन $f(x)$ है

$\therefore f(x)=\int 4 x^{3}-\dfrac{3}{x^{4}} d x$

$f(x)=4 \int x^{3} d x-3 \int(x^{-4}) d x$

$f(x)=4(\dfrac{x^{4}}{4})-3(\dfrac{x^{-3}}{-3})+C$

$\therefore f(x)=x^{4}+\dfrac{1}{x^{3}}+C$

इसके अतिरिक्त,

$f(2)=0$

$\therefore f(2)=(2)^{4}+\dfrac{1}{(2)^{3}}+C=0$

$\Rightarrow 16+\dfrac{1}{8}+C=0$

$\Rightarrow C=-(16+\dfrac{1}{8})$

$\Rightarrow C=\dfrac{-129}{8}$

$\therefore f(x)=x^{4}+\dfrac{1}{x^{3}}-\dfrac{129}{8}$

इसलिए, सही उत्तर A है।

हल

मान लीजिए $1+x^{2}=t$

$\therefore 2 x d x=d t$

$\Rightarrow \int \dfrac{2 x}{1+x^{2}} d x=\int \dfrac{1}{t} d t$

$=\log |t|+C$

$=\log |1+x^{2}|+C$

$=\log (1+x^{2})+C$

हल

मान लीजिए $\log |x|=t$

$\therefore \dfrac{1}{x} d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{(\log |x|)^{2}}{x} d x & =\int t^{2} d t \\ & =\dfrac{t^{3}}{3}+C \\ & =\dfrac{(\log |x|)^{3}}{3}+C \end{aligned} $

हल

$\dfrac{1}{x+x \log x}=\dfrac{1}{x(1+\log x)}$

मान लीजिए $1+\log x=t$

$\therefore \dfrac{1}{x} d x=d t$

$\Rightarrow \int \dfrac{1}{x(1+\log x)} d x=\int \dfrac{1}{t} d t$

$=\log |t|+C$

$=\log |1+\log x|+C$

हल

$\sin x \cdot \sin (\cos x)$

मान लीजिए $\cos x=t$

$\therefore-\sin x d x=d t$

$ \begin{aligned} \Rightarrow \int \sin x \cdot \sin (\cos x) d x & =-\int \sin t d t \\ & =-[-\cos t]+C \\ & =\cos t+C \\ & =\cos (\cos x)+C \end{aligned} $

हल

$\sin (a x+b) \cos (a x+b)=\dfrac{2 \sin (a x+b) \cos (a x+b)}{2}=\dfrac{\sin 2(a x+b)}{2}$

मान लीजिए $2(a x+b)=t$

$\therefore 2 a d x=d t$

$\Rightarrow \int \dfrac{\sin 2(a x+b)}{2} d x=\dfrac{1}{2} \int \dfrac{\sin t d t}{2 a}$

$ \begin{aligned} & =\dfrac{1}{4 a}[-\cos t]+C \\ & =\dfrac{-1}{4 a} \cos 2(a x+b)+C \end{aligned} $

हल

मान लीजिए $a x+b=t$

$\Rightarrow a d x=d t$

$\therefore d x=\dfrac{1}{a} d t$

$\Rightarrow \int(a x+b)^{\dfrac{1}{2}} d x=\dfrac{1}{a} \int t^{\dfrac{1}{2}} d t$

$=\dfrac{1}{a}(\dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}})+C$

$=\dfrac{2}{3 a}(a x+b)^{\dfrac{3}{2}}+C$

Solution

Let $(x+2)=t$

$\therefore d x=d t$

$ \begin{aligned} \Rightarrow \int x \sqrt{x+2} d x & =\int(t-2) \sqrt{t} d t \\ & =\int(t^{\dfrac{3}{2}}-2 t^{\dfrac{1}{2}}) d t \\ & =\int t^{\dfrac{3}{2}} d t-2 \int t^{\dfrac{1}{2}} d t \\ & =\dfrac{t^{\dfrac{5}{2}}}{\dfrac{5}{2}}-2(\dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}})+C \\ & =\dfrac{2}{5} t^{\dfrac{5}{2}}-\dfrac{4}{3} t^{\dfrac{3}{2}}+C \\ & =\dfrac{2}{5}(x+2)^{\dfrac{5}{2}}-\dfrac{4}{3}(x+2)^{\dfrac{3}{2}}+C \end{aligned} $

Solution

Let $1+2 x^{2}=t$

$\therefore 4 x d x=d t$

$\Rightarrow \int x \sqrt{1+2 x^{2}} d x=\int \dfrac{\sqrt{t} d t}{4}$

$=\dfrac{1}{4} \int t^{\dfrac{1}{2}} d t$

$=\dfrac{1}{4}(\dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}})+C$

$=\dfrac{1}{6}(1+2 x^{2})^{\dfrac{3}{2}}+C$

Solution

Let $x^{2}+x+1=t$

$\therefore(2 x+1) d x=d t$

$\int(4 x+2) \sqrt{x^{2}+x+1} d x$

$=\int 2 \sqrt{t} d t$

$=2 \int \sqrt{t} d t$

$=2(\dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}})+C$

$=\dfrac{4}{3}(x^{2}+x+1)^{\dfrac{3}{2}}+C$

Solution

$\dfrac{1}{x-\sqrt{x}}=\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}$

Let $(\sqrt{x}-1)=t$

$\therefore \dfrac{1}{2 \sqrt{x}} d x=d t$

$\Rightarrow \int \dfrac{1}{\sqrt{x}(\sqrt{x}-1)} d x=\int \dfrac{2}{t} d t$

$=2 \log |t|+C$

$=2 \log |\sqrt{x}-1|+C$

Solution

Let $x+4=t$

$\therefore d x=d t$

$ \begin{aligned} \int \dfrac{x}{\sqrt{x+4}} d x & =\int \dfrac{(t-4)}{\sqrt{t}} d t \\ & =\int(\sqrt{t}-\dfrac{4}{\sqrt{t}}) d t \\ & =\dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}}-4(\dfrac{t^{\dfrac{1}{2}}}{\dfrac{1}{2}})+C \\ `

$$ \begin{aligned} & =\dfrac{2}{3}(t)^{\dfrac{3}{2}}-8(t)^{\dfrac{1}{2}}+C \\ & =\dfrac{2}{3} t \cdot t^{\dfrac{1}{2}}-8 t^{\dfrac{1}{2}}+C \\ & =\dfrac{2}{3} t^{\dfrac{1}{2}}(t-12)+C \\ & =\dfrac{2}{3}(x+4)^{\dfrac{1}{2}}(x+4-12)+C \\ & =\dfrac{2}{3} \sqrt{x+4}(x-8)+C \end{aligned} $$

Solution

Let $x^{3}-1=t$

$\therefore 3 x^{2} d x=d t$ $\Rightarrow \int(x^{3}-1)^{\dfrac{1}{3}} x^{5} d x=\int(x^{3}-1)^{\dfrac{1}{3}} x^{3} \cdot x^{2} d x$

$=\int t^{\dfrac{1}{3}}(t+1) \dfrac{d t}{3}$

$=\dfrac{1}{3} \int(t^{\dfrac{4}{3}}+t^{\dfrac{1}{3}}) d t$

$=\dfrac{1}{3}[\dfrac{t^{\dfrac{7}{3}}}{\dfrac{7}{3}}+\dfrac{t^{\dfrac{4}{3}}}{\dfrac{4}{3}}]+C$

$=\dfrac{1}{3}[\dfrac{3}{7} t^{\dfrac{7}{3}}+\dfrac{3}{4} t^{\dfrac{4}{3}}]+C$

$=\dfrac{1}{7}(x^{3}-1)^{\dfrac{7}{3}}+\dfrac{1}{4}(x^{3}-1)^{\dfrac{4}{3}}+C$

Solution

Let $2+3 x^{3}=t$

$\therefore 9 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{x^{2}}{(2+3 x^{3})^{3}} d x & =\dfrac{1}{9} \int \dfrac{d t}{(t)^{3}} \\ & =\dfrac{1}{9}[\dfrac{t^{-2}}{-2}]+C \\ & =\dfrac{-1}{18}(\dfrac{1}{t^{2}})+C \\ & =\dfrac{-1}{18(2+3 x^{3})^{2}}+C \end{aligned} $

Solution

Let $\log x=t$

$\therefore \dfrac{1}{x} d x=d t$

$\Rightarrow \int \dfrac{1}{x(\log x)^{m}} d x=\int \dfrac{d t}{(t)^{m}}$

$=(\dfrac{t^{-m+1}}{1-m})+C$

$=\dfrac{(\log x)^{1-m}}{(1-m)}+C$

Solution

Let $9-4 x^{2}=t$

$\therefore-8 x d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{x}{9-4 x^{2}} d x & =\dfrac{-1}{8} \int_t^{1} d t \\ & =\dfrac{-1}{8} \log |t|+C \\ & =\dfrac{-1}{8} \log |9-4 x^{2}|+C \end{aligned} $

हल

मान लीजिए $2 x+3=t$

$\therefore 2 d x=d t$

$ \begin{aligned} \Rightarrow \int e^{2 x+3} d x & =\dfrac{1}{2} \int e^{t} d t \\ & =\dfrac{1}{2}(e^{t})+C \\ & =\dfrac{1}{2} e^{(2 x+3)}+C \end{aligned} $

हल

मान लीजिए $x^{2}=t$

$\therefore 2 x d x=d t$

$\Rightarrow \int \dfrac{x}{e^{x^{2}}} d x=\dfrac{1}{2} \int \dfrac{1}{e^{t}} d t$

$=\dfrac{1}{2} \int e^{-t} d t$

$=\dfrac{1}{2}(\dfrac{e^{-t}}{-1})+C$

$=-\dfrac{1}{2} e^{-x^{2}}+C$

$=\dfrac{-1}{2 e^{x^{2}}}+C$

हल

मान लीजिए $\tan ^{-1} x=t$

$\therefore \dfrac{1}{1+x^{2}} d x=d t$

$\Rightarrow \int \dfrac{e^{\tan ^{-1} x}}{1+x^{2}} d x=\int e^{t} d t$

$=e^{t}+C$

$=e^{\tan ^{-1} x}+C$

हल

$\dfrac{e^{2 x}-1}{e^{2 x}+1}$

अंश और हर को $e^{x}$ से विभाजित करने पर, हम प्राप्त करते हैं

$ \dfrac{\dfrac{(e^{2 x}-1)}{e^{x}}}{(e^{2 x}+1)}=\dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}} $

मान लीजिए $e^{x}+e^{-x}=t$

$\therefore(e^{x}-e^{-x}) d x=d t$

$\Rightarrow \int \dfrac{e^{2 x}-1}{e^{2 x}+1} d x=\int \dfrac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x$

$=\int \dfrac{d t}{t}$

$=\log |t|+C$

$=\log |e^{x}+e^{-x}|+C$

हल

मान लीजिए $e^{2 x}+e^{-2 x}=t$

$\therefore(2 e^{2 x}-2 e^{-2 x}) d x=d t$

$\Rightarrow 2(e^{2 x}-e^{-2 x}) d x=d t$

$\Rightarrow \int(\dfrac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}) d x=\int \dfrac{d t}{2 t}$

$=\dfrac{1}{2} \int_t^{1} d t$

$=\dfrac{1}{2} \log |t|+C$

$=\dfrac{1}{2} \log |e^{2 x}+e^{-2 x}|+C$

हल

$\tan ^{2}(2 x-3)=\sec ^{2}(2 x-3)-1$

मान लीजिए $2 x-3=t$

$\therefore 2 d x=d t$

$ \begin{aligned} \Rightarrow \int \tan ^{2}(2 x-3) d x & =\int[(\sec ^{2}(2 x-3))-1] d x \\

& =\dfrac{1}{2} \int(\sec ^{2} t) d t-\int 1 d x \\ & =\dfrac{1}{2} \int \sec ^{2} t d t-\int 1 d x \\ & =\dfrac{1}{2} \tan t-x+C \\ & =\dfrac{1}{2} \tan (2 x-3)-x+C \end{aligned} $

Solution

मान लीजिए $7-4 x=t$

$\therefore-4 d x=d t$

$ \begin{aligned} \therefore \int \sec ^{2}(7-4 x) d x & =\dfrac{-1}{4} \int \sec ^{2} t d t \\ & =\dfrac{-1}{4}(\tan t)+C \\ & =\dfrac{-1}{4} \tan (7-4 x)+C \end{aligned} $

Solution

मान लीजिए $\sin ^{-1} x=t$

$\therefore \dfrac{1}{\sqrt{1-x^{2}}} d x=d t$

$\Rightarrow \int \dfrac{\sin ^{-1} x}{\sqrt{1-x^{2}}} d x=\int t d t$

$=\dfrac{t^{2}}{2}+C$

$=\dfrac{(\sin ^{-1} x)^{2}}{2}+C$

Solution

$\dfrac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}=\dfrac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)}$

मान लीजिए $3 \cos x+2 \sin x=t$

$\therefore(-3 \sin x+2 \cos x) d x=d t$

$ \begin{aligned} \int \dfrac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} d x & =\int \dfrac{d t}{2 t} \\ & =\dfrac{1}{2} \int \dfrac{1}{t} d t \\ & =\dfrac{1}{2} \log |t|+C \\ & =\dfrac{1}{2} \log |2 \sin x+3 \cos x|+C \end{aligned} $

Solution

$\dfrac{1}{\cos ^{2} x(1-\tan x)^{2}}=\dfrac{\sec ^{2} x}{(1-\tan x)^{2}}$

मान लीजिए $(1-\tan x)=t$

$\therefore-\sec ^{2} x d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{\sec ^{2} x}{(1-\tan x)^{2}} d x & =\int \dfrac{-d t}{t^{2}} \\ & =-\int t^{-2} d t \\ & =+\dfrac{1}{t}+C \\ & =\dfrac{1}{(1-\tan x)}+C \end{aligned} $

Solution

मान लीजिए $\sqrt{x}=t$

$\therefore \dfrac{1}{2 \sqrt{x}} d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{\cos \sqrt{x}}{\sqrt{x}} d x & =2 \int \cos t d t \\ `

& =2 \sin t+C \\ & =2 \sin \sqrt{x}+C \end{aligned} $

Solution

Let $\sin 2 x=t$

$\therefore 2 \cos 2 x d x=d t$

$ \begin{aligned} \Rightarrow \int \sqrt{\sin 2 x} \cos 2 x d x & =\dfrac{1}{2} \int \sqrt{t} d t \\ & =\dfrac{1}{2}(\dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}})+C \\ & =\dfrac{1}{3} t^{\dfrac{3}{2}}+C \\ & =\dfrac{1}{3}(\sin 2 x)^{\dfrac{3}{2}}+C \end{aligned} $

Solution

Let $1+\sin x=t$

$\therefore \cos x d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{\cos x}{\sqrt{1+\sin x}} d x & =\int \dfrac{d t}{\sqrt{t}} \\ & =\dfrac{t^{\dfrac{1}{2}}}{\dfrac{1}{2}}+C \\ & =2 \sqrt{t}+C \\ & =2 \sqrt{1+\sin x}+C \end{aligned} $

Solution

Let $\log \sin x=t$

$\Rightarrow \dfrac{1}{\sin x} \cdot \cos x d x=d t$

$\therefore \cot x d x=d t$

$\Rightarrow \int \cot x \log \sin x d x=\int t d t$

$ \begin{aligned} & =\dfrac{t^{2}}{2}+C \\ & =\dfrac{1}{2}(\log \sin x)^{2}+C \end{aligned} $

Solution

Let $1+\cos x=t$

$\therefore-\sin x d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{\sin x}{1+\cos x} d x & =\int-\dfrac{d t}{t} \\ & =-\log |t|+C \\ & =-\log |1+\cos x|+C \end{aligned} $

Solution

Let $1+\cos x=t$

$\therefore-\sin x d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{\sin x}{(1+\cos x)^{2}} d x & =\int-\dfrac{d t}{t^{2}} \\ & =-\int t^{-2} d t \\ & =\dfrac{1}{t}+C \\ & =\dfrac{1}{1+\cos x}+\mathbf{C} \end{aligned} $

Solution

$ \begin{aligned} \text{ Let } I & =\int \dfrac{1}{1+\cot x} d x \\ & =\int \dfrac{1}{1+\dfrac{\cos x}{\sin x}} d x \\

& =\int \dfrac{\sin x}{\sin x+\cos x} d x \\ & =\dfrac{1}{2} \int \dfrac{2 \sin x}{\sin x+\cos x} d x \\ & =\dfrac{1}{2} \int \dfrac{(\sin x+\cos x)+(\sin x-\cos x)}{(\sin x+\cos x)} d x \\ & =\dfrac{1}{2} \int 1 d x+\dfrac{1}{2} \int \dfrac{\sin x-\cos x}{\sin x+\cos x} d x \\ & =\dfrac{1}{2}(x)+\dfrac{1}{2} \int \dfrac{\sin x-\cos x}{\sin x+\cos x} d x \end{aligned} $

मान लीजिए $\sin x+\cos x=t \Rightarrow(\cos x-\sin x) d x=d t$

$ \begin{aligned} \therefore I & =\dfrac{x}{2}+\dfrac{1}{2} \int \dfrac{-(d t)}{t} \\ & =\dfrac{x}{2}-\dfrac{1}{2} \log |t|+C \\ & =\dfrac{x}{2}-\dfrac{1}{2} \log |\sin x+\cos x|+C \end{aligned} $

Solution

$ \begin{aligned} \text{ मान लीजिए } I & =\int \dfrac{1}{1-\tan x} d x \\ & =\int \dfrac{1}{1-\dfrac{\sin x}{\cos x}} d x \\ & =\int \dfrac{\cos x}{\cos x-\sin x} d x \\ & =\dfrac{1}{2} \int \dfrac{2 \cos x}{\cos x-\sin x} d x \\ & =\dfrac{1}{2} \int \dfrac{(\cos x-\sin x)+(\cos x+\sin x)}{(\cos x-\sin x)} d x \\ & =\dfrac{1}{2} \int 1 d x+\dfrac{1}{2} \int \dfrac{\cos x+\sin x}{\cos x-\sin x} d x \\ & =\dfrac{x}{2}+\dfrac{1}{2} \int \dfrac{\cos x+\sin x}{\cos x-\sin x} d x \end{aligned} $

मान लीजिए $\cos x-\sin x=t \Rightarrow(-\sin x-\cos x) d x=d t$

$ \begin{aligned} \therefore I & =\dfrac{x}{2}+\dfrac{1}{2} \int \dfrac{-(d t)}{t} \\ & =\dfrac{x}{2}-\dfrac{1}{2} \log |t|+C \\ & =\dfrac{x}{2}-\dfrac{1}{2} \log |\cos x-\sin x|+C \end{aligned} $

Solution

$ \begin{aligned} \text{ मान लीजिए } I & =\int \dfrac{\sqrt{\tan x}}{\sin x \cos x} d x \\ & =\int \dfrac{\sqrt{\tan x} \times \cos x}{\sin x \cos x \times \cos x} d x \\ & =\int \dfrac{\sqrt{\tan x}}{\tan x \cos ^{2} x} d x \\ & =\int \dfrac{\sec ^{2} x d x}{\sqrt{\tan x}} \end{aligned} $

मान लीजिए $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$ \begin{aligned} \therefore I & =\int \dfrac{d t}{\sqrt{t}} \\

& =2 \sqrt{t}+C \\ & =2 \sqrt{\tan x}+C \end{aligned} $

Solution

$\dfrac{(x+1)(x+\log x)^{2}}{x}=(\dfrac{x+1}{x})(x+\log x)^{2}=(1+\dfrac{1}{x})(x+\log x)^{2}$

मान लीजिए $(x+\log x)=t$

$\therefore(1+\dfrac{1}{x}) d x=d t$

$\Rightarrow \int(1+\dfrac{1}{x})(x+\log x)^{2} d x=\int t^{2} d t$

$ \begin{aligned} & =\dfrac{t^{3}}{3}+C \\ & =\dfrac{1}{3}(x+\log x)^{3}+C \end{aligned} $

Solution

$\dfrac{(x+1)(x+\log x)^{2}}{x}=(\dfrac{x+1}{x})(x+\log x)^{2}=(1+\dfrac{1}{x})(x+\log x)^{2}$

मान लीजिए $(x+\log x)=t$

$\therefore(1+\dfrac{1}{x}) d x=d t$

$\Rightarrow \int(1+\dfrac{1}{x})(x+\log x)^{2} d x=\int t^{2} d t$

$ \begin{aligned} & =\dfrac{t^{3}}{3}+C \\ & =\dfrac{1}{3}(x+\log x)^{3}+C \end{aligned} $

Solution

मान लीजिए $x^{4}=t$

$\therefore 4 x^{3} d x=d t$

$\Rightarrow \int \dfrac{x^{3} \sin (\tan ^{-1} x^{4})}{1+x^{8}} d x=\dfrac{1}{4} \int \dfrac{\sin (\tan ^{-1} t)}{1+t^{2}} d t \qquad …(1)$

मान लीजिए $\tan ^{-1} t=u$

$\therefore \dfrac{1}{1+t^{2}} d t=d u$

समीकरण (1) से, हम प्राप्त करते हैं

$ \begin{aligned} & \begin{aligned} & \int \dfrac{x^{3} \sin (\tan ^{-1} x^{4}) d x}{1+x^{8}}=\dfrac{1}{4} \int \sin u d u \\ &=\dfrac{1}{4}(-\cos u)+C \\ &=\dfrac{-1}{4} \cos (\tan ^{-1} t)+C \\ &= \dfrac{-1}{4} \cos (\tan ^{-1} x^{4})+C \end{aligned} \end{aligned} $

Solution

मान लीजिए $x^{10}+10^{x}=t$

$\therefore(10 x^{9}+10^{x} \log _{e} 10) d x=d t$

$\Rightarrow \int \dfrac{10 x^{9}+10^{x} \log _{e} 10}{x^{10}+10^{x}} d x=\int \dfrac{d t}{t}$

$=\log t+C$

$=\log (10^{x}+x^{10})+C$

इसलिए, सही उत्तर D है।

हल

मान लीजिए $I=\int \dfrac{d x}{\sin ^{2} x \cos ^{2} x}$

$=\int \dfrac{1}{\sin ^{2} x \cos ^{2} x} d x$

$=\int \dfrac{\sin ^{2} x+\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$

$=\int \dfrac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x} d x+\int \dfrac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$

$=\int \sec ^{2} x d x+\int cosec^{2} x d x$

$=\tan x-\cot x+C$

इसलिए, सही उत्तर B है।

हल

$ \begin{aligned} & \sin ^{2}(2 x+5)=\dfrac{1-\cos 2(2 x+5)}{2}=\dfrac{1-\cos (4 x+10)}{2} \\ & \Rightarrow \int \sin ^{2}(2 x+5) d x=\int \dfrac{1-\cos (4 x+10)}{2} d x \\ & =\dfrac{1}{2} \int 1 d x-\dfrac{1}{2} \int \cos (4 x+10) d x \\ & =\dfrac{1}{2} x-\dfrac{1}{2}(\dfrac{\sin (4 x+10)}{4})+C \\ & =\dfrac{1}{2} x-\dfrac{1}{8} \sin (4 x+10)+C \end{aligned} $

हल

ज्ञात है कि, $\sin A \cos B=\dfrac{1}{2}{\sin (A+B)+\sin (A-B)}$

$\therefore \int \sin 3 x \cos 4 x d x=\dfrac{1}{2} \int{\sin (3 x+4 x)+\sin (3 x-4 x)} d x$

$ =\dfrac{1}{2} \int{\sin 7 x+\sin (-x)} d x $

$ \begin{aligned} & =\dfrac{1}{2} \int{\sin 7 x-\sin x} d x \\ & =\dfrac{1}{2} \int \sin 7 x d x-\dfrac{1}{2} \int \sin x d x \\ & =\dfrac{1}{2}(\dfrac{-\cos 7 x}{7})-\dfrac{1}{2}(-\cos x)+C \\ & =\dfrac{-\cos 7 x}{14}+\dfrac{\cos x}{2}+C \end{aligned} $

हल

ज्ञात है कि, $\cos A \cos B=\dfrac{1}{2}{\cos (A+B)+\cos (A-B)}$

$ \begin{aligned} \therefore \int \cos 2 x(\cos 4 x \cos 6 x) d x & =\int \cos 2 x[\dfrac{1}{2}{\cos (4 x+6 x)+\cos (4 x-6 x)}] d x \\ & =\dfrac{1}{2} \int{\cos 2 x \cos 10 x+\cos 2 x \cos (-2 x)} d x \\ & =\dfrac{1}{2} \int{\cos 2 x \cos 10 x+\cos ^{2} 2 x} d x \\ & =\dfrac{1}{2} \int[\dfrac{1}{2} {\cos (2 x+10 x)+\cos (2 x-10 x)}+(\dfrac{1+\cos 4 x}{2})] d x \\ & =\dfrac{1}{4} \int(\cos 12 x+\cos 8 x+1+\cos 4 x) d x \\ & =\dfrac{1}{4}[\dfrac{\sin 12 x}{12}+\dfrac{\sin 8 x}{8}+x+\dfrac{\sin 4 x}{4}]+C \end{aligned} $

हल

$ \begin{aligned} & \text{ मान लीजिए } I=\int \sin ^{3}(2 x+1) \\

$$ & \Rightarrow \int \sin ^{3}(2 x+1) d x=\int \sin ^{2}(2 x+1) \cdot \sin (2 x+1) d x \\ & =\int(1-\cos ^{2}(2 x+1)) \sin (2 x+1) d x \end{aligned} $

Let $\cos (2 x+1)=t$

$\Rightarrow-2 \sin (2 x+1) d x=d t$

$\Rightarrow \sin (2 x+1) d x=\dfrac{-d t}{2}$

$ \begin{aligned} \Rightarrow I & =\dfrac{-1}{2} \int(1-t^{2}) d t \\ & =\dfrac{-1}{2}{t-\dfrac{t^{3}}{3}} \\ & =\dfrac{-1}{2}{\cos (2 x+1)-\dfrac{\cos ^{3}(2 x+1)}{3}} \\ & =\dfrac{-\cos (2 x+1)}{2}+\dfrac{\cos ^{3}(2 x+1)}{6}+C \end{aligned} $

Solution

$ \text{ Let } \begin{aligned} I & =\int \sin ^{3} x \cos ^{3} x \cdot d x \\ & =\int \cos ^{3} x \cdot \sin ^{2} x \cdot \sin x \cdot d x \\ & =\int \cos ^{3} x(1-\cos ^{2} x) \sin x \cdot d x \end{aligned} $

Let $\cos x=t$

$ \begin{aligned} \Rightarrow & -\sin x \cdot d x=d t \\ \Rightarrow I & =-\int t^{3}(1-t^{2}) d t \\ & =-\int(t^{3}-t^{5}) d t \\ & =-{\dfrac{t^{4}}{4}-\dfrac{t^{6}}{6}}+C \\ & =-{\dfrac{\cos ^{4} x}{4}-\dfrac{\cos ^{6} x}{6}}+C \\ & =\dfrac{\cos ^{6} x}{6}-\dfrac{\cos ^{4} x}{4}+C \end{aligned} $

Solution

It is known that, $\sin A \sin B=\dfrac{1}{2}{\cos (A-B)-\cos (A+B)}$

$\therefore \int \sin x \sin 2 x \sin 3 x d x=\int[\sin x \cdot \dfrac{1}{2}{\cos (2 x-3 x)-\cos (2 x+3 x)}] d x$

$=\dfrac{1}{2} \int(\sin x \cos (-x)-\sin x \cos 5 x) d x$

$=\dfrac{1}{2} \int(\sin x \cos x-\sin x \cos 5 x) d x$

$=\dfrac{1}{2} \int \dfrac{\sin 2 x}{2} d x-\dfrac{1}{2} \int \sin x \cos 5 x d x$

$=\dfrac{1}{4}[\dfrac{-\cos 2 x}{2}]-\dfrac{1}{2} \int{\dfrac{1}{2} \sin (x+5 x)+\sin (x-5 x)} d x$

$=\dfrac{-\cos 2 x}{8}-\dfrac{1}{4} \int(\sin 6 x+\sin (-4 x)) d x$

$=\dfrac{-\cos 2 x}{8}-\dfrac{1}{4}[\dfrac{-\cos 6 x}{3}+\dfrac{\cos 4 x}{4}]+C$

$=\dfrac{-\cos 2 x}{8}-\dfrac{1}{8}[\dfrac{-\cos 6 x}{3}+\dfrac{\cos 4 x}{2}]+C$

$=\dfrac{1}{8}[\dfrac{\cos 6 x}{3}-\dfrac{\cos 4 x}{2}-\cos 2 x]+C$

हल

यह ज्ञात है कि, $\sin A \sin B=\dfrac{1}{2}{ \cos (A-B)-\cos (A+B)}$

$\therefore \int \sin 4 x \sin 8 x d x=\int\dfrac{1}{2}{ \cos (4 x-8 x)-\cos (4 x+8 x)} d x$

$ \begin{aligned} & =\dfrac{1}{2} \int(\cos (-4 x)-\cos 12 x) d x \\ & =\dfrac{1}{2} \int(\cos 4 x-\cos 12 x) d x \\ & =\dfrac{1}{2}[\dfrac{\sin 4 x}{4}-\dfrac{\sin 12 x}{12}] \end{aligned} $

हल

$ \begin{aligned} & \dfrac{1-\cos x}{1+\cos x}=\dfrac{2 \sin ^{2} \dfrac{x}{2}}{2 \cos ^{2} x} \\ & {[2 \sin ^{2} \dfrac{x}{2}=1-\cos x \text{ और } 2 \cos ^{2} \dfrac{x}{2}=1+\cos x]} \\ & =\tan ^{2} \dfrac{x}{2} \\ & =(\sec ^{2} \dfrac{x}{2}-1) \\ & \therefore \int \dfrac{1-\cos x}{1+\cos x} d x=\int(\sec ^{2} \dfrac{x}{2}-1) d x \\ & =[\dfrac{\tan \dfrac{x}{2}}{\dfrac{1}{2}}-x]+C \\ & =2 \tan \dfrac{x}{2}-x+C \end{aligned} $

हल

$ \begin{aligned} & \dfrac{\cos x}{1+\cos x}=\dfrac{\cos ^{2} \dfrac{x}{2}-\sin ^{2} \dfrac{x}{2}}{2 \cos ^{2} \dfrac{x}{2}} \quad[\because \cos x=\cos ^{2} \dfrac{x}{2}-\sin ^{2} \dfrac{x}{2} \text{ और } \cos x=2 \cos ^{2} \dfrac{x}{2}-1] \\ & =\dfrac{1}{2}[1-\tan ^{2} \dfrac{x}{2}] \\ & \begin{aligned} \therefore \int \dfrac{\cos x}{1+\cos x} d x & =\dfrac{1}{2} \int(1-\tan ^{2} \dfrac{x}{2}) d x \\ & =\dfrac{1}{2} \int(1-\sec ^{2} \dfrac{x}{2}+1) d x \\ & =\dfrac{1}{2} \int(2-\sec ^{2} \dfrac{x}{2}) d x \\ & =\dfrac{1}{2}[2 x-\dfrac{\tan {\dfrac{x}{2}}}{\dfrac{1}{2}}]+C \\ & =x-\tan \dfrac{x}{2}+C \end{aligned} \end{aligned} $

हल

$ \begin{aligned} & \sin ^{4} x=\sin ^{2} x \sin ^{2} x \\ & =(\dfrac{1-\cos 2 x}{2})(\dfrac{1-\cos 2 x}{2}) \\ & =\dfrac{1}{4}(1-\cos 2 x)^{2} \\ & =\dfrac{1}{4}[1+\cos ^{2} 2 x-2 \cos 2 x] \\

$$ \begin{aligned} & =\dfrac{-2 \sin (x+\alpha) \sin (x-\alpha)}{-2 \sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2} \sin \dfrac{x-\alpha}{2}} \\ & =\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin \dfrac{x+\alpha}{2

&=\dfrac{\sin (x+\alpha) \sin (x-\alpha)}{\sin (\dfrac{x+\alpha}{2}) \sin (\dfrac{x-\alpha}{2})} \\ &=\dfrac{[2 \sin (\dfrac{x+\alpha}{2}) \cos (\dfrac{x+\alpha}{2})][2 \sin (\dfrac{x-\alpha}{2}) \cos (\dfrac{x-\alpha}{2})]}{\sin (\dfrac{x+\alpha}{2}) \sin (\dfrac{x-\alpha}{2})} \\ &=4 \cos (\dfrac{x+\alpha}{2}) \cos (\dfrac{x-\alpha}{2}) \\ &=2[\cos (\dfrac{x+\alpha}{2}+\dfrac{x-\alpha}{2})+\cos \dfrac{x+\alpha}{2}-\dfrac{x-\alpha}{2}] \\ &=2[\cos (x)+\cos \alpha] \\ &=2 \cos x+2 \cos \alpha \\ & \therefore \int \dfrac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha} d x=\int 2 \cos x+2 \cos \alpha \\ &=2[\sin x+x \cos \alpha]+C \end{aligned} $

Solution

$ \begin{aligned} \dfrac{\cos x-\sin x}{1+\sin 2 x} & =\dfrac{\cos x-\sin x}{(\sin ^{2} x+\cos ^{2} x)+2 \sin x \cos x} \\ & =\dfrac{\cos x-\sin x}{(\sin x+\cos x)^{2}}\ [\because \sin^{2}x+\cos ^{2} x=1 ; \sin 2 x=2 \sin x \cos x] \end{aligned} $

मान लीजिए $\sin x+\cos x=t$

$\therefore(\cos x-\sin x) d x=d t$

$\Rightarrow \int \dfrac{\cos x-\sin x}{1+\sin 2 x} d x=\int \dfrac{\cos x-\sin x}{(\sin x+\cos x)^{2}} d x$

$ =\int \dfrac{d t}{t^{2}} $

$ \begin{aligned} & =\int t^{-2} d t \\ & =-t^{-1}+C \\ & =-\dfrac{1}{t}+C \\ & =\dfrac{-1}{\sin x+\cos x}+C \end{aligned} $

Solution

$ \begin{aligned} & \tan ^{3} 2 x \sec 2 x=\tan ^{2} 2 x \tan 2 x \sec 2 x \\ & =(\sec ^{2} 2 x-1) \tan 2 x \sec 2 x \\ & =\sec ^{2} 2 x \cdot \tan 2 x \sec 2 x-\tan 2 x \sec 2 x \\ & \therefore \int \tan ^{3} 2 x \sec 2 x d x=\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\int \tan 2 x \sec 2 x d x \\ & =\int \sec ^{2} 2 x \tan 2 x \sec 2 x d x-\dfrac{\sec 2 x}{2}+C \end{aligned} $

मान लीजिए $\sec 2 x=t$

$\therefore 2 \sec 2 x \tan 2 x d x=d t$

$ \begin{aligned} \therefore \int \tan ^{3} 2 x \sec 2 x d x & =\dfrac{1}{2} \int t^{2} d t-\dfrac{\sec 2 x}{2}+C \\

& =\dfrac{t^{3}}{6}-\dfrac{\sec 2 x}{2}+C \\ & =\dfrac{(\sec 2 x)^{3}}{6}-\dfrac{\sec 2 x}{2}+C \end{aligned} $

Solution

$\tan ^{4} x$

$=\tan ^{2} x \cdot \tan ^{2} x$

$=(\sec ^{2} x-1) \tan ^{2} x$

$=\sec ^{2} x \tan ^{2} x-\tan ^{2} x$

$=\sec ^{2} x \tan ^{2} x-(\sec ^{2} x-1)$

$=\sec ^{2} x \tan ^{2} x-\sec ^{2} x+1$

$\therefore \int \tan ^{4} x d x=\int \sec ^{2} x \tan ^{2} x d x-\int \sec ^{2} x d x+\int 1 \cdot d x$

$=\int \sec ^{2} x \tan ^{2} x d x-\tan x+x+C \qquad …(1)$

Consider $\int \sec ^{2} x \tan ^{2} x d x$

Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$\Rightarrow \int \sec ^{2} x \tan ^{2} x d x=\int t^{2} d t=\dfrac{t^{3}}{3}=\dfrac{\tan ^{3} x}{3}$

From equation (1), we obtain

$\int \tan ^{4} x d x=\dfrac{1}{3} \tan ^{3} x-\tan x+x+C$

Solution

$ \begin{aligned} & \dfrac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}=\dfrac{\sin ^{3} x}{\sin ^{2} x \cos ^{2} x}+\dfrac{\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} \\ & =\dfrac{\sin x}{\cos ^{2} x}+\dfrac{\cos x}{\sin ^{2} x} \\ & =\tan x \sec x+\cot x cosec x \\ & \therefore \quad \int \dfrac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x} d x=\int(\tan x \sec x+\cot x cosec x) d x \\ & =\sec x-cosec x+C \end{aligned} $

Solution

$ \begin{aligned} & \dfrac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} \\ & =\dfrac{\cos 2 x+(1-\cos 2 x)}{\cos ^{2} x} \quad[\because\cos 2 x=1-2 \sin ^{2} x] \\ & =\dfrac{1}{\cos ^{2} x} \\ & =\sec ^{2} x \\ & \therefore \int \dfrac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x=\int \sec ^{2} x d x=\tan x+C \end{aligned} $

Solution

$ \begin{aligned} \dfrac{1}{\sin x \cos ^{3} x} & =\dfrac{\sin ^{2} x+\cos ^{2} x}{\sin x \cos ^{3} x} \\

& =\dfrac{\sin x}{\cos ^{3} x}+\dfrac{1}{\sin x \cos x} \\ & =\tan x \sec ^{2} x+\dfrac{1 \cos ^{2} x}{\dfrac{\sin x \cos x}{\cos ^{2} x}} \\ & =\tan x \sec ^{2} x+\dfrac{\sec ^{2} x}{\tan x} \end{aligned} $

$\therefore \int \dfrac{1}{\sin x \cos ^{3} x} d x=\int \tan x \sec ^{2} x d x+\int \dfrac{\sec ^{2} x}{\tan x} d x$

Let $\tan x=t \Rightarrow \sec ^{2} x d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{1}{\sin x \cos ^{3} x} d x & =\int t d t+\int \dfrac{1} {t} d t \\ & =\dfrac{t^{2}}{2}+\log |t|+C \\ & =\dfrac{1}{2} \tan ^{2} x+\log |\tan x|+C \end{aligned} $

Solution

$\begin{aligned} & \dfrac{\cos 2 \mathrm{x}}{(\cos \mathrm{x}+\sin \mathrm{x})^2}=\dfrac{\cos 2 \mathrm{x}}{\cos ^2 \mathrm{x}+\sin ^2 \mathrm{x}+2 \sin \mathrm{x} \cos \mathrm{x}}=\dfrac{\cos 2 \mathrm{x}}{1+\sin 2 \mathrm{x}} \\ & \therefore \int \dfrac{\cos 2 \mathrm{x}}{(\cos \mathrm{x}+\sin \mathrm{x})^2} \mathrm{dx}=\int \dfrac{\cos 2 \mathrm{x}}{(1+\sin 2 \mathrm{x})} \mathrm{dx} \end{aligned}$

Let $1+\sin 2 x=t \Rightarrow 2 \cos 2 x d x=d t$

$\begin{aligned} & \therefore \int \dfrac{\cos 2 x}{(\cos x+\sin x)^2} d x=\dfrac{1}{2} \int \dfrac{1}{t} d t \\ & =\dfrac{1}{2} \log |t|+C \\ & =\dfrac{1}{2} \log |1+\sin 2 x|+C \\ & =\dfrac{1}{2} \log \left|(\sin x+\cos x)^2\right|+C \\ & =\log |\sin x+\cos x|+C \\ \end{aligned}$

Solution

$\sin ^{-1}(\cos x)$

Let $\cos x=t$

$\Rightarrow(-\sin x) d x=d t$

$d x=\dfrac{-d t}{\sin x}$

$d x=\dfrac{-d t}{\sqrt{1-t^{2}}}\ [\because \sin x=\sqrt{1-t^{2}}]$

$\therefore \int \sin ^{-1}(\cos x) d x=\int \sin ^{-1} t(\dfrac{-d t}{\sqrt{1-t^{2}}})$

$=-\int \dfrac{\sin ^{-1} t}{\sqrt{1-t^{2}}} d t$

Let $\sin ^{-1} t=u$

$\Rightarrow \dfrac{1}{\sqrt{1-t^{2}}} d t=d u$

$\therefore \int \sin ^{-1}(\cos x) d x\\ =-\int u d u $

$ \begin{aligned}

& =-\dfrac{u^{2}}{2}+C \\ & =\dfrac{-(\sin ^{-1} t)^{2}}{2}+C \\ & =\dfrac{-[\sin ^{-1}(\cos x)]^{2}}{2}+C \qquad …(1) \end{aligned} $

यह ज्ञात है कि,

$ \begin{aligned} & \sin ^{-1} x+\cos ^{-1} x=\dfrac{\pi}{2} \\ & \therefore \sin ^{-1}(\cos x)=\dfrac{\pi}{2}-\cos ^{-1}(\cos x)=(\dfrac{\pi}{2}-x) \end{aligned} $

समीकरण (1) में प्रतिस्थापित करने पर, हम प्राप्त करते हैं

$ \begin{aligned} \int \sin ^{-1}(\cos x) d x & =\dfrac{-[\dfrac{\pi}{2}-x]^{2}}{2}+C \\ & =-\dfrac{1}{2}(\dfrac{\pi^{2}}{4}+x^{2}-\pi x)+C \\ & =-\dfrac{\pi^{2}}{8}-\dfrac{x^{2}}{2}+\dfrac{1}{2} \pi x+C \\ & =\dfrac{\pi x}{2}-\dfrac{x^{2}}{2}+(C-\dfrac{\pi^{2}}{8}) \\ & =\dfrac{\pi x}{2}-\dfrac{x^{2}}{2}+C_1 \end{aligned} $

Solution

$ \begin{aligned} \dfrac{1}{\cos (x-a) \cos (x-b)} & =\dfrac{1}{\sin (a-b)}[\dfrac{\sin (a-b)}{\cos (x-a) \cos (x-b)}] \\ & =\dfrac{1}{\sin (a-b)}[\dfrac{\sin [(x-b)-(x-a)]}{\cos (x-a) \cos (x-b)}] \\ & =\dfrac{1}{\sin (a-b)} \dfrac{[\sin (x-b) \cos (x-a)-\cos (x-b) \sin (x-a)]}{\cos (x-a) \cos (x-b)} \\ & =\dfrac{1}{\sin (a-b)}[\tan (x-b)-\tan (x-a)] \\ \Rightarrow \int \dfrac{1}{\cos (x-a) \cos (x-b)} d x & =\dfrac{1}{\sin (a-b)} \int[\tan (x-b)-\tan (x-a)] d x \\ & =\dfrac{1}{\sin (a-b)}[-\log |\cos (x-b)|+\log |\cos (x-a)|] \\ & =\dfrac{1}{\sin (a-b)}[\log |\dfrac{\cos (x-a)}{\cos (x-b)}|]+C \end{aligned} $

Solution

$ \begin{aligned} \int \dfrac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x & =\int(\dfrac{\sin ^{2} x}{\sin ^{2} x \cos ^{2} x}-\dfrac{\cos ^{2} x}{\sin ^{2} x \cos ^{2} x}) d x \\ & =\int(\sec ^{2} x-cosec^{2} x) d x \\ & =\tan x+\cot x+C \end{aligned} $

इसलिए, सही उत्तर A है।

हल

$\int \dfrac{e^{x}(1+x)}{\cos ^{2}(e^{x} x)} d x$

मान लीजिए $e x^{x}=t$

$ \begin{aligned} & \Rightarrow(e^{x} \cdot x+e^{x} \cdot 1) d x=d t \\ & e^{x}(x+1) d x=d t \\ & \begin{aligned} \therefore \int \dfrac{e^{x}(1+x)}{\cos ^{2}(e^{x} x)} d x & =\int \dfrac{d t}{\cos ^{2} t} \\ & =\int \sec ^{2} t d t \\ & =\tan t+C \\ & =\tan (e^{x} \cdot x)+C \end{aligned} \end{aligned} $

इसलिए, सही उत्तर B है।

हल

मान लीजिए $x^{3}=t$

$\therefore 3 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{3 x^{2}}{x^{6}+1} d x & =\int \dfrac{d t}{t^{2}+1} \\ & =\tan ^{-1} t+C \\ & =\tan ^{-1}(x^{3})+C \end{aligned} $

हल

मान लीजिए $2 x=t$

$\therefore 2 d x=d t$

$\Rightarrow \int \dfrac{1}{\sqrt{1+4 x^{2}}} d x=\dfrac{1}{2} \int \dfrac{d t}{\sqrt{1+t^{2}}}$

$ \begin{matrix} =\dfrac{1}{2}[\log |t+\sqrt{t^{2}+1}|]+C & {[\because \int \dfrac{1}{\sqrt{x^{2}+a^{2}}} d t=\log |x+\sqrt{x^{2}+a^{2}}|]} \\ =\dfrac{1}{2} \log |2 x+\sqrt{4 x^{2}+1}|+C & \end{matrix} $

हल

मान लीजिए $2-x=t$

$\Rightarrow-d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{1}{\sqrt{(2-x)^{2}+1}} d x & =-\int \dfrac{1}{\sqrt{t^{2}+1}} d t \\ & =-\log |t+\sqrt{t^{2}+1}|+C \quad[\because\int \dfrac{1}{\sqrt{x^{2}+a^{2}}} d t=\log |x+\sqrt{x^{2}+a^{2}}|+ C] \\ & =-\log |2-x+\sqrt{(2-x)^{2}+1}|+C \\ & =\log |\dfrac{1}{(2-x)+\sqrt{x^{2}-4 x+5}}|+C \end{aligned} $

हल

मान लीजिए $5 x=t$

$\therefore 5 d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{1}{\sqrt{9-25 x^{2}}} d x & =\dfrac{1}{5} \int \dfrac{1}{9-t^{2}} d t \\ & =\dfrac{1}{5} \int \dfrac{1}{\sqrt{3^{2}-t^{2}}} d t \\ & =\dfrac{1}{5} \sin ^{-1}(\dfrac{t}{3})+C \\ & =\dfrac{1}{5} \sin ^{-1}(\dfrac{5 x}{3})+C \end{aligned} $

हल

मान लीजिए $\sqrt{2} x^{2}=t$

$\therefore 2 \sqrt{2} x d x=d t$

$\Rightarrow \int \dfrac{3 x}{1+2 x^{4}} d x=\dfrac{3}{2 \sqrt{2}} \int \dfrac{d t}{1+t^{2}}$

$ =\dfrac{3}{2 \sqrt{2}}[\tan ^{-1} t]+C $

$=\dfrac{3}{2 \sqrt{2}} \tan ^{-1}(\sqrt{2} x^{2})+C$

Solution

मान लीजिए $x^{3}=t$

$\therefore 3 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{x^{2}}{1-x^{6}} d x & =\dfrac{1}{3} \int \dfrac{d t}{1-t^{2}} \\ & =\dfrac{1}{3}[\dfrac{1}{2} \log |\dfrac{1+t}{1-t}|]+C \\ & =\dfrac{1}{6} \log |\dfrac{1+x^{3}}{1-x^{3}}|+C \end{aligned} $

Solution

$ \begin{aligned} \int \dfrac{x-1}{\sqrt{x^{2}-1}} d x=\int \dfrac{x}{\sqrt{x^{2}-1}} d x-\int \dfrac{1}{\sqrt{x^{2}-1}} d x \qquad…(1) \end{aligned} $

$\int \dfrac{x}{\sqrt{x^{2}-1}} d x$ के लिए, मान लीजिए $x^{2}-1=t \Rightarrow 2 x d x=d t$

$\therefore \int \dfrac{x}{\sqrt{x^{2}-1}} d x=\dfrac{1}{2} \int \dfrac{d t}{\sqrt{t}}$

$=\dfrac{1}{2} \int t^{-\dfrac{1}{2}} d t$

$=\dfrac{1}{2}[2 t^{\dfrac{1}{2}}]$

$=\sqrt{t}$

$=\sqrt{x^{2}-1}$

(1) से, हम प्राप्त करते हैं

$ \begin{aligned} \int \dfrac{x-1}{\sqrt{x^{2}-1}} d x & =\int \dfrac{x}{\sqrt{x^{2}-1}} d x-\int \dfrac{1}{\sqrt{x^{2}-1}} d x \quad \quad[\because \int \dfrac{1}{\sqrt{x^{2}-a^{2}}} d t=\log |x+\sqrt{x^{2}-a^{2}}|] \\ & =\sqrt{x^{2}-1}-\log |x+\sqrt{x^{2}-1}|+C \end{aligned} $

Solution

मान लीजिए $x^{3}=t$

$\Rightarrow 3 x^{2} d x=d t$

$ \begin{aligned} \therefore \int \dfrac{x^{2}}{\sqrt{x^{6}+a^{6}}} d x & =\dfrac{1}{3} \int \dfrac{d t}{\sqrt{t^{2}+(a^{3})^{2}}} \\ & =\dfrac{1}{3} \log |t+\sqrt{t^{2}+a^{6}}|+C \\ & =\dfrac{1}{3} \log |x^{3}+\sqrt{x^{6}+a^{6}}|+C \end{aligned} $

Solution

मान लीजिए $\tan x=t$

$\therefore \sec ^{2} x d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}} d x & =\int \dfrac{d t}{\sqrt{t^{2}+2^{2}}} \\

& =\log |t+\sqrt{t^{2}+4}|+C \\ & =\log |\tan x+\sqrt{\tan ^{2} x+4}|+C \end{aligned} $

हल

$ \int \dfrac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \dfrac{1}{\sqrt{(x+1)^{2}+(1)^{2}}} d x $

मान लीजिए $x+1=t$

$\therefore d x=d t$

$\Rightarrow \int \dfrac{1}{\sqrt{x^{2}+2 x+2}} d x=\int \dfrac{1}{\sqrt{t^{2}+1}} d t$

$ \begin{aligned} & =\log |t+\sqrt{t^{2}+1}|+C \\ & =\log |(x+1)+\sqrt{(x+1)^{2}+1}|+C \\ & =\log |(x+1)+\sqrt{x^{2}+2 x+2}|+C \end{aligned} $

हल

$ \int \dfrac{1}{9 x^{2}+6 x+5} d x=\int \dfrac{1}{(3 x+1)^{2}+(2)^{2}} d x $

मान लीजिए $(3 x+1)=t$

$\therefore 3 d x=d t$

$\Rightarrow \int \dfrac{1}{(3 x+1)^{2}+(2)^{2}} d x=\dfrac{1}{3} \int \dfrac{1}{t^{2}+2^{2}} d t$

$ \begin{aligned} & =\dfrac{1}{3}[\dfrac{1}{2} \tan ^{-1}(\dfrac{t}{2})]+C \\ & =\dfrac{1}{6} \tan ^{-1}(\dfrac{3 x+1}{2})+C \end{aligned} $

फलन $\dfrac{1}{\sqrt{7-6x-x^2}}$ के समाकलन के लिए, हम निर्माण कर सकते हैं धनात्मक वर्ग के निर्माण के अंश में।

पहले, वर्ग व्यंजक $7 - 6x - x^2$ को निम्नलिखित रूप में लिखा जा सकता है:

$7 - 6x - x^2 = -((x + 3)^2 - 16) = 16 - (x + 3)^2$

इसलिए, समाकलन निम्नलिखित रूप में बन जाता है:

$ \int \dfrac{1}{\sqrt{16 - (x + 3)^2}} , dx $

इस समाकलन के रूप $\int \dfrac{1}{\sqrt{a^2 - u^2}} , du$ है, जो एक मानक समाकलन है जो $\sin^{-1}\left(\dfrac{u}{a}\right) + C$ के रूप में मूल्यांकित होता है।

यहाँ, $a = 4$ और $u = x + 3$ है। इसलिए, समाकलन है:

$ \int \dfrac{1}{\sqrt{16 - (x + 3)^2}} , dx = \sin^{-1} \left(\dfrac{x + 3}{4}\right) + C $

जहाँ $C$ समाकलन के नियतांक है।

इसलिए, अंतिम उत्तर है:

$ \int \dfrac{1}{\sqrt{7-6x-x^2}} , dx = \sin^{-1}\left(\dfrac{x + 3}{4}\right) + C $

हल

$(x-1)(x-2)$ को $x^{2}-3 x+2$ के रूप में लिखा जा सकता है।

इसलिए,

$x^{2}-3 x+2$

$=x^{2}-3 x+\dfrac{9}{4}-\dfrac{9}{4}+2$

$=(x-\dfrac{3}{2})^{2}-\dfrac{1}{4}$

$=(x-\dfrac{3}{2})^{2}-(\dfrac{1}{2})^{2}$

$\therefore \int \dfrac{1}{\sqrt{(x-1)(x-2)}} d x=\int \dfrac{1}{\sqrt{(x-\dfrac{3}{2})^{2}-(\dfrac{1}{2})^{2}}} d x$

मान लीजिए $x-\dfrac{3}{2}=t$

$\therefore d x=d t$

$\Rightarrow \int \dfrac{1}{\sqrt{(x-\dfrac{3}{2})^{2}-(\dfrac{1}{2})^{2}}} d x=\int \dfrac{1}{\sqrt{t^{2}-(\dfrac{1}{2})^{2}}} d t$

$=\log |t+\sqrt{t^{2}-(\dfrac{1}{2})^{2}}|+C$

$=\log |(x-\dfrac{3}{2})+\sqrt{x^{2}-3 x+2}|+C$

हल

$8+3 x-x^{2}$ को $8-(x^{2}-3 x+\dfrac{9}{4}-\dfrac{9}{4})$ के रूप में लिखा जा सकता है।

इसलिए,

$8-(x^{2}-3 x+\dfrac{9}{4}-\dfrac{9}{4})$

$=\dfrac{41}{4}-(x-\dfrac{3}{2})^{2}$

$\Rightarrow \int \dfrac{1}{\sqrt{8+3 x-x^{2}}} d x=\int \dfrac{1}{\sqrt{\dfrac{41}{4}-(x-\dfrac{3}{2})^{2}}} d x$

मान लीजिए $x-\dfrac{3}{2}=t$

$\therefore d x=d t$

$\Rightarrow \int \dfrac{1}{\sqrt{\dfrac{41}{4}-(x-\dfrac{3}{2})^{2}}} d x=\int \dfrac{1}{\sqrt{(\dfrac{\sqrt{41}}{2})^{2}-t^{2}}} d t$

$=\sin ^{-1}(\dfrac{t}{\dfrac{\sqrt{41}}{2}})+C$

$=\sin ^{-1}(\dfrac{x-\dfrac{3}{2}}{\dfrac{\sqrt{41}}{2}})+C$

$=\sin ^{-1}(\dfrac{2 x-3}{\sqrt{41}})+C$

हल

$\begin{aligned} (x - a)(x - b) &= x^2 - (a + b)x + ab = x^2 - (a + b)x + \dfrac{(a + b)^2}{4} - \dfrac{(a + b)^2}{4} + ab \ &= \left[ x - \dfrac{a + b}{2} \right]^2 - \dfrac{(a - b)^2}{4} \ &\Rightarrow \int \dfrac{1}{\sqrt{(x - a)(x - b)}} , dx = \int \dfrac{1}{\sqrt{\left[ x - \dfrac{a + b}{2} \right]^2 - \dfrac{(a - b)^2}{4}}} , dx \ \text{मान लीजिए } x - \dfrac{a + b}{2} &= t \Rightarrow dx = dt \ &\Rightarrow \int \dfrac{1}{\sqrt{\left[ x - \dfrac{a + b}{2} \right]^2 - \dfrac{(a - b)^2}{4}}} , dx = \int \dfrac{1}{\sqrt{t^2 - \left( \dfrac{a - b}{2} \right)^2}} , dt \

&= \log \left| t + \sqrt{t^2 - \left( \dfrac{a - b}{2} \right)^2} \right| + C \ &= \log \left| x - \dfrac{a + b}{2} + \sqrt{(x - a)(x - b)} \right| + C \end{aligned}$

Solution

मान लीजिए $4 x+1=A \dfrac{d}{d x}(2 x^{2}+x-3)+B$

$\Rightarrow 4 x+1=A(4 x+1)+B$

$\Rightarrow 4 x+1=4 A x+A+B$

दोनों तरफ $x$ के गुणांक और स्थिरांक की तुलना करने पर, हम प्राप्त करते हैं

$4 A=4 \Rightarrow A=1$

$A+B=1 \Rightarrow B=0$

मान लीजिए $2 x^{2}+x-3=t$

$\therefore(4 x+1) d x=d t$

$ \begin{aligned} \Rightarrow \int \dfrac{4 x+1}{\sqrt{2 x^{2}+x-3}} d x & =\int \dfrac{1}{\sqrt{t}} d t \\ & =2 \sqrt{t}+C \\ & =2 \sqrt{2 x^{2}+x-3}+C \end{aligned} $

Solution

मान लीजिए $x+2=A \dfrac{d}{d x}(x^{2}-1)+B$

$\Rightarrow x+2=A(2 x)+B \qquad …(1)$

दोनों तरफ $x$ के गुणांक और स्थिरांक की तुलना करने पर, हम प्राप्त करते हैं

$2 A=1 \Rightarrow A=\dfrac{1}{2}$

$B=2$

समीकरण (1) से, हम प्राप्त करते हैं

$(x+2)=\dfrac{1}{2}(2 x)+2$

फिर, $\int \dfrac{x+2}{\sqrt{x^{2}-1}} d x=\int \dfrac{\dfrac{1}{2}(2 x)+2}{\sqrt{x^{2}-1}} d x$

$ \begin{aligned} =\dfrac{1}{2} \int \dfrac{2 x}{\sqrt{x^{2}-1}} d x+\int \dfrac{2}{\sqrt{x^{2}-1}} d x \qquad …(2) \end{aligned} $

$\dfrac{1}{2} \int \dfrac{2 x}{\sqrt{x^{2}-1}} d x$ में, मान लीजिए $x^{2}-1=t \Rightarrow 2 x d x=d t$

$ \begin{aligned} \dfrac{1}{2} \int \dfrac{2 x}{\sqrt{x^{2}-1}} d x & =\dfrac{1}{2} \int \dfrac{d t}{\sqrt{t}} \\ & =\dfrac{1}{2}[2 \sqrt{t}] \\ & =\sqrt{t} \\ & =\sqrt{x^{2}-1} \end{aligned} $

फिर, $\int \dfrac{2}{\sqrt{x^{2}-1}} d x=2 \int \dfrac{1}{\sqrt{x^{2}-1}} d x=2 \log |x+\sqrt{x^{2}-1}|$

समीकरण (2) से, हम प्राप्त करते हैं

$ \int \dfrac{x+2}{\sqrt{x^{2}-1}} d x=\sqrt{x^{2}-1}+2 \log |x+\sqrt{x^{2}-1}|+C $

Solution

$\begin{aligned} &\int \dfrac{5x - 2}{1 + 2x + 3x^2} dx = I \quad (\text{say}) \

&\text{मान लीजिए } 5x - 2 = A \cdot \dfrac{d}{dx} (1 + 2x + 3x^2) + B \ &\Rightarrow 5x - 2 = A (2 + 6x) + B \ &\text{x के गुणांक की तुलना करने पर,} \ &5 = 6A \Rightarrow A = \dfrac{5}{6} \ &\text{स्थिरांक पदों की तुलना करने पर} \ &-2 = 2A + B \ &\Rightarrow -2 = 2 \times \dfrac{5}{6} + B \ &\Rightarrow B = -\dfrac{11}{3} \ &I = \dfrac{5}{6} \int \dfrac{2 + 6x}{1 + 2x + 3x^2} dx - \dfrac{11}{3} \int \dfrac{1}{1 + 2x + 3x^2} dx \ &I = \dfrac{5}{6} \int \dfrac{d(1 + 2x + 3x^2)}{1 + 2x + 3x^2} - \dfrac{11}{3} \int \dfrac{1}{1 + 2x + 3x^2} dx \ &I = \dfrac{5}{6} \log |1 + 2x + 3x^2| - \dfrac{11}{3} \int \dfrac{1}{1 + 2x + 3x^2} dx \ &\text{पहले समाकल के लिए} \ &\text{मान लीजिए } 1 + 2x + 3x^2 = t \ &\Rightarrow (2 + 6x) dx = dt \ &I = \dfrac{5}{6} \int \dfrac{dt}{t} - \dfrac{11}{3} \int \dfrac{1}{\sqrt{(x + \dfrac{1}{3})^2 + (\dfrac{\sqrt{2}}{3})^2}} dx \ &I = \dfrac{5}{6} \log |t| - \dfrac{11}{3} \left( \dfrac{1}{\dfrac{\sqrt{2}}{3}} \right) \tan^{-1} \left( \dfrac{x + \dfrac{1}{3}}{\dfrac{\sqrt{2}}{3}} \right) + c \ &I = \dfrac{5}{6} \log |1 + 2x + 3x^2| - \dfrac{11}{3} \cdot \dfrac{3}{\sqrt{2}} \tan^{-1} \left( \dfrac{3x + 1}{\sqrt{2}} \right) + c \ &I = \dfrac{5}{6} \log |1 + 2x + 3x^2| - \dfrac{11}{\sqrt{2}} \tan^{-1} \left( \dfrac{3x + 1}{\sqrt{2}} \right) + c \end{aligned} $

Solution

$\dfrac{6 x+7}{\sqrt{(x-5)(x-4)}}=\dfrac{6 x+7}{\sqrt{x^{2}-9 x+20}}$

Let $6 x+7=A \dfrac{d}{d x}(x^{2}-9 x+20)+B$

$\Rightarrow 6 x+7=A(2 x-9)+B$

x के गुणांक और स्थिरांक पदों की तुलना करने पर, हम प्राप्त करते हैं

$2 A=6 \Rightarrow A=3$

$-9 A+B=7 \Rightarrow B=34$

$\therefore 6 x+7=3(2 x-9)+34$

$ \begin{aligned} \int \dfrac{6 x+7}{\sqrt{x^{2}-9 x+20}} & =\int \dfrac{3(2 x-9)+34}{\sqrt{x^{2}-9 x+20}} d x \\ & =3 \int \dfrac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x+34 \int \dfrac{1}{\sqrt{x^{2}-9 x+20}} d x \end{aligned} $

Let $I_1=\int \dfrac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x$ and $I_2=\int \dfrac{1}{\sqrt{x^{2}-9 x+20}} d x$

$\therefore \int \dfrac{6 x+7}{\sqrt{x^{2}-9 x+20}}=3 I_1+34 I_2 \qquad …(1)$

तब,

$I_1=\int \dfrac{2 x-9}{\sqrt{x^{2}-9 x+20}} d x$

मान लीजिए $x^{2}-9 x+20=t$

$\Rightarrow(2 x-9) d x=d t$

$\Rightarrow I_1=\dfrac{d t}{\sqrt{t}}$

$I_1=2 \sqrt{t}$

$I_1=2 \sqrt{x^{2}-9 x+20} \qquad …(2)$

और $I_2=\int \dfrac{1}{\sqrt{x^{2}-9 x+20}} d x$ $x^{2}-9 x+20$ को $x^{2}-9 x+20+\dfrac{81}{4}-\dfrac{81}{4}$ के रूप में लिखा जा सकता है।

इसलिए,

$x^{2}-9 x+20+\dfrac{81}{4}-\dfrac{81}{4}$

$=(x-\dfrac{9}{2})^{2}-\dfrac{1}{4}$

$=(x-\dfrac{9}{2})^{2}-(\dfrac{1}{2})^{2}$

$\Rightarrow I_2=\int \dfrac{1}{\sqrt{(x-\dfrac{9}{2})^{2}-(\dfrac{1}{2})^{2}}} d x$

$I_2=\log |(x-\dfrac{9}{2})+\sqrt{x^{2}-9 x+20}| \qquad …(3)$

समीकरण (2) और (3) को (1) में बदल देने पर, हम प्राप्त करते हैं

$ \begin{aligned} \int \dfrac{6 x+7}{\sqrt{x^{2}-9 x+20}} d x & =3[2 \sqrt{x^{2}-9 x+20}]+34 \log [(x-\dfrac{9}{2})+\sqrt{x^{2}-9 x+20}]+C \\ & =6 \sqrt{x^{2}-9 x+20}+34 \log [(x-\dfrac{9}{2})+\sqrt{x^{2}-9 x+20}]+C \end{aligned} $

Solution

मान लीजिए $x+2=A \dfrac{d}{d x}(4 x-x^{2})+B$

$\Rightarrow x+2=A(4-2 x)+B$

दोनों तरफ $x$ के गुणांक और अचर पद की तुलना करने पर, हम प्राप्त करते हैं

$-2 A=1 \Rightarrow A=-\dfrac{1}{2}$

$4 A+B=2 \Rightarrow B=4$

$\Rightarrow(x+2)=-\dfrac{1}{2}(4-2 x)+4$

$\therefore \int \dfrac{x+2}{\sqrt{4 x-x^{2}}} d x=\int \dfrac{-\dfrac{1}{2}(4-2 x)+4}{\sqrt{4 x-x^{2}}} d x$ $=-\dfrac{1}{2} \int \dfrac{4-2 x}{\sqrt{4 x-x^{2}}} d x+4 \int \dfrac{1}{\sqrt{4 x-x^{2}}} d x$

मान लीजिए $I_1=\int \dfrac{4-2 x}{\sqrt{4 x-x^{2}}} d x$ और $I_2= \int \dfrac{1}{\sqrt{4 x-x^{2}}} d x$

$\therefore \int \dfrac{x+2}{\sqrt{4 x-x^{2}}} d x=-\dfrac{1}{2} I_1+4 I_2 \qquad …(1)$

तब, $I_1=\int \dfrac{4-2 x}{\sqrt{4 x-x^{2}}} d x$

मान लीजिए $4 x-x^{2}=t$

$\Rightarrow(4-2 x) d x=d t$

$\Rightarrow I_1=\int{\dfrac {d t} {\sqrt{t}}}=2 \sqrt{t}=2 \sqrt{4 x-x^{2}} \qquad …(2)$

$I_2=\int \dfrac{1}{\sqrt{4 x-x^{2}}} d x$

$\Rightarrow 4 x-x^{2}=-(-4 x+x^{2})$

$=(-4 x+x^{2}+4-4)$

$=4-(x-2)^{2}$

$=(2)^{2}-(x-2)^{2}$

$\therefore I_2=\int \dfrac{1}{\sqrt{(2)^{2}-(x-2)^{2}}} d x=\sin ^{-1}(\dfrac{x-2}{2}) \qquad …(3)$

समीकरण (2) और (3) को समीकरण (1) में उपयोग करने पर, हम प्राप्त करते हैं

$ \begin{aligned} \int \dfrac{x+2}{\sqrt{4 x-x^{2}}} d x & =-\dfrac{1}{2}(2 \sqrt{4 x-x^{2}})+4 \sin ^{-1}(\dfrac{x-2}{2})+C \\ & =-\sqrt{4 x-x^{2}}+4 \sin ^{-1}(\dfrac{x-2}{2})+C \end{aligned} $

Solution

$\begin{aligned} &\int \dfrac{(x+2)}{\sqrt{x^2+2x+3}} , dx = \dfrac{1}{2} \int \dfrac{2(x+2)}{\sqrt{x^2+2x+3}} , dx \ &= \dfrac{1}{2} \int \dfrac{2x+4}{\sqrt{x^2+2x+3}} , dx \ &= \dfrac{1}{2} \int \dfrac{2x+2+2}{\sqrt{x^2+2x+3}} , dx \ &= \dfrac{1}{2} \int \dfrac{2x+2}{\sqrt{x^2+2x+3}} , dx + \dfrac{1}{2} \int \dfrac{2}{\sqrt{x^2+2x+3}} , dx \ &= \dfrac{1}{2} \int \dfrac{2x+2}{\sqrt{x^2+2x+3}} , dx + \int \dfrac{1}{\sqrt{x^2+2x+3}} , dx \ &\text{Let } I_1 = \int \dfrac{2x+2}{\sqrt{x^2+2x+3}} , dx \text{ and } I_2 = \int \dfrac{1}{\sqrt{x^2+2x+3}} , dx \ &\therefore \int \dfrac{x+2}{\sqrt{x^2+2x+3}} , dx = \dfrac{1}{2} I_1 + I_2 \quad \text{……..(1)} \ &\text{Then, } I_1 = \int \dfrac{2x+2}{\sqrt{x^2+2x+3}} , dx \ &\text{Let } x^2 + 2x + 3 = t \ &\Rightarrow (2x+2) , dx = dt \ &I_1 = \int \dfrac{dt}{\sqrt{t}} = 2\sqrt{t} = 2\sqrt{x^2+2x+3} \quad \text{……..(2)} \ &I_2 = \int \dfrac{1}{\sqrt{x^2+2x+3}} , dx \ &\Rightarrow x^2 + 2x + 3 = (x+1)^2 + (\sqrt{2})^2 \ &\Rightarrow I_2 = \int \dfrac{1}{\sqrt{(x+1)^2 + (\sqrt{2})^2}} , dx = \log \left| (x+1) + \sqrt{x^2+2x+3} \right| \quad \text{……..(3)} \ &\text{Using equation (2) and (3) in (1), we obtain} \ &\int \dfrac{x+2}{\sqrt{x^2+2x+3}} , dx = \dfrac{1}{2} \left[ 2\sqrt{x^2+2x+3} \right] + \log \left| (x+1) + \sqrt{x^2+2x+3} \right| + C \ &= \sqrt{x^2+2x+3} + \log \left| (x+1) + \sqrt{x^2+2x+3} \right| + C \end{aligned} $

हल

मान लीजिए $(x+3)=A \dfrac{d}{d x}(x^{2}-2 x-5)+B$

$(x+3)=A(2 x-2)+B$

दोनों ओर $x$ के गुणांक और अचर पद के गुणांक की तुलना करने पर, हम प्राप्त करते हैं

$ \begin{aligned} & 2 A=1 \Rightarrow A=\dfrac{1}{2} \\ & -2 A+B=3 \Rightarrow B=4 \\ & \therefore(x+3)=\dfrac{1}{2}(2 x-2)+4 \\ & \Rightarrow \int \dfrac{x+3}{x^{2}-2 x-5} d x=\int \dfrac{\dfrac{1}{2}(2 x-2)+4}{x^{2}-2 x-5} d x \\ & \quad=\dfrac{1}{2} \int \dfrac{2 x-2}{x^{2}-2 x-5} d x+4 \int \dfrac{1}{x^{2}-2 x-5} d x \end{aligned} $

मान लीजिए $I_1=\int \dfrac{2 x-2}{x^{2}-2 x-5} d x$ और $I_2=\int \dfrac{1}{x^{2}-2 x-5} d x$

$\therefore \int \dfrac{x+3}{(x^{2}-2 x-5)} d x=\dfrac{1}{2} I_1+4 I_2 \qquad …(1)$

तब, $I_1=\int \dfrac{2 x-2}{x^{2}-2 x-5} d x$

मान लीजिए $x^{2}-2 x-5=t$

$\Rightarrow(2 x-2) d x=d t$

$\Rightarrow I_1=\int \dfrac{d t}{t}=\log |t|=\log |x^{2}-2 x-5| \qquad …(2)$

$I_2=\int \dfrac{1}{x^{2}-2 x-5} d x$

$=\int \dfrac{1}{(x^{2}-2 x+1)-6} d x$

$=\int \dfrac{1}{(x-1)^{2}+(\sqrt{6})^{2}} d x$

$=\dfrac{1}{2 \sqrt{6}} \log (\dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}}) \qquad …(3)$

(2) और (3) को (1) में समावेशित करने पर, हम प्राप्त करते हैं

$ \begin{aligned} \int \dfrac{x+3}{x^{2}-2 x-5} d x & =\dfrac{1}{2} \log |x^{2}-2 x-5|+\dfrac{4}{2 \sqrt{6}} \log |\dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C \\ & =\dfrac{1}{2} \log |x^{2}-2 x-5|+\dfrac{2}{\sqrt{6}} \log |\dfrac{x-1-\sqrt{6}}{x-1+\sqrt{6}}|+C \end{aligned} $

हल

मान लीजिए $5 x+3=A \dfrac{d}{d x}(x^{2}+4 x+10)+B$

$\Rightarrow 5 x+3=A(2 x+4)+B$

$x$ के गुणांक और अचर पद के गुणांक की तुलना करने पर, हम प्राप्त करते हैं

$ \begin{aligned} & 2 A=5 \Rightarrow A=\dfrac{5}{2} \\ & 4 A+B=3 \Rightarrow B=-7 \\ & \therefore 5 x+3=\dfrac{5}{2}(2 x+4)-7 \\ & \Rightarrow \int \dfrac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\int \dfrac{\dfrac{5}{2}(2 x+4)-7}{\sqrt{x^{2}+4 x+10}} d x \\ & \quad=\dfrac{5}{2} \int \dfrac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x-7 \int \dfrac{1}{\sqrt{x^{2}+4 x+10}} d x $

\end{aligned} $

मान लीजिए $I_1=\int \dfrac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$ और $I_2=\int \dfrac{1}{\sqrt{x^{2}+4 x+10}} d x$

$\therefore \int \dfrac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x=\dfrac{5}{2} I_1-7 I_2 \qquad …(1)$

फिर, $I_1=\int \dfrac{2 x+4}{\sqrt{x^{2}+4 x+10}} d x$

मान लीजिए $x^{2}+4 x+10=t$

$\therefore(2 x+4) d x=d t$

$\Rightarrow I_1=\int \dfrac{d t}{t}=2 \sqrt{t}=2 \sqrt{x^{2}+4 x+10} \qquad …(2)$

$I_2=\int \dfrac{1}{\sqrt{x^{2}+4 x+10}} d x$

$=\int \dfrac{1}{\sqrt{(x^{2}+4 x+4)+6}} d x$

$=\int \dfrac{1}{(x+2)^{2}+(\sqrt{6})^{2}} d x$

$=\log |(x+2) \sqrt{x^{2}+4 x+10}| \qquad …(3)$

समीकरण (2) और (3) का उपयोग करके (1) में, हम प्राप्त करते हैं

$ \begin{aligned} \int \dfrac{5 x+3}{\sqrt{x^{2}+4 x+10}} d x & =\dfrac{5}{2}[2 \sqrt{x^{2}+4 x+10}]-7 \log |(x+2)+\sqrt{x^{2}+4 x+10}|+C \\ & =5 \sqrt{x^{2}+4 x+10}-7 \log (x+2)+\sqrt{x^{2}+4 x+10} \mid+C \end{aligned} $

समाधान

$ \begin{aligned} \int \dfrac{d x}{x^{2}+2 x+2} & =\int \dfrac{d x}{(x^{2}+2 x+1)+1} \\ & =\int \dfrac{1}{(x+1)^{2}+(1)^{2}} d x \\ & =[\tan ^{-1}(x+1)]+C \end{aligned} $

अतः, सही उत्तर B है।

समाधान

$\int \dfrac{d x}{\sqrt{9 x-4 x^{2}}}$

$=\int \dfrac{1}{\sqrt{-4(x^{2}-\dfrac{9}{4} x)}} d x$

$=\int \dfrac{1}{-4(x^{2}-\dfrac{9}{4} x+\dfrac{81}{64}-\dfrac{81}{64})} d x$

$=\int \dfrac{1}{\sqrt{-4[(x-\dfrac{9}{8})^{2}-(\dfrac{9}{8})^{2}]}} d x$

$=\dfrac{1}{2} \int \dfrac{1}{\sqrt{(\dfrac{9}{8})^{2}-(x-\dfrac{9}{8})^{2}}} d x$

$=\dfrac{1}{2}[\sin ^{-1}(\dfrac{x-\dfrac{9}{8}}{\dfrac{9}{8}})]+C\qquad (\because \int \dfrac{d y}{\sqrt{a^{2}-y^{2}}}=\sin ^{-1} \dfrac{y}{a}+C)$

$=\dfrac{1}{2} \sin ^{-1}(\dfrac{8 x-9}{9})+C$

इसलिए, सही उत्तर B है।

हल

मान लीजिए $\dfrac{x}{(x+1)(x+2)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}$

$\Rightarrow x=A(x+2)+B(x+1)$

$ x $ के गुणांक और स्थिरांक की तुलना करने पर, हम प्राप्त करते हैं

$ A + B = 1 $

$ 2A + B = 0 $

हल करने पर, हम प्राप्त करते हैं

$ A = -1 $ और $ B = 2 $

$\therefore \dfrac{x}{(x+1)(x+2)}=\dfrac{-1}{(x+1)}+\dfrac{2}{(x+2)}$

$\Rightarrow \int \dfrac{x}{(x+1)(x+2)} d x=\int \dfrac{-1}{(x+1)}+\dfrac{2}{(x+2)} d x$

$=-\log |x+1|+2 \log |x+2|+C$

$=\log (x+2)^{2}-\log |x+1|+C$

$=\log \dfrac{(x+2)^{2}}{(x+1)}+C$

हल

मान लीजिए $\dfrac{1}{(x+3)(x-3)}=\dfrac{A}{(x+3)}+\dfrac{B}{(x-3)}$

$1=A(x-3)+B(x+3)$

$ x $ के गुणांक और स्थिरांक की तुलना करने पर, हम प्राप्त करते हैं

$ A + B = 0 $

$ -3A + 3B = 1 $

हल करने पर, हम प्राप्त करते हैं

$ A = -\dfrac{1}{6} $ और $ B = \dfrac{1}{6} $

$\therefore \dfrac{1}{(x+3)(x-3)}=\dfrac{-1}{6(x+3)}+\dfrac{1}{6(x-3)}$

$\Rightarrow \int \dfrac{1}{(x^{2}-9)} d x=\int(\dfrac{-1}{6(x+3)}+\dfrac{1}{6(x-3)}) d x$

$ =-\dfrac{1}{6} \log |x+3|+\dfrac{1}{6} \log |x-3|+C $

$ =\dfrac{1}{6} \log |\dfrac{(x-3)}{(x+3)}|+C $

हल

मान लीजिए $\dfrac{3 x-1}{(x-1)(x-2)(x-3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-3)}$

$3 x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \qquad…(1)$

समीकरण (1) में क्रमशः $x=1,2$, और 3 को प्रतिस्थापित करने पर, हम प्राप्त करते हैं

$A=1, B=-5$, और $C=4$

$\therefore \dfrac{3 x-1}{(x-1)(x-2)(x-3)}=\dfrac{1}{(x-1)}-\dfrac{5}{(x-2)}+\dfrac{4}{(x-3)}$

$\Rightarrow \int \dfrac{3 x-1}{(x-1)(x-2)(x-3)} d x=\int{\dfrac{1}{(x-1)}-\dfrac{5}{(x-2)}+\dfrac{4}{(x-3)}} d x$

$=\log |x-1|-5 \log |x-2|+4 \log |x-3|+C$

हल

मान लीजिए $\dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}+\dfrac{C}{(x-3)}$

$$x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \qquad…(1)$$

समीकरण (1) में क्रमशः $x=1,2$, और 3 को रखने पर, हम प्राप्त करते हैं

$A=\dfrac{1}{2}, B=-2$, और $C=\dfrac{3}{2}$

$\therefore \dfrac{x}{(x-1)(x-2)(x-3)}=\dfrac{1}{2(x-1)}-\dfrac{2}{(x-2)}+\dfrac{3}{2(x-3)}$

$\Rightarrow \int \dfrac{x}{(x-1)(x-2)(x-3)} d x=\int{\dfrac{1}{2(x-1)}-\dfrac{2}{(x-2)}+\dfrac{3}{2(x-3)}} d x$

$ =\dfrac{1}{2} \log |x-1|-2 \log |x-2|+\dfrac{3}{2} \log |x-3|+C $

Solution

मान लीजिए $\dfrac{2 x}{x^{2}+3 x+2}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}$

$2 x=A(x+2)+B(x+1) \qquad…(1)$

समीकरण (1) में क्रमशः $x=-1$ और -2 को रखने पर, हम प्राप्त करते हैं

$A=-2$ और $B=4$ $\therefore \dfrac{2 x}{(x+1)(x+2)}=\dfrac{-2}{(x+1)}+\dfrac{4}{(x+2)}$

$\Rightarrow \int \dfrac{2 x}{(x+1)(x+2)} d x=\int{\dfrac{4}{(x+2)}-\dfrac{2}{(x+1)}} d x$

$ =4 \log |x+2|-2 \log |x+1|+C $

Solution

दिए गए समाकलन के अंश को एक अच्छी भिन्न नहीं कहा जा सकता।

इसलिए, $(1-x^{2})$ को $x(1-2 x)$ से विभाजित करने पर, हम प्राप्त करते हैं

$\dfrac{1-x^{2}}{x(1-2 x)}=\dfrac{1}{2}+\dfrac{1}{2}(\dfrac{2-x}{x(1-2 x)}) \qquad…(1)$

मान लीजिए $\dfrac{2-x}{x(1-2 x)}=\dfrac{A}{x}+\dfrac{B}{(1-2 x)}$

$\Rightarrow(2-x)=A(1-2 x)+B x $

समीकरण (1) में क्रमशः $x=0$ और $\dfrac{1}{2}$ को रखने पर, हम प्राप्त करते हैं

$A=2$ और $B=3$

$\therefore \dfrac{2-x}{x(1-2 x)}=\dfrac{2}{x}+\dfrac{3}{1-2 x}$

समीकरण (1) में प्रतिस्थापित करने पर, हम प्राप्त करते हैं

$ \begin{aligned} & \dfrac{1-x^{2}}{x(1-2 x)}=\dfrac{1}{2}+\dfrac{1}{2}{\dfrac{2}{x}+\dfrac{3}{(1-2 x)}} \\ & \Rightarrow \int \dfrac{1-x^{2}}{x(1-2 x)} d x=\int{\dfrac{1}{2}+\dfrac{1}{2}(\dfrac{2}{x}+\dfrac{3}{1-2 x})} d x \\ & =\dfrac{x}{2}+\log |x|+\dfrac{3}{2(-2)} \log |1-2 x|+C \\ & =\dfrac{x}{2}+\log |x|-\dfrac{3}{4} \log |1-2 x|+C \end{aligned} $

Solution

मान लीजिए $\dfrac{x}{(x^{2}+1)(x-1)}=\dfrac{A x+B}{(x^{2}+1)}+\dfrac{C}{(x-1)} \qquad…(1)$

$x=(A x+B)(x-1)+C(x^{2}+1)$

$x=A x^{2}-A x+B x-B+C x^{2}+C $

$ x^{2} $, $ x $, और अचर पद के गुणांक के तुलना करने पर हम प्राप्त करते हैं

$ A+C=0 $

$ -A+B=1 $

$ -B+C=0 $

इन समीकरणों को हल करने पर हम प्राप्त करते हैं

$ A=-\dfrac{1}{2}, B=\dfrac{1}{2} $, और $ C=\dfrac{1}{2} $

समीकरण (1) से हम प्राप्त करते हैं

$ \begin{aligned} & \therefore \dfrac{x}{(x^{2}+1)(x-1)}=\dfrac{(-\dfrac{1}{2} x+\dfrac{1}{2})}{x^{2}+1}+\dfrac{\dfrac{1}{2}}{(x-1)} \\ & \Rightarrow \int \dfrac{x}{(x^{2}+1)(x-1)}=-\dfrac{1}{2} \int \dfrac{x}{x^{2}+1} d x+\dfrac{1}{2} \int \dfrac{1}{x^{2}+1} d x+\dfrac{1}{2} \int \dfrac{1}{x-1} d x \\ & \quad=-\dfrac{1}{4} \int \dfrac{2 x}{x^{2}+1} d x+\dfrac{1}{2} \tan ^{-1} x+\dfrac{1}{2} \log |x-1|+C \end{aligned} $

$ \int \dfrac{2 x}{x^{2}+1} d x $ को विचार करें, मान लीजिए $ (x^{2}+1)=t \Rightarrow 2 x d x=d t $

$ \Rightarrow \int \dfrac{2 x}{x^{2}+1} d x=\int \dfrac{d t}{t}=\log |t|=\log |x^{2}+1| $

$ \begin{aligned} \therefore \int \dfrac{x}{(x^{2}+1)(x-1)} & =-\dfrac{1}{4} \log |x^{2}+1|+\dfrac{1}{2} \tan ^{-1} x+\dfrac{1}{2} \log |x-1|+C \\ & =\dfrac{1}{2} \log |x-1|-\dfrac{1}{4} \log |x^{2}+1|+\dfrac{1}{2} \tan ^{-1} x+C \end{aligned} $

Solution

मान लीजिए $\dfrac{x}{(x-1)^{2}(x+2)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-1)^{2}}+\dfrac{C}{(x+2)}$

$x=A(x-1)(x+2)+B(x+2)+C(x-1)^{2}$

$ x=1 $ के लिए प्रतिस्थापन करने पर हम प्राप्त करते हैं

$ B=\dfrac{1}{3} $

$ x^{2} $ और अचर पद के गुणांक के तुलना करने पर हम प्राप्त करते हैं

$ A+C=0 $

$ -2 A+2 B+C=0 $

हल करने पर हम प्राप्त करते हैं

$ \begin{aligned} & A=\dfrac{2}{9} \text{ and } C=\dfrac{-2}{9} \\ & \begin{aligned} & \therefore \dfrac{x}{(x-1)^{2}(x+2)}=\dfrac{2}{9(x-1)}+\dfrac{1}{3(x-1)^{2}}-\dfrac{2}{9(x+2)} \\ & \Rightarrow \int \dfrac{x}{(x-1)^{2}(x+2)} d x=\dfrac{2}{9} \int \dfrac{1}{(x-1)} d x+\dfrac{1}{3} \int \dfrac{1}{(x-1)^{2}} d x-\dfrac{2}{9} \int \dfrac{1}{(x+2)} d x \\ &=\dfrac{2}{9} \log |x-1|+\dfrac{1}{3}(\dfrac{-1}{x-1})-\dfrac{2}{9} \log |x+2|+C \\ &=\dfrac{2}{9} \log |\dfrac{x-1}{x+2}|-\dfrac{1}{3(x-1)}+C

\end{aligned} \end{aligned} $

Solution

$\dfrac{3 x+5}{x^{3}-x^{2}-x+1}=\dfrac{3 x+5}{(x-1)^{2}(x+1)}$

Let $\dfrac{3 x+5}{(x-1)^{2}(x+1)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-1)^{2}}+\dfrac{C}{(x+1)}$

$3 x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2}$

$3 x+5=A(x^{2}-1)+B(x+1)+C(x^{2}+1-2 x) \qquad…(1)$

$ x=1 $ के समीकरण (1) में प्रतिस्थापित करने पर, हम प्राप्त करते हैं

$ B=4 $

$ x^{2} $ और $ x $ के गुणांक के तुलना करने पर, हम प्राप्त करते हैं

$ A+C=0 $

$ B-2 C=3 $

हल करने पर, हम प्राप्त करते हैं

$ A=-\dfrac{1}{2} $ और $ C=\dfrac{1}{2} $

$\therefore \dfrac{3 x+5}{(x-1)^{2}(x+1)}=\dfrac{-1}{2(x-1)}+\dfrac{4}{(x-1)^{2}}+\dfrac{1}{2(x+1)}$

$\Rightarrow \int \dfrac{3 x+5}{(x-1)^{2}(x+1)} d x\=-\dfrac{1}{2} \int \dfrac{1}{x-1} d x+4 \int \dfrac{1}{(x-1)^{2}} d x+\dfrac{1}{2} \int \dfrac{1}{(x+1)} d x$

$=-\dfrac{1}{2} \log |x-1|+4(\dfrac{-1}{x-1})+\dfrac{1}{2} \log |x+1|+C$

$=\dfrac{1}{2} \log |\dfrac{x+1}{x-1}|-\dfrac{4}{(x-1)}+C$

Solution

$\dfrac{2 x-3}{(x^{2}-1)(2 x+3)}=\dfrac{2 x-3}{(x+1)(x-1)(2 x+3)}$

Let $\dfrac{2 x-3}{(x+1)(x-1)(2 x+3)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}+\dfrac{C}{(2 x+3)}$

$\Rightarrow(2 x-3)=A(x-1)(2 x+3)+B(x+1)(2 x+3)+C(x+1)(x-1)$

$\Rightarrow(2 x-3)=A(2 x^{2}+x-3)+B(2 x^{2}+5 x+3)+C(x^{2}-1)$

$\Rightarrow(2 x-3)=(2 A+2 B+C) x^{2}+(A+5 B) x+(-3 A+3 B-C)$

$ x^{2} $ और $ x $ के गुणांक के तुलना करने पर, हम प्राप्त करते हैं

$ B=-\dfrac{1}{10}, A=\dfrac{5}{2} $, और $ C=-\dfrac{24}{5} $

$\therefore \dfrac{2 x-3}{(x+1)(x-1)(2 x+3)}=\dfrac{5}{2(x+1)}-\dfrac{1}{10(x-1)}-\dfrac{24}{5(2 x+3)}$

$\Rightarrow \int \dfrac{2 x-3}{(x^{2}-1)(2 x+3)} d x=\dfrac{5}{2} \int \dfrac{1}{(x+1)} d x-\dfrac{1}{10} \int \dfrac{1}{x-1} d x-\dfrac{24}{5} \int \dfrac{1}{(2 x+3)} d x$

$=\dfrac{5}{2} \log |x+1|-\dfrac{1}{10} \log |x-1|-\dfrac{24}{5 \times 2} \log |2 x+3|$

$=\dfrac{5}{2} \log |x+1|-\dfrac{1}{10} \log |x-1|-\dfrac{12}{5} \log |2 x+3|+C$

हल

$\dfrac{5 x}{(x+1)(x^{2}-4)}=\dfrac{5 x}{(x+1)(x+2)(x-2)}$

मान लीजिए $\dfrac{5 x}{(x+1)(x+2)(x-2)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+2)}+\dfrac{C}{(x-2)}$

$5 x=A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2) \qquad…(1)$

समीकरण (1) में क्रमशः $x=-1,-2$, और 2 को रखने पर, हम प्राप्त करते हैं

$ \begin{aligned} & A=\dfrac{5}{3}, B=-\dfrac{5}{2}, \text{ और } C=\dfrac{5}{6} \\ & \begin{aligned} \therefore \dfrac{5 x}{(x+1)(x+2)(x-2)} & =\dfrac{5}{3(x+1)}-\dfrac{5}{2(x+2)}+\dfrac{5}{6(x-2)} \\ \Rightarrow \int \dfrac{5 x}{(x+1)(x^{2}-4)} d x & =\dfrac{5}{3} \int \dfrac{1}{(x+1)} d x-\dfrac{5}{2} \int \dfrac{1}{(x+2)} d x+\dfrac{5}{6} \int \dfrac{1}{(x-2)} d x \\ & =\dfrac{5}{3} \log |x+1|-\dfrac{5}{2} \log |x+2|+\dfrac{5}{6} \log |x-2|+C \end{aligned} \end{aligned} $

हल

दिया गया समकारक एक निर्मित भिन्न नहीं है।

इसलिए, $(x^{3}+x+1)$ को $x^{2}-1$ से विभाजित करने पर, हम प्राप्त करते हैं

$\dfrac{x^{3}+x+1}{x^{2}-1}=x+\dfrac{2 x+1}{x^{2}-1}$

मान लीजिए $\dfrac{2 x+1}{x^{2}-1}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}$

$2 x+1=A(x-1)+B(x+1) \qquad…(1)$

समीकरण (1) में क्रमशः $x=1$ और -1 को रखने पर, हम प्राप्त करते हैं

$ A=\dfrac{1}{2} \text{ और } B=\dfrac{3}{2} $

$\therefore \dfrac{x^{3}+x+1}{x^{2}-1}=x+\dfrac{1}{2(x+1)}+\dfrac{3}{2(x-1)}$

$\Rightarrow \int \dfrac{x^{3}+x+1}{x^{2}-1} d x=\int x d x+\dfrac{1}{2} \int \dfrac{1}{(x+1)} d x+\dfrac{3}{2} \int \dfrac{1}{(x-1)} d x$

$ =\dfrac{x^{2}}{2}+\dfrac{1}{2} \log |x+1|+\dfrac{3}{2} \log |x-1|+C $

हल

मान लीजिए $\dfrac{2}{(1-x)(1+x^{2})}=\dfrac{A}{(1-x)}+\dfrac{B x+C}{(1+x^{2})}$

$2=A(1+x^{2})+(B x+C)(1-x)$

$2=A+A x^{2}+B x-B x^{2}+C-C x$

$ x^{2}, x $, और अचर पद के गुणांक के बराबर करने पर, हम प्राप्त करते हैं

$A-B=0$

$B-C=0$

$A+C=2$

इन समीकरणों को हल करने पर, हम प्राप्त करते हैं

$A=1, B=1$, और $C=1$

$\therefore \dfrac{2}{(1-x)(1+x^{2})}=\dfrac{1}{1-x}+\dfrac{x+1}{1+x^{2}}$

$\Rightarrow \int \dfrac{2}{(1-x)(1+x^{2})} d x=\int \dfrac{1}{1-x} d x+\int \dfrac{x}{1+x^{2}} d x+\int \dfrac{1}{1+x^{2}} d x$

$=-\int \dfrac{1}{x-1} d x+\dfrac{1}{2} \int \dfrac{2 x}{1+x^{2}} d x+\int \dfrac{1}{1+x^{2}} d x$

$=-\log |x-1|+\dfrac{1}{2} \log |1+x^{2}|+\tan ^{-1} x+C$

Solution

Let $\dfrac{3 x-1}{(x+2)^{2}}=\dfrac{A}{(x+2)}+\dfrac{B}{(x+2)^{2}}$

$\Rightarrow 3 x-1=A(x+2)+B$

Coefficient of $x$ और constant term के तुलना करने पर, हम प्राप्त करते हैं $A=3$

$2 A+B=-1 \Rightarrow B=-7$

$\therefore \dfrac{3 x-1}{(x+2)^{2}}=\dfrac{3}{(x+2)}-\dfrac{7}{(x+2)^{2}}$

$\Rightarrow \int \dfrac{3 x-1}{(x+2)^{2}} d x=3 \int \dfrac{1}{(x+2)} d x-7 \int \dfrac{x}{(x+2)^{2}} d x$

$=3 \log |x+2|-7(\dfrac{-1}{(x+2)})+C$

$=3 \log |x+2|+\dfrac{7}{(x+2)}+C$

Solution

$\dfrac{1}{(x^{4}-1)}=\dfrac{1}{(x^{2}-1)(x^{2}+1)}=\dfrac{1}{(x+1)(x-1)(1+x^{2})}$

Let $\dfrac{1}{(x+1)(x-1)(1+x^{2})}=\dfrac{A}{(x+1)}+\dfrac{B}{(x-1)}+\dfrac{C x+D}{(x^{2}+1)}$

$1=A(x-1)(x^{2}+1)+B(x+1)(x^{2}+1)+(C x+D)(x^{2}-1)$

$1=A(x^{3}+x-x^{2}-1)+B(x^{3}+x+x^{2}+1)+C x^{3}+D x^{2}-C x-D$

$1=(A+B+C) x^{3}+(-A+B+D) x^{2}+(A+B-C) x+(-A+B-D)$

$x^{3}, x^{2}, x$ और constant term के गुणांक के तुलना करने पर, हम प्राप्त करते हैं

$A+B+C=0$

$-A+B+D=0$

$A+B-C=0$

$-A+B-D=1$

इन समीकरणों को हल करने पर, हम प्राप्त करते हैं

$A=-\dfrac{1}{4}, B=\dfrac{1}{4}, C=0$, और $D=-\dfrac{1}{2}$

$\therefore \dfrac{1}{x^{4}-1}=\dfrac{-1}{4(x+1)}+\dfrac{1}{4(x-1)}-\dfrac{1}{2(x^{2}+1)}$

$\Rightarrow \int \dfrac{1}{x^{4}-1} d x=-\dfrac{1}{4} \log |x-1|+\dfrac{1}{4} \log |x-1|-\dfrac{1}{2} \tan ^{-1} x+C$

$ =\dfrac{1}{4} \log |\dfrac{x-1}{x+1}|-\dfrac{1}{2} \tan ^{-1} x+C $

हल

$\dfrac{1}{x(x^{n}+1)}$

अंश और हर में $x^{n-1}$ से गुणा करने पर, हम प्राप्त करते हैं

$\dfrac{1}{x(x^{n}+1)}=\dfrac{x^{n-1}}{x^{n-1} x(x^{n}+1)}=\dfrac{x^{n-1}}{x^{n}(x^{n}+1)}$

मान लीजिए $x^{n}=t \Rightarrow x^{n-1} d x=d t$

$\therefore \int \dfrac{1}{x(x^{n}+1)} d x=\int \dfrac{x^{n-1}}{x^{n}(x^{n}+1)} d x=\dfrac{1}{n} \int \dfrac{1}{t(t+1)} d t$

मान लीजिए $\dfrac{1}{t(t+1)}=\dfrac{A}{t}+\dfrac{B}{(t+1)}$

$1=A(1+t)+B t$

समीकरण (1) में $t=0,-1$ के लिए प्रतिस्थापित करने पर, हम प्राप्त करते हैं

$A=1$ और $B=-1$

$\therefore \dfrac{1}{t(t+1)}=\dfrac{1}{t}-\dfrac{1}{(1+t)}$ $\Rightarrow \int \dfrac{1}{x(x^{n}+1)} d x=\dfrac{1}{n} \int{\dfrac{1}{t}-\dfrac{1}{(t+1)}} d x$

$=\dfrac{1}{n}[\log |t|-\log |t+1|]+C$

$=-\dfrac{1}{n}[\log |x^{n}|-\log |x^{n}+1|]+C$

$=\dfrac{1}{n} \log |\dfrac{x^{n}}{x^{n}+1}|+C$

हल

$\dfrac{\cos x}{(1-\sin x)(2-\sin x)}$

मान लीजिए $\sin x=t \Rightarrow \cos x d x=d t$

$\therefore \int \dfrac{\cos x}{(1-\sin x)(2-\sin x)} d x=\int \dfrac{d t}{(1-t)(2-t)}$

मान लीजिए $\dfrac{1}{(1-t)(2-t)}=\dfrac{A}{(1-t)}+\dfrac{B}{(2-t)}$

$1=A(2-t)+B(1-t)$

समीकरण (1) में $t=2$ और फिर $t=1$ के लिए प्रतिस्थापित करने पर, हम प्राप्त करते हैं

$A=1$ और $B=-1$

$\therefore \dfrac{1}{(1-t)(2-t)}=\dfrac{1}{(1-t)}-\dfrac{1}{(2-t)}$

$ \begin{aligned} \Rightarrow \int \dfrac{\cos x}{(1-\sin x)(2-\sin x)} d x & =\int {\dfrac{1}{1-t}-\dfrac{1}{(2-t)}} d t \\ & =-\log |1-t|+\log |2-t|+C \\ & =\log |\dfrac{2-t}{1-t}|+C \\ & =\log |\dfrac{2-\sin x}{1-\sin x}|+C \end{aligned} $

हल

$\dfrac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)}=1-\dfrac{(4 x^{2}+10)}{(x^{2}+3)(x^{2}+4)}$

मान लीजिए $\dfrac{4 x^{2}+10}{(x^{2}+3)(x^{2}+4)}=\dfrac{A x+B}{(x^{2}+3)}+\dfrac{C x+D}{(x^{2}+4)}$

$4 x^{2}+10=(A x+B)(x^{2}+4)+(C x+D)(x^{2}+3)$

$4 x^{2}+10=A x^{3}+4 A x+B x^{2}+4 B+C x^{3}+3 C x+D x^{2}+3 D$

$4 x^{2}+10=(A+C) x^{3}+(B+D) x^{2}+(4 A+3 C) x+(4 B+3 D)$

$ x^{3}, x^{2}, x $ और स्थिरांक के गुणांक के तुलना करने पर हम प्राप्त करते हैं

$ A+C=0 $

$ B+D=4 $

$ 4 A+3 C=0 $

$ 4 B+3 D=10 $

इन समीकरणों को हल करने पर हम प्राप्त करते हैं

$ A=0, B=-2, C=0 $, और $ D=6 $

$\therefore \dfrac{4 x^{2}+10}{(x^{2}+3)(x^{2}+4)}=\dfrac{-2}{(x^{2}+3)}+\dfrac{6}{(x^{2}+4)}$

$\dfrac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)}=1-(\dfrac{-2}{(x^{2}+3)}+\dfrac{6}{(x^{2}+4)})$

$\Rightarrow \int \dfrac{(x^{2}+1)(x^{2}+2)}{(x^{2}+3)(x^{2}+4)} d x=\int{1+\dfrac{2}{(x^{2}+3)}-\dfrac{6}{(x^{2}+4)}} d x$

$={1+\dfrac{2}{x^{2}+(\sqrt{3})^{2}}-\dfrac{6}{x^{2}+2^{2}}}$

$=x+2(\dfrac{1}{\sqrt{3}} \tan ^{-1} \dfrac{x}{\sqrt{3}})-6(\dfrac{1}{2} \tan ^{-1} \dfrac{x}{2})+C$

$=x+\dfrac{2}{\sqrt{3}} \tan ^{-1} \dfrac{x}{\sqrt{3}}-3 \tan ^{-1} \dfrac{x}{2}+C$

Solution

$\dfrac{2 x}{(x^{2}+1)(x^{2}+3)}$

मान लीजिए $x^{2}=t \Rightarrow 2 x d x=d t$

$\therefore \int \dfrac{2 x}{(x^{2}+1)(x^{2}+3)} d x=\int \dfrac{d t}{(t+1)(t+3)}$

मान लीजिए $\dfrac{1}{(t+1)(t+3)}=\dfrac{A}{(t+1)}+\dfrac{B}{(t+3)}$

$1=A(t+3)+B(t+1) \qquad …(1)$

समीकरण (1) में $t=-3$ और $t=-1$ के लिए प्रतिस्थापन करने पर हम प्राप्त करते हैं

$A=\dfrac{1}{2}$ और $B=-\dfrac{1}{2}$

$\therefore \dfrac{1}{(t+1)(t+3)}=\dfrac{1}{2(t+1)}-\dfrac{1}{2(t+3)}$

$\Rightarrow \int \dfrac{2 x}{(x^{2}+1)(x^{2}+3)} d x=\int{\dfrac{1}{2(t+1)}-\dfrac{1}{2(t+3)}} d t$

$=\dfrac{1}{2} \log |(t+1)|-\dfrac{1}{2} \log |t+3|+C$

$=\dfrac{1}{2} \log |\dfrac{t+1}{t+3}|+C$

$=\dfrac{1}{2} \log |\dfrac{x^{2}+1}{x^{2}+3}|+C$

Solution

$\dfrac{1}{x(x^{4}-1)}$

अंश और हर में $x^{3}$ से गुणा करने पर हम प्राप्त करते हैं

$ \begin{aligned} & \dfrac{1}{x(x^{4}-1)}=\dfrac{x^{3}}{x^{4}(x^{4}-1)} \\ & \therefore \int \dfrac{1}{x(x^{4}-1)} d x=\int \dfrac{x^{3}}{x^{4}(x^{4}-1)} d x \end{aligned} $

मान लीजिए $x^{4}=t \Rightarrow 4 x^{3} d x=d t$

$\therefore \int \dfrac{1}{x(x^{4}-1)} d x=\dfrac{1}{4} \int \dfrac{d t}{t(t-1)}$

मान लीजिए $\dfrac{1}{t(t-1)}=\dfrac{A}{t}+\dfrac{B}{(t-1)}$

$1=A(t-1)+B t$

समीकरण (1) में $t=0$ और 1 के लिए प्रतिस्थापित करने पर, हम प्राप्त करते हैं

$A=-1$ और $B=1$

$\Rightarrow \dfrac{1}{t(t+1)}=\dfrac{-1}{t}+\dfrac{1}{t-1}$

$\Rightarrow \int \dfrac{1}{x(x^{4}-1)} d x=\dfrac{1}{4} \int{\dfrac{-1}{t}+\dfrac{1}{t-1}} d t$

$=\dfrac{1}{4}[-\log |t|+\log |t-1|]+C$

$=\dfrac{1}{4} \log |\dfrac{t-1}{t}|+C$

$=\dfrac{1}{4} \log |\dfrac{x^{4}-1}{x^{4}}|+C$

Solution

$\dfrac{1}{(e^{x}-1)}$

मान लीजिए $e^{x}=t \Rightarrow e^{x} d x=d t$

$\Rightarrow \int \dfrac{1}{e^{x}-1} d x=\int \dfrac{1}{t-1} \times \dfrac{d t}{t}=\int \dfrac{1}{t(t-1)} d t$

मान लीजिए $\dfrac{1}{t(t-1)}=\dfrac{A}{t}+\dfrac{B}{t-1}$

$1=A(t-1)+B t$

समीकरण (1) में $t=1$ और $t=0$ के लिए प्रतिस्थापित करने पर, हम प्राप्त करते हैं

$A=-1$ और $B=1$

$\therefore \dfrac{1}{t(t-1)}=\dfrac{-1}{t}+\dfrac{1}{t-1}$

$\Rightarrow \int \dfrac{1}{t(t-1)} d t=\log |\dfrac{t-1}{t}|+C$

$ =\log |\dfrac{e^{x}-1}{e^{x}}|+C $

Solution

मान लीजिए $\dfrac{x}{(x-1)(x-2)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x-2)}$

$x=A(x-2)+B(x-1) \qquad…(1)$

समीकरण (1) में $x=1$ और 2 के लिए प्रतिस्थापित करने पर, हम प्राप्त करते हैं

$A=-1$ और $B=2$

$\therefore \dfrac{x}{(x-1)(x-2)}=-\dfrac{1}{(x-1)}+\dfrac{2}{(x-2)}$

$\Rightarrow \int \dfrac{x}{(x-1)(x-2)} d x=\int{\dfrac{-1}{(x-1)}+\dfrac{2}{(x-2)}} d x$

$ =-\log |x-1|+2 \log |x-2|+C $

$ =\log |\dfrac{(x-2)^{2}}{x-1}|+C $

अतः, सही उत्तर B है।

हल

मान लीजिए $\dfrac{1}{x(x^{2}+1)}=\dfrac{A}{x}+\dfrac{B x+C}{x^{2}+1}$

$1=A(x^{2}+1)+(B x+C) x$

$ x^{2}, x $, और अचर पद के गुणांक के तुलना करने पर, हम प्राप्त करते हैं

$ A+B=0 $

$ C=0 $

$ A=1 $

इन समीकरणों को हल करने पर, हम प्राप्त करते हैं

$ A=1, B=-1 $, और $ C=0 $

$\therefore \dfrac{1}{x(x^{2}+1)}=\dfrac{1}{x}+\dfrac{-x}{x^{2}+1}$

$\Rightarrow \int \dfrac{1}{x(x^{2}+1)} d x=\int{\dfrac{1}{x}-\dfrac{x}{x^{2}+1}} d x$

$ =\log |x|-\dfrac{1}{2} \log |x^{2}+1|+C $

इसलिए, सही उत्तर A है।

हल

मान लीजिए $I=\int x \sin x d x$

$ x $ को पहला फलन और $ \sin x $ को दूसरा फलन मानकर समाकलन द्वारा एक बार इंटीग्रेट करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =x \int \sin x d x - \int {\dfrac{d}{d x} (x) \int \sin x d x} d x \\ & =x(-\cos x)-\int 1 \cdot(-\cos x) d x \\ & =-x \cos x+\sin x+C \end{aligned} $

हल

मान लीजिए $I=\int x \sin 3 x d x$

$ x $ को पहला फलन और $ \sin 3 x $ को दूसरा फलन मानकर समाकलन द्वारा एक बार इंटीग्रेट करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =x \int \sin 3 x d x-\int\ {(\dfrac{d}{d x} x) \int \sin 3 x d x}dx \\ & =x(\dfrac{-\cos 3 x}{3})-\int 1 \cdot(\dfrac{-\cos 3 x}{3}) d x \\ & =\dfrac{-x \cos 3 x}{3}+\dfrac{1}{3} \int \cos 3 x d x \\ & =\dfrac{-x \cos 3 x}{3}+\dfrac{1}{9} \sin 3 x+C \end{aligned} $

हल

$x^{2} e^{x}$

मान लीजिए $I=\int x^{2} e^{x} d x$

$ x^{2} $ को पहला फलन और $ e^{x} $ को दूसरा फलन मानकर समाकलन द्वारा एक बार इंटीग्रेट करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =x^{2} \int e^{x} d x-\int {(\dfrac{d}{d x} x^{2}) \int e^{x} d x} d x \\ & =x^{2} e^{x}-\int 2 x \cdot e^{x} d x \\ & =x^{2} e^{x}-2 \int x \cdot e^{x} d x \end{aligned} $

फिर एक बार इंटीग्रेट करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} & =x^{2} e^{x}-2[x \cdot \int e^{x} d x-\int{(\dfrac{d}{d x} x) \cdot \int e^{x} d x} d x] \\ & =x^{2} e^{x}-2[x e^{x}-\int e^{x} d x] \\ & =x^{2} e^{x}-2[x e^{x}-e^{x}] \\ & =x^{2} e^{x}-2 x e^{x}+2 e^{x}+C \\ & =e^{x}(x^{2}-2 x+2)+C \end{aligned} $

हल

मान लीजिए $I=\int x \log x d x$

$ \log x $ को पहला फलन और $ x $ को दूसरा फलन मानकर समाकलन द्वारा एक बार इंटीग्रेट करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =\log x \int x d x-\int{(\dfrac{d}{d x} \log x) \int x d x} d x \\

& =\log x \cdot \dfrac{x^{2}}{2}-\int \dfrac{1}{x} \cdot \dfrac{x^{2}}{2} d x \\ & =\dfrac{x^{2} \log x}{2}-\int \dfrac{x}{2} d x \\ & =\dfrac{x^{2} \log x}{2}-\dfrac{x^{2}}{4}+C \end{aligned} $

Solution

$ I=\int x \log 2 x d x $ लेते हुए, $ \log 2 x $ को पहला फलन और $ x $ को दूसरा फलन मानकर समाकलन द्वारा बर्णन करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =\log 2 x \int x d x-\int{\dfrac{d}{d x} ( \log2 x) \int x d x} d x \\ & =\log 2 x \cdot \dfrac{x^{2}}{2}-\int \dfrac{2}{2 x} \cdot \dfrac{x^{2}}{2} d x \\ & =\dfrac{x^{2} \log 2 x}{2}-\int \dfrac{x}{2} d x \\ & =\dfrac{x^{2} \log 2 x}{2}-\dfrac{x^{2}}{4}+C \end{aligned} $

Solution

$ I=\int x^{2} \log x d x $ लेते हुए, $ \log x $ को पहला फलन और $ x^{2} $ को दूसरा फलन मानकर समाकलन द्वारा बर्णन करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =\log x \int x^{2} d x-\int{(\dfrac{d}{d x} \log x) \int x^{2} d x} d x \\ & =\log x(\dfrac{x^{3}}{3})-\int \dfrac{1}{x} \cdot \dfrac{x^{3}}{3} d x \\ & =\dfrac{x^{3} \log x}{3}-\int \dfrac{x^{2}}{3} d x \\ & =\dfrac{x^{3} \log x}{3}-\dfrac{x^{3}}{9}+C \end{aligned} $

Solution

$ I=\int x \sin ^{-1} x d x $ लेते हुए, $ \sin ^{-1} x $ को पहला फलन और $ x $ को दूसरा फलन मानकर समाकलन द्वारा बर्णन करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =\sin ^{-1} x \int x d x-\int{(\dfrac{d}{d x} \sin ^{-1} x) \int x d x} d x \\ & =\sin ^{-1} x(\dfrac{x^{2}}{2})-\int \dfrac{1}{\sqrt{1-x^{2}}} \cdot \dfrac{x^{2}}{2} d x \\ & =\dfrac{x^{2} \sin ^{-1} x}{2}+\dfrac{1}{2} \int \dfrac{-x^{2}}{\sqrt{1-x^{2}}} d x \\ & =\dfrac{x^{2} \sin ^{-1} x}{2}+\dfrac{1}{2} \int\ {\dfrac{1-x^{2}}{\sqrt{1-x^{2}}}-\dfrac{1}{\sqrt{1-x^{2}}}\} d x \\ & =\dfrac{x^{2} \sin ^{-1} x}{2}+\dfrac{1}{2} \int\ {\sqrt{1-x^{2}}-\dfrac{1}{\sqrt{1-x^{2}}}\} d x \\ & =\dfrac{x^{2} \sin ^{-1} x}{2}+\dfrac{1}{2}\ {\int \sqrt{1-x^{2}} d x-\int \dfrac{1}{\sqrt{1-x^{2}}} d x\} \\

& =\dfrac{x^{2} \sin ^{-1} x}{2}+\dfrac{1}{2}\ {\dfrac{x}{2} \sqrt{1-x^{2}}+\dfrac{1}{2} \sin ^{-1} x-\sin ^{-1} x\}+C \\ & =\dfrac{x^{2} \sin ^{-1} x}{2}+\dfrac{x}{4} \sqrt{1-x^{2}}+\dfrac{1}{4} \sin ^{-1} x-\dfrac{1}{2} \sin ^{-1} x+C \\ & =\dfrac{1}{4}(2 x^{2}-1) \sin ^{-1} x+\dfrac{x}{4} \sqrt{1-x^{2}}+C \end{aligned} $

Solution

मान लीजिए $I=\int x \tan ^{-1} x d x$

$ \tan ^{-1} x $ को पहला फलन और $x$ को दूसरा फलन मानकर समाकलन द्वारा अंतरगत विधि का उपयोग करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =\tan ^{-1} x \int x d x-\int{\dfrac{d}{d x} (\tan ^{-1} x) \int x d x} d x \\ & =\tan ^{-1} x(\dfrac{x^{2}}{2})-\int \dfrac{1}{1+x^{2}} \cdot \dfrac{x^{2}}{2} d x \\ & =\dfrac{x^{2} \tan ^{-1} x}{2}-\dfrac{1}{2} \int \dfrac{x^{2}}{1+x^{2}} d x \\ & =\dfrac{x^{2} \tan ^{-1} x}{2}-\dfrac{1}{2} \int(\dfrac{x^{2}+1}{1+x^{2}}-\dfrac{1}{1+x^{2}}) d x \\ & =\dfrac{x^{2} \tan ^{-1} x}{2}-\dfrac{1}{2} \int(1-\dfrac{1}{1+x^{2}}) d x \\ & =\dfrac{x^{2} \tan ^{-1} x}{2}-\dfrac{1}{2}(x-\tan ^{-1} x)+C \\ & =\dfrac{x^{2}}{2} \tan ^{-1} x-\dfrac{x}{2}+\dfrac{1}{2} \tan ^{-1} x+C \end{aligned} $

Solution

मान लीजिए $I=\int x \cos ^{-1} x d x$

$\cos ^{-1} x$ को पहला फलन और $x$ को दूसरा फलन मानकर समाकलन द्वारा अंतरगत विधि का उपयोग करते हुए, हम प्राप्त करते हैं

$$ \begin{align*} I & =\cos ^{-1} x \int x d x-\int{\dfrac{d}{d x} (\cos ^{-1} x) \int x d x} d x \\ & =\cos ^{-1} x \dfrac{x^{2}}{2}-\int \dfrac{-1}{\sqrt{1-x^{2}}} \cdot \dfrac{x^{2}}{2} d x \\ & =\dfrac{x^{2} \cos ^{-1} x}{2}-\dfrac{1}{2} \int \dfrac{1-x^{2}-1}{\sqrt{1-x^{2}}} d x \\ & =\dfrac{x^{2} \cos ^{-1} x}{2}-\dfrac{1}{2} \int\ {\sqrt{1-x^{2}}+(\dfrac{-1}{\sqrt{1-x^{2}}})\} d x \\ & =\dfrac{x^{2} \cos ^{-1} x}{2}-\dfrac{1}{2} \int \sqrt{1-x^{2}} d x-\dfrac{1}{2} \int(\dfrac{-1}{\sqrt{1-x^{2}}}) d x \\ & =\dfrac{x^{2} \cos ^{-1} x}{2}-\dfrac{1}{2} I_1-\dfrac{1}{2} \cos ^{-1} x \qquad …(1)

\end{align*} $$

जहाँ, $I_1=\int \sqrt{1-x^{2}} d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int {\dfrac{d}{d x} (\sqrt{1-x^{2})} \int x d x}dx$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int \dfrac{-2 x}{2 \sqrt{1-x^{2}}} \cdot x d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int \dfrac{-x^{2}}{\sqrt{1-x^{2}}} d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int \dfrac{1-x^{2}-1}{\sqrt{1-x^{2}}} d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-{\int \sqrt{1-x^{2}} d x+\int \dfrac{-d x}{\sqrt{1-x^{2}}}}$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-{I_1+\cos ^{-1} x}$

$\Rightarrow 2 I_1=x \sqrt{1-x^{2}}-\cos ^{-1} x$

$\therefore I_1=\dfrac{x}{2} \sqrt{1-x^{2}}-\dfrac{1}{2} \cos ^{-1} x$

(1) में समाकलन करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =\dfrac{x \cos ^{-1} x}{2}-\dfrac{1}{2}(\dfrac{x}{2} \sqrt{1-x^{2}}-\dfrac{1}{2} \cos ^{-1} x)-\dfrac{1}{2} \cos ^{-1} x \\ & =\dfrac{(2 x^{2}-1)}{4} \cos ^{-1} x-\dfrac{x}{4} \sqrt{1-x^{2}}+C \end{aligned} $

हल

मान लीजिए $I=\int(\sin ^{-1} x)^{2} \cdot 1 d x$

$(\sin ^{-1} x)^{2}$ को पहला फलन और 1 को दूसरा फलन मानकर समाकलन द्वारा अंतरगत करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =(\sin ^{-1} x)^2 \int 1 d x-\int{\dfrac{d}{d x}(\sin ^{-1} x)^{2} \cdot \int 1 \cdot d x} d x \\ & =(\sin ^{-1} x)^{2} \cdot x-\int \dfrac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}} \cdot x d x \\ & =x(\sin ^{-1} x)^{2}+\int \sin ^{-1} x \cdot(\dfrac{-2 x}{\sqrt{1-x^{2}}}) d x \\ & =x(\sin ^{-1} x)^{2}+[\sin ^{-1} x \int \dfrac{-2 x}{\sqrt{1-x^{2}}} d x-\int\ {(\dfrac{d}{d x} \sin ^{-1} x) \int (\dfrac{-2 x}{\sqrt{1-x^{2}}} d x)\} d x] \\ & =x(\sin ^{-1} x)^{2}+[\sin ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \dfrac{1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x] \\ & =x(\sin ^{-1} x)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-\int 2 d x \\ & =x(\sin ^{-1} x)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-2 x+C \end{aligned} $

हल

मान लीजिए $I=\int \dfrac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$

$\Rightarrow I=\dfrac{-1}{2} \int \dfrac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x d x$

$\cos ^{-1} x$ को पहला फलन और $(\dfrac{-2 x}{\sqrt{1-x^{2}}})$ को दूसरा फलन मानकर समाकलन द्वारा अंतरगत बर्बाद करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =\dfrac{-1}{2}[\cos ^{-1} x \int \dfrac{-2 x}{\sqrt{1-x^{2}}} d x-\int{\dfrac{d}{d x} (\cos ^{-1} x) \int \dfrac{-2 x}{\sqrt{1-x^{2}}} d x} d x] \\ & =\dfrac{-1}{2}[\cos ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \dfrac{-1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x] \\ & =\dfrac{-1}{2}[2 \sqrt{1-x^{2}} \cos ^{-1} x+\int 2 d x] \\ & =\dfrac{-1}{2}[2 \sqrt{1-x^{2}} \cos ^{-1} x+2 x]+C \\ & =-[\sqrt{1-x^{2}} \cos ^{-1} x+x]+C \end{aligned} $

हल

मान लीजिए $I=\int x \sec ^{2} x d x$

$x$ को पहला फलन और $\sec ^{2} x$ को दूसरा फलन मानकर समाकलन द्वारा अंतरगत बर्बाद करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =x \int \sec ^{2} x d x-\int{\dfrac{d}{d x} (x) \int (\sec ^{2} x) d x} d x \\ & =x \tan x-\int 1 \cdot \tan x d x \\ & =x \tan x+\log |\cos x|+C \end{aligned} $

हल

मान लीजिए $I=\int 1 \cdot \tan ^{-1} x d x$

$\tan ^{-1} x$ को पहला फलन और 1 को दूसरा फलन मानकर समाकलन द्वारा अंतरगत बर्बाद करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =\tan ^{-1} x \int 1 d x-\int{\dfrac{d}{d x} (\tan ^{-1} x) \int 1 \cdot d x} d x \\ & =\tan ^{-1} x \cdot x-\int \dfrac{1}{1+x^{2}} \cdot x d x \\ & =x \tan ^{-1} x-\dfrac{1}{2} \int \dfrac{2 x}{1+x^{2}} d x \\ & =x \tan ^{-1} x-\dfrac{1}{2} \log |1+x^{2}|+C \\ & =x \tan ^{-1} x-\dfrac{1}{2} \log (1+x^{2})+C \end{aligned} $

हल

$I=\int x(\log x)^{2} d x$

$(\log x)^{2}$ को पहला फलन और 1 को दूसरा फलन मानकर समाकलन द्वारा अंतरगत बर्बाद करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =(\log x)^{2} \int x d x-\int[\ {\dfrac{d}{d x} (\log x)^{2}\} \int x d x] d x \\

& =\dfrac{x^{2}}{2}(\log x)^{2}-\int 2 \log x \cdot \dfrac{1}{x} \cdot \dfrac{x^{2}}{2} d x \\ & =\dfrac{x^{2}}{2}(\log x)^{2}-\int x \log x d x \end{aligned} $

फिर भी एक बार इंटीग्रेट करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =\dfrac{x^{2}}{2}(\log x)^{2}-[\log x \int x d x-\int{\dfrac{d}{d x} (\log x) \int x d x} d x] \\ & =\dfrac{x^{2}}{2}(\log x)^{2}-[\dfrac{x^{2}}{2}-\log x-\int \dfrac{1}{x} \cdot \dfrac{x^{2}}{2} d x] \\ & =\dfrac{x^{2}}{2}(\log x)^{2}-\dfrac{x^{2}}{2} \log x+\dfrac{1}{2} \int x d x \\ & =\dfrac{x^{2}}{2}(\log x)^{2}-\dfrac{x^{2}}{2} \log x+\dfrac{x^{2}}{4}+C \end{aligned} $

हल

मान लीजिए $I=\int(x^{2}+1) \log x d x=\int x^{2} \log x d x+\int \log x d x$

मान लीजिए $I=I_1+I_2 \qquad…(1)$

जहाँ, $I_1=\int x^{2} \log x d x$ और $I_2=\int \log x d x$

$I_1=\int x^{2} \log x d x$

$\log x$ को पहला फ़ंक्शन और $x^{2}$ को दूसरा फ़ंक्शन मानकर इंटीग्रेट बाट करते हुए, हम प्राप्त करते हैं

$$ \begin{align*} I_1 & =\log x.\int x^{2} d x-\int{\dfrac{d}{d x} (\log x) \int x^{2} d x} d x \\ & =\log x \cdot \dfrac{x^{3}}{3}-\int \dfrac{1}{x} \cdot \dfrac{x^{3}}{3} d x \\ & =\dfrac{x^{3}}{3} \log x-\dfrac{1}{3}\int x^{2} d x \\ & =\dfrac{x^{3}}{3} \log x-\dfrac{x^{3}}{9}+C_1 \qquad…(2)\\ I_2 & =\int \log x d x \end{align*} $$

$\log x$ को पहला फ़ंक्शन और 1 को दूसरा फ़ंक्शन मानकर इंटीग्रेट बाट करते हुए, हम प्राप्त करते हैं

$$ \begin{align*} I_2 & =\log x \int 1 \cdot d x-\int{\dfrac{d}{d x} (\log x) \int 1 \cdot d x}dx \\ & =\log x \cdot x-\int \dfrac{1}{x} \cdot x d x \\ & =x \log x-\int 1 d x \\ & =x \log x-x+C_2 \qquad…(3) \end{align*} $$

समीकरण (2) और (3) को (1) में उपयोग करते हुए, हम प्राप्त करते हैं

$ \begin{aligned} I & =\dfrac{x^{3}}{3} \log x-\dfrac{x^{3}}{9}+C_1+x \log x-x+C_2 \\ & =\dfrac{x^{3}}{3} \log x-\dfrac{x^{3}}{9}+x \log x-x+(C_1+C_2) \\ & =(\dfrac{x^{3}}{3}+x) \log x-\dfrac{x^{3}}{9}-x+C \end{aligned} $

हल

मान लीजिए $I=\int e^{x}(\sin x+\cos x) d x$

मान लीजिए $f(x)=\sin x$

$\Rightarrow f^{’}(x)=\cos x$

ज्ञात है कि, $\int e^{x}{f(x)+f^{\prime}(x)} d x=e^{x} f(x)+C$

$\therefore I=e^{x} \sin x+C$

हल

मान लीजिए $I=\int \dfrac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}{\dfrac{x}{(1+x)^{2}}\} d x$

$=\int e^{x}{\dfrac{1+x-1}{(1+x)^{2}}\} d x$

$=\int e^{x}{\dfrac{1}{1+x}-\dfrac{1}{(1+x)^{2}}\} d x$

मान लीजिए $f(x)=\dfrac{1}{1+x} $

$\Rightarrow f^{\prime}(x)=\dfrac{-1}{(1+x)^{2}}$

ज्ञात है कि, $\int e^{x}{f(x)+f^{\prime}(x)} d x=e^{x} f(x)+C$

$\therefore \int \dfrac{x e^{x}}{(1+x)^{2}} d x=\dfrac{e^{x}}{1+x}+C$

हल

$e^{x}(\dfrac{1+\sin x}{1+\cos x})$

$=e^{x}(\dfrac{\sin ^{2} \dfrac{x}{2}+\cos ^{2} \dfrac{x}{2}+2 \sin \dfrac{x}{2} \cos \dfrac{x}{2}}{2 \cos ^{2} \dfrac{x}{2}})$

$=\dfrac{e^{x}(\sin \dfrac{x}{2}+\cos \dfrac{x}{2})^{2}}{2 \cos ^{2} \dfrac{x}{2}}$

$=\dfrac{1}{2} e^{x} \cdot \Bigg(\dfrac{\sin \dfrac{x}{2}+\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}}\Bigg)^{2}$

$=\dfrac{1}{2} e^{x}[\tan \dfrac{x}{2}+1]^{2}$

$=\dfrac{1}{2} e^{2}(1+\tan \dfrac{x}{2})^{2}$

$=\dfrac{1}{2} e^{x}[1+\tan ^{2} \dfrac{x}{2}+2 \tan \dfrac{x}{2}]$

$=\dfrac{1}{2} e^{x}[\sec ^{2} \dfrac{x}{2}+2 \tan \dfrac{x}{2}]$

$\dfrac{e^{x}(1+\sin x) d x}{(1+\cos x)}=e^{x}[\dfrac{1}{2} \sec ^{2} \dfrac{x}{2}+\tan \dfrac{x}{2}]\qquad…(1)$

मान लीजिए $\tan \dfrac{x}{2}=f(x) \Rightarrow f^{\prime}(x)=\dfrac{1}{2} \sec ^{2} \dfrac{x}{2} $

ज्ञात है कि, $\int e^{x}{f(x)+f^{\prime}(x)\} d x=e^{x} f(x)+C$

समीकरण (1) से, हम प्राप्त करते हैं

$\int \dfrac{e^{x}(1+\sin x)}{(1+\cos x)} d x=e^{x} \tan \dfrac{x}{2}+C$

हल

मान लीजिए $I=\int e^{x}[\dfrac{1}{x}-\dfrac{1}{x^{2}}] d x$

इसके अतिरिक्त, मान लीजिए $\dfrac{1}{x}=f(x) \Rightarrow f^{\prime}(x)=\dfrac{-1}{x^{2}}$

It is known that, $\int e^{x}{f(x)+f^{\prime}(x)\} d x=e^{x} f(x)+C$

$\therefore I=\dfrac{e^{x}}{x}+C$

Solution

$ \begin{aligned} & \begin{aligned} \int e^{x}{\dfrac{x-3}{(x-1)^{3}}\} d x & =\int e^{x}{\dfrac{x-1-2}{(x-1)^{3}}\} d x \\ & =\int e^{x}{\dfrac{1}{(x-1)^{2}}-\dfrac{2}{(x-1)^{3}}\} d x \end{aligned} \\ & \text{ Let } f(x)=\dfrac{1}{(x-1)^{2}}\ \Rightarrow f^{\prime}(x)=\dfrac{-2}{(x-1)^{3}} \end{aligned} $

It is known that, $\int e^x\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+C$

$\therefore \int e^{x}{\dfrac{(x-3)}{(x-1)^{2}}\} d x=\dfrac{e^{x}}{(x-1)^{2}}+C$

Solution

Let $I=\int e^{2 x} \sin x d x$

Integrating by parts, we obtain

$I=\sin x \int e^{2 x} d x-\int{\dfrac{d}{d x} (\sin x) \int e^{2 x} d x} d x$

$\Rightarrow I=\sin x \cdot \dfrac{e^{2 x}}{2}-\int \cos x \cdot \dfrac{e^{2 x}}{2} d x$

$\Rightarrow I=\dfrac{e^{2 x} \sin x}{2}-\dfrac{1}{2} \int e^{2 x} \cos x d x$

Again integrating by parts, we obtain

$ \begin{aligned} & I=\dfrac{e^{2 x} \cdot \sin x}{2}-\dfrac{1}{2}[\cos x \int e^{2 x} d x-\int{\dfrac{d}{d x} (\cos x) \int e^{2 x} d x} d x] \\ & \Rightarrow I=\dfrac{e^{2 x} \sin x}{2}-\dfrac{1}{2}[\cos x \cdot \dfrac{e^{2 x}}{2}-\int(-\sin x) \dfrac{e^{2 x}}{2} d x] \\ & \Rightarrow I=\dfrac{e^{2 x} \cdot \sin x}{2}-\dfrac{1}{2}[\dfrac{e^{2 x} \cos x}{2}+\dfrac{1}{2} \int e^{2 x} \sin x d x] \\ & \Rightarrow I=\dfrac{e^{2 x} \sin x}{2}-\dfrac{e^{2 x} \cos x}{4}-\dfrac{1}{4} I \\ & \Rightarrow I+\dfrac{1}{4} I=\dfrac{e^{2 x} \cdot \sin x}{2}-\dfrac{e^{2 x} \cos x}{4} \\ & \Rightarrow \dfrac{5}{4} I=\dfrac{e^{2 x} \sin x}{2}-\dfrac{e^{2 x} \cos x}{4} \\ & \Rightarrow I=\dfrac{4}{5}[\dfrac{e^{2 x} \sin x}{2}-\dfrac{e^{2 x} \cos x}{4}]+C \\ & \Rightarrow I=\dfrac{e^{2 x}}{5}[2 \sin x-\cos x]+C \end{aligned} $

हल

मान लीजिए $x=\tan \theta \quad \Rightarrow d x=\sec ^{2} \theta d \theta$

$\therefore \sin ^{-1}(\dfrac{2 x}{1+x^{2}})=\sin ^{-1}(\dfrac{2 \tan \theta}{1+\tan ^{2} \theta})=\sin ^{-1}(\sin 2 \theta)=2 \theta$

$\int \sin^{-1} (\dfrac{2x}{1+x^2})dx=\int 2 \theta \cdot \sec^2 \theta d \theta=2 \int \theta \cdot \sec^2 \theta d \theta$

भाग विधि से समाकलन करने पर, हम प्राप्त करते हैं

$ \begin{aligned} & 2[\theta \cdot \int \sec ^{2} \theta d \theta-\int {\dfrac{d}{d \theta} (\theta) \int (\sec ^{2} \theta d \theta)} d \theta] \\ & =2[\theta \cdot \tan \theta-\int \tan \theta d \theta] \\ & =2[\theta \tan \theta+\log |\cos \theta|]+C \\ & =2[x \tan ^{-1} x+\log |\dfrac{1}{\sqrt{1+x^{2}}}|]+C \\ & =2 x \tan ^{-1} x+2 \log (1+x^{2})^{-\dfrac{1}{2}}+C \\ & =2 x \tan ^{-1} x+2[-\dfrac{1}{2} \log (1+x^{2})]+C \\ & =2 x \tan ^{-1} x-\log (1+x^{2})+C \end{aligned} $

हल

मान लीजिए $I=\int x^{2} e^{x^{3}} d x$

इसके अतिरिक्त, मान लीजिए $x^{3}=t \Rightarrow 3 x^{2} d x=d t$

$ \begin{aligned} \Rightarrow I & =\dfrac{1}{3} \int e^{t} d t \\ & =\dfrac{1}{3}(e^{t})+C \\ & =\dfrac{1}{3} e^{x^{3}}+C \end{aligned} $

अतः, सही उत्तर A है।

हल

$\int e^{x} \sec x(1+\tan x) d x$

मान लीजिए $I=\int e^{x} \sec x(1+\tan x) d x=\int e^{x}(\sec x+\sec x \tan x) d x$

इसके अतिरिक्त, मान लीजिए $\sec x=f(x) \Rightarrow {\sec x \tan x=f^{\prime}(x)}$

ज्ञात है कि, $\int e^{x}{f(x)+f^{\prime}(x)} d x=e^{x} f(x)+C$

$\therefore I=e^{x} \sec x+C$

अतः, सही उत्तर B है।

हल

मान लीजिए $I=\int \sqrt{4-x^{2}} d x=\int \sqrt{(2)^{2}-(x)^{2}} d x$

ज्ञात है कि, $\int \sqrt{a^{2}-x^{2}} d x=\dfrac{x}{2} \sqrt{a^{2}-x^{2}} +\dfrac{a^{2}}{2} \sin ^{-1} \dfrac{x}{a}+C$

$ \begin{aligned} \therefore I & =\dfrac{x}{2} \sqrt{4-x^{2}}+\dfrac{4}{2} \sin ^{-1} \dfrac{x}{2}+C \\ & =\dfrac{x}{2} \sqrt{4-x^{2}}+2 \sin ^{-1} \dfrac{x}{2}+C \end{aligned} $

हल

मान लीजिए $I=\int \sqrt{1-4 x^{2}} d x=\int \sqrt{(1)^{2}-(2 x)^{2}} d x$

मान लीजिए $2 x=t \Rightarrow 2 d x=d t$

$\therefore I=\dfrac{1}{2} \int \sqrt{(1)^{2}-(t)^{2}} d t$

ज्ञात है कि, $\int \sqrt{a^{2}-x^{2}} d x=\dfrac{x}{2} \sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2} \sin ^{-1} \dfrac{x}{a}+C$

$ \begin{aligned} \Rightarrow I & =\dfrac{1}{2}[\dfrac{t}{2} \sqrt{1-t^{2}}+\dfrac{1}{2} \sin ^{-1} t]+C \\ & =\dfrac{t}{4} \sqrt{1-t^{2}}+\dfrac{1}{4} \sin ^{-1} t+C \\ & =\dfrac{2 x}{4} \sqrt{1-4 x^{2}}+\dfrac{1}{4} \sin ^{-1} 2 x+C \\ & =\dfrac{x}{2} \sqrt{1-4 x^{2}}+\dfrac{1}{4} \sin ^{-1} 2 x+C \end{aligned} $

हल

मान लीजिए $I=\int \sqrt{x^{2}+4 x+6} d x$

$ \begin{aligned} & =\int \sqrt{x^{2}+4 x+4+2} d x \\ & =\int \sqrt{(x^{2}+4 x+4)+2} d x \\ & =\int \sqrt{(x+2)^{2}+(\sqrt{2})^{2}} d x \end{aligned} $

ज्ञात है कि, $\int \sqrt{x^{2}+a^{2}} d x=\dfrac{x}{2} \sqrt{x^{2}+a^{2}}+\dfrac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$

$ \begin{aligned} \therefore I & =\dfrac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\dfrac{2}{2} \log |(x+2)+\sqrt{x^{2}+4 x+6}|+C \\ & =\dfrac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\log |(x+2)+\sqrt{x^{2}+4 x+6}|+C \end{aligned} $

हल

मान लीजिए $I=\int \sqrt{x^{2}+4 x+1} d x$

$ \begin{aligned} & =\int \sqrt{(x^{2}+4 x+4)-3} d x \\ & =\int \sqrt{(x+2)^{2}-(\sqrt{3})^{2}} d x \end{aligned} $

ज्ञात है कि, $\int \sqrt{x^{2}-a^{2}} d x=\dfrac{x}{2} \sqrt{x^{2}-a^{2}}-\dfrac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$\therefore I=\dfrac{(x+2)}{2} \sqrt{x^{2}+4 x+1}-\dfrac{3}{2} \log |(x+2)+\sqrt{x^{2}+4 x+1}|+C$

Solution

मान लीजिए $I=\int \sqrt{1-4 x-x^{2}} d x$

$ \begin{aligned} & =\int \sqrt{1-(x^{2}+4 x+4-4)} d x \\ & =\int \sqrt{1+4-(x+2)^{2}} d x \\ & =\int \sqrt{(\sqrt{5})^{2}-(x+2)^{2}} d x \end{aligned} $

ज्ञात है कि, $\int \sqrt{a^{2}-x^{2}} d x=\dfrac{x}{2} \sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2} \sin ^{-1} \dfrac{x}{a}+C$

$\therefore I=\dfrac{(x+2)}{2} \sqrt{1-4 x-x^{2}}+\dfrac{5}{2} \sin ^{-1}(\dfrac{x+2}{\sqrt{5}})+C$

Solution

मान लीजिए $I=\int \sqrt{x^{2}+4 x-5} d x$

$ \begin{aligned} & =\int \sqrt{(x^{2}+4 x+4)-9} d x \\ & =\int \sqrt{(x+2)^{2}-(3)^{2}} d x \end{aligned} $

ज्ञात है कि, $\int \sqrt{x^{2}-a^{2}} d x=\dfrac{x}{2} \sqrt{x^{2}-a^{2}}-\dfrac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$\therefore I=\dfrac{(x+2)}{2} \sqrt{x^{2}+4 x-5}-\dfrac{9}{2} \log |(x+2)+\sqrt{x^{2}+4 x-5}|+C$

Solution

मान लीजिए $I=\int \sqrt{1+3 x-x^{2}} d x$

$ \begin{aligned} & =\int \sqrt{1-(x^{2}-3 x+\dfrac{9}{4}-\dfrac{9}{4})} d x \\ & =\int \sqrt{(1+\dfrac{9}{4})-(x-\dfrac{3}{2})^{2}} d x \\ & =\int \sqrt{(\dfrac{\sqrt{13}}{2})^{2}-(x-\dfrac{3}{2})^{2}} d x \end{aligned} $

ज्ञात है कि, $\int \sqrt{a^{2}-x^{2}} d x=\dfrac{x}{2} \sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2} \sin ^{-1} \dfrac{x}{a}+C$

$ \begin{aligned} \therefore I & =\dfrac{x-\dfrac{3}{2}}{2} \sqrt{1+3 x-x^{2}}+\dfrac{13}{4 \times 2} \sin ^{-1}(\dfrac{x-\dfrac{3}{2}}{\dfrac{\sqrt{13}}{2}})+C \\ & =\dfrac{2 x-3}{4} \sqrt{1+3 x-x^{2}}+\dfrac{13}{8} \sin ^{-1}(\dfrac{2 x-3}{\sqrt{13}})+C

\end{aligned} $

हल

मान लीजिए $I=\int \sqrt{x^{2}+3 x} d x$

$ \begin{aligned} & =\int \sqrt{x^{2}+3 x+\dfrac{9}{4}-\dfrac{9}{4}} d x \\ & =\int \sqrt{(x+\dfrac{3}{2})^{2}-(\dfrac{3}{2})^{2}} d x \end{aligned} $

ज्ञात है कि, $\int \sqrt{x^{2}-a^{2}} d x=\dfrac{x}{2} \sqrt{x^{2}-a^{2}}-\dfrac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$ \begin{aligned} \therefore I & =\dfrac{(x+\dfrac{3}{2})}{2} \sqrt{x^{2}+3 x}-\dfrac{9}{2} \log |(x+\dfrac{3}{2})+\sqrt{x^{2}+3 x}|+C \\ & =\dfrac{(2 x+3)}{4} \sqrt{x^{2}+3 x}-\dfrac{9}{8} \log |(x+\dfrac{3}{2})+\sqrt{x^{2}+3 x}|+C \end{aligned} $

हल

मान लीजिए $I=\int \sqrt{1+\dfrac{x^{2}}{9}} d x=\dfrac{1}{3} \int \sqrt{9+x^{2}} d x=\dfrac{1}{3} \int \sqrt{(3)^{2}+x^{2}} d x$

ज्ञात है कि, $\int \sqrt{x^{2}+a^{2}} d x=\dfrac{x}{2} \sqrt{x^{2}+a^{2}}+\dfrac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$

$ \begin{aligned} \therefore I & =\dfrac{1}{3}[\dfrac{x}{2} \sqrt{x^{2}+9}+\dfrac{9}{2} \log |x+\sqrt{x^{2}+9}|]+C \\ & =\dfrac{x}{6} \sqrt{x^{2}+9}+\dfrac{3}{2} \log |x+\sqrt{x^{2}+9}|+C \end{aligned} $

हल

ज्ञात है कि, $\int \sqrt{a^{2}+x^{2}} d x=\dfrac{x}{2} \sqrt{a^{2}+x^{2}}+\dfrac{a^{2}}{2} \log |x+\sqrt{x^{2}+a^{2}}|+C$

$ \therefore \int \sqrt{1+x^{2}} d x=\dfrac{x}{2} \sqrt{1+x^{2}}+\dfrac{1}{2} \log |x+\sqrt{1+x^{2}}|+C $

अतः, सही उत्तर A है।

हल

मान लीजिए $I=\int \sqrt{x^{2}-8 x+7} d x$

$ \begin{aligned} & =\int \sqrt{(x^{2}-8 x+16)-9} d x \\ & =\int \sqrt{(x-4)^{2}-(3)^{2}} d x \end{aligned} $

यह ज्ञात है कि, $\int \sqrt{x^{2}-a^{2}} d x=\dfrac{x}{2} \sqrt{x^{2}-a^{2}}-\dfrac{a^{2}}{2} \log |x+\sqrt{x^{2}-a^{2}}|+C$

$ \therefore I=\dfrac{(x-4)}{2} \sqrt{x^{2}-8 x+7}-\dfrac{9}{2} \log |(x-4)+\sqrt{x^{2}-8 x+7}|+C $

अतः, सही उत्तर D है।

हल

मान लीजिए $I=\int _{-1}^{1}(x+1) d x$

$\int(x+1) d x=\dfrac{x^{2}}{2}+x=F(x)$

समाकलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(1)-F(-1) \\ & =(\dfrac{1}{2}+1)-(\dfrac{1}{2}-1) \\ & =\dfrac{1}{2}+1-\dfrac{1}{2}+1 \\ & =2 \end{aligned} $

हल

मान लीजिए $I=\int_2^{3} \dfrac{1}{x} d x$

$\dfrac{1}{x} d x=\log |x|=F(x)$

समाकलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(3)-F(2) \\ & =\log |3|-\log |2|=\log \dfrac{3}{2} \end{aligned} $

हल

मान लीजिए $I=\int_1^{2}(4 x^{3}-5 x^{2}+6 x+9) d x$

$\int(4 x^{3}-5 x^{2}+6 x+9) d x=4(\dfrac{x^{4}}{4})-5(\dfrac{x^{3}}{3})+6(\dfrac{x^{2}}{2})+9(x)$

$=x^{4}-\dfrac{5 x^{3}}{3}+3 x^{2}+9 x=F(x)$

समाकलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(2)-F(1) \\ I & ={2^{4}-\dfrac{5 \cdot(2)^{3}}{3}+3(2)^{2}+9(2)}-{(1)^{4}-\dfrac{5(1)^{3}}{3}+3(1)^{2}+9(1)} \\ & =(16-\dfrac{40}{3}+12+18)-(1-\dfrac{5}{3}+3+9) \\ & =16-\dfrac{40}{3}+12+18-1+\dfrac{5}{3}-3-9 \\ & =33-\dfrac{35}{3} \\ & =\dfrac{99-35}{3} \\ & =\dfrac{64}{3} \end{aligned} $

हल

मान लीजिए $I=\int_0^{\pi} \sin 2 x d x$

$\int \sin 2 x d x=(\dfrac{-\cos 2 x}{2})=F(x)$

समाकलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ I =F(\dfrac{\pi}{4})-F(0) $

$ =-\dfrac{1}{2}[\cos 2(\dfrac{\pi}{4})-\cos 0] $

$ =-\dfrac{1}{2}[\cos (\mathrm{\pi} / 2)-1] $

$ =-\dfrac{1}{2}[0-1] $

$ =\dfrac{1}{2} $

हल

मान लीजिए $I=\int_0^{\dfrac{\pi}{2}} \cos 2 x d x$

$\int \cos 2 x d x=(\dfrac{\sin 2 x}{2})=F(x)$

गणितीय कलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(\dfrac{\pi}{2})-F(0) \\ & =\dfrac{1}{2}[\sin 2(\dfrac{\pi}{2})-\sin 0] \\ & =\dfrac{1}{2}[\sin \pi-\sin 0] \\ & =\dfrac{1}{2}[0-0]=0 \end{aligned} $

हल

मान लीजिए $I=\int_4^{5} e^{x} d x$

$\int e^{x} d x=e^{x}=F(x)$

गणितीय कलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(5)-F(4) \\ & =e^{5}-e^{4} \\ & =e^{4}(e-1) \end{aligned} $

हल

मान लीजिए $I=\int_0^{\dfrac{\pi}{4}} \tan x d x$

$\int \tan x d x=-\log |\cos x|=F(x)$

गणितीय कलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(\dfrac{\pi}{4})-F(0) \\ & =-\log |\cos \dfrac{\pi}{4}|+\log |\cos 0| \\ & =-\log |\dfrac{1}{\sqrt{2}}|+\log |1| \\ & =-\log (2)^{-\dfrac{1}{2}} \\ & =\dfrac{1}{2} \log 2 \end{aligned} $

हल

मान लीजिए $I=\int _{\dfrac{\pi}{6}}^{\dfrac{\pi}{4}} cosec x d x$

$\int cosec x d x=\log |cosec x-\cot x|=F(x)$

गणितीय कलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(\dfrac{\pi}{4})-F(\dfrac{\pi}{6}) \\ & =\log |cosec \dfrac{\pi}{4}-\cot \dfrac{\pi}{4}|-\log |cosec \dfrac{\pi}{6}-\cot \dfrac{\pi}{6}| \\ & =\log |\sqrt{2}-1|-\log |2-\sqrt{3}| \\ & =\log (\dfrac{\sqrt{2}-1}{2-\sqrt{3}}) \end{aligned} $

हल

मान लीजिए $I=\int_0^{1} \dfrac{d x}{\sqrt{1-x^{2}}}$

$\int \dfrac{d x}{\sqrt{1-x^{2}}}=\sin ^{-1} x=F(x)$

गणितीय कलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(1)-F(0) \\ & =\sin ^{-1}(1)-\sin ^{-1}(0) \\ & =\dfrac{\pi}{2}-0 \\

& =\dfrac{\pi}{2} \end{aligned} $

Solution

Let $I=\int_0^{1} \dfrac{d x}{1+x^{2}}$

$\int \dfrac{d x}{1+x^{2}}=\tan ^{-1} x=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & =\tan ^{-1}(1)-\tan ^{-1}(0) \\ & =\dfrac{\pi}{4} \end{aligned} $

Solution

Let $I=\int_2^{3} \dfrac{d x}{x^{2}-1}$

$\int \dfrac{d x}{x^{2}-1}=\dfrac{1}{2} \log |\dfrac{x-1}{x+1}|=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(3)-F(2) \\ & =\dfrac{1}{2}[\log |\dfrac{3-1}{3+1}|-\log |\dfrac{2-1}{2+1}|] \\ & =\dfrac{1}{2}[\log |\dfrac{2}{4}|-\log |\dfrac{1}{3}|] \\ & =\dfrac{1}{2}[\log \dfrac{1}{2}-\log \dfrac{1}{3}] \\ & =\dfrac{1}{2}[\log \dfrac{3}{2}] \end{aligned} $

Solution

Let $I=\int_0^{\dfrac{\pi}{2}} \cos ^{2} x d x$

$\int \cos ^{2} x d x=\int(\dfrac{1+\cos 2 x}{2}) d x=\dfrac{x}{2}+\dfrac{\sin 2 x}{4}=\dfrac{1}{2}(x+\dfrac{\sin 2 x}{2})=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =[F(\dfrac{\pi}{2})-F(0)] \\ & =\dfrac{1}{2}[(\dfrac{\pi}{2}-\dfrac{\sin \pi}{2})-(0+\dfrac{\sin 0}{2})] \\ & =\dfrac{1}{2}[\dfrac{\pi}{2}+0-0-0] \\ & =\dfrac{\pi}{4} \end{aligned} $

Solution

Let $I=\int_2^{3} \dfrac{x}{x^{2}+1} d x$

$\int \dfrac{x}{x^{2}+1} d x=\dfrac{1}{2} \int \dfrac{2 x}{x^{2}+1} d x=\dfrac{1}{2} \log (1+x^{2})=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(3)-F(2) \\ & =\dfrac{1}{2}[\log (1+(3)^{2})-\log (1+(2)^{2})] \\ & =\dfrac{1}{2}[\log (10)-\log (5)] \\ & =\dfrac{1}{2} \log (\dfrac{10}{5})=\dfrac{1}{2} \log 2

\end{aligned} $

Solution

Let $I=\int_0^{1} \dfrac{2 x+3}{5 x^{2}+1} d x$

$\int \dfrac{2 x+3}{5 x^{2}+1} d x=\dfrac{1}{5} \int \dfrac{5(2 x+3)}{5 x^{2}+1} d x$

$=\dfrac{1}{5} \int \dfrac{10 x+15}{5 x^{2}+1} d x$

$=\dfrac{1}{5} \int \dfrac{10 x}{5 x^{2}+1} d x+3 \int \dfrac{1}{5 x^{2}+1} d x$

$=\dfrac{1}{5} \int \dfrac{10 x}{5 x^{2}+1} d x+3 \int \dfrac{1}{5(x^{2}+\dfrac{1}{5})} d x$

$=\dfrac{1}{5} \log (5 x^{2}+1)+\dfrac{3}{5} \cdot \dfrac{1}{\dfrac{1}{\sqrt{5}}} \tan ^{-1} \dfrac{x}{\dfrac{1}{\sqrt{5}}}$

$=\dfrac{1}{5} \log (5 x^{2}+1)+\dfrac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5} x)$

$=F(x)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & ={\dfrac{1}{5} \log (5+1)+\dfrac{3}{\sqrt{5}} \tan ^{-1}(\sqrt{5})}-{\dfrac{1}{5} \log (1)+\dfrac{3}{\sqrt{5}} \tan ^{-1}(0)} \\ & =\dfrac{1}{5} \log 6+\dfrac{3}{\sqrt{5}} \tan ^{-1} \sqrt{5} \end{aligned} $

Solution

Let $I=\int_0^{1} x e^{x^{2}} d x$

Put $x^{2}=t \Rightarrow 2 x d x=d t$

As $x \to 0, t \to 0$ and as $x \to 1, t \to 1$,

$\therefore I=\dfrac{1}{2} \int_0^{1} e^{t} d t$

$\dfrac{1}{2} \int e^{t} d t=\dfrac{1}{2} e^{t}=F(t)$

By second fundamental theorem of calculus, we obtain

$ \begin{aligned} I & =F(1)-F(0) \\ & =\dfrac{1}{2} e-\dfrac{1}{2} e^{0} \\ & =\dfrac{1}{2}(e-1) \end{aligned} $

Solution

Let $I=\int_1^{2} \dfrac{5 x^{2}}{x^{2}+4 x+3} d x$

Dividing $5 x^{2}$ by $x^{2}+4 x+3$, we obtain

$I=\int_1^{2}{5-\dfrac{20 x+15}{x^{2}+4 x+3}} d x$

$=\int_1^{2} 5 d x-\int_1^{2} \dfrac{20 x+15}{x^{2}+4 x+3} d x$

$=[5 x]_1^{2}-\int_1^{2} \dfrac{20 x+15}{x^{2}+4 x+3} d x$

$I=5-I_1$, where $I_1=\int_1^{2} \dfrac{20 x+15}{x^{2}+4 x+3} d x \qquad …(1)$

Let $20 x+15=A \dfrac{d}{d x}(x^{2}+4 x+3)+B$

$ =2 A x+(4 A+B) $

$ x $ के गुणांक और स्थिरांक के गुणांक के तुलना करने पर, हम प्राप्त करते हैं

$ A=10 $ और $ B=-25 $

$\Rightarrow I_1=10 \int_1^{2} \dfrac{2 x+4}{x^{2}+4 x+3} d x-25 \int_1^{2} \dfrac{d x}{x^{2}+4 x+3}$

मान लीजिए $ x^{2}+4 x+3=t $

$\Rightarrow(2 x+4) d x=d t$

$\Rightarrow I_1=10 \int \dfrac{d t}{t}-25 \int \dfrac{d x}{(x+2)^{2}-1^{2}}$

$=10 \log t-25[\dfrac{1}{2} \log (\dfrac{x+2-1}{x+2+1})]$

$=[10 \log (x^{2}+4 x+3)]_1^{2}-25[\dfrac{1}{2} \log (\dfrac{x+1}{x+3})]_1^{2}$

$=[10 \log 15-10 \log 8]-25[\dfrac{1}{2} \log \dfrac{3}{5}-\dfrac{1}{2} \log \dfrac{2}{4}]$

$=[10 \log (5 \times 3)-10 \log (4 \times 2)]-\dfrac{25}{2}[\log 3-\log 5-\log 2+\log 4]$

$=[10 \log 5+10 \log 3-10 \log 4-10 \log 2]-\dfrac{25}{2}[\log 3-\log 5-\log 2+\log 4]$

$=[10+\dfrac{25}{2}] \log 5+[-10-\dfrac{25}{2}] \log 4+[10-\dfrac{25}{2}] \log 3+[-10+\dfrac{25}{2}] \log 2$

$=\dfrac{45}{2} \log 5-\dfrac{45}{2} \log 4-\dfrac{5}{2} \log 3+\dfrac{5}{2} \log 2$

$=\dfrac{45}{2} \log \dfrac{5}{4}-\dfrac{5}{2} \log \dfrac{3}{2}$

(1) में $ I_1 $ के मान को रखने पर, हम प्राप्त करते हैं

$ \begin{aligned} I & =5-[\dfrac{45}{2} \log \dfrac{5}{4}-\dfrac{5}{2} \log \dfrac{3}{2}] \\ & =5-\dfrac{5}{2}[9 \log \dfrac{5}{4}-\log \dfrac{3}{2}] \end{aligned} $

Solution

मान लीजिए $ I=\int_0^{\dfrac{\pi}{4}}(2 \sec ^{2} x+x^{3}+2) d x $

$\int(2 \sec ^{2} x+x^{3}+2) d x=2 \tan x+\dfrac{x^{4}}{4}+2 x=F(x)$

गणित के द्वितीय मूल उपप्रमेय के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(\dfrac{\pi}{4})-F(0) \\ & ={(2 \tan \dfrac{\pi}{4}+\dfrac{1}{4}(\dfrac{\pi}{4})^{4}+2(\dfrac{\pi}{4}))-(2 \tan 0+0+0)} \\ & =2 \tan \dfrac{\pi}{4}+\dfrac{\pi^{4}}{4^{5}}+\dfrac{\pi}{2} \\ & =2+\dfrac{\pi}{2}+\dfrac{\pi^{4}}{1024} \end{aligned} $

Solution

मान लीजिए $ I=\int_0^{\pi}(\sin ^{2} \dfrac{x}{2}-\cos ^{2} \dfrac{x}{2}) d x $

$ \begin{aligned} & =-\int_0^{\pi}(\cos ^{2} \dfrac{x}{2}-\sin ^{2} \dfrac{x}{2}) d x \\ & =-\int_0^{\pi} \cos x d x \end{aligned} $

$\int \cos x d x=\sin x=F(x)$

कलन के द्वितीय मूल तत्व प्रमेय के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(\pi)-F(0) \\ & =\sin \pi-\sin 0 \\ & =0 \end{aligned} $

हल

मान लीजिए $I=\int_0^{2} \dfrac{6 x+3}{x^{2}+4} d x$

$ \begin{aligned} \int \dfrac{6 x+3}{x^{2}+4} d x & =3 \int \dfrac{2 x+1}{x^{2}+4} d x \\ & =3 \int \dfrac{2 x}{x^{2}+4} d x+3 \int \dfrac{1}{x^{2}+4} d x \\ & =3 \log (x^{2}+4)+\dfrac{3}{2} \tan ^{-1} \dfrac{x}{2}=F(x) \end{aligned} $

कलन के द्वितीय मूल तत्व प्रमेय के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(2)-F(0) \\ & ={3 \log (2^{2}+4)+\dfrac{3}{2} \tan ^{-1}(\dfrac{2}{2})}-{3 \log (0+4)+\dfrac{3}{2} \tan ^{-1}(\dfrac{0}{2})} \\ & =3 \log 8+\dfrac{3}{2} \tan ^{-1} 1-3 \log 4-\dfrac{3}{2} \tan ^{-1} 0 \\ & =3 \log 8+\dfrac{3}{2}(\dfrac{\pi}{4})-3 \log 4-0 \\ & =3 \log (\dfrac{8}{4})+\dfrac{3 \pi}{8} \\ & =3 \log 2+\dfrac{3 \pi}{8} \end{aligned} $

हल

मान लीजिए $I=\int_0^{1}(x e^{x}+\sin \dfrac{\pi x}{4}) d x$

$ \begin{aligned} \int(x e^{x}+\sin \dfrac{\pi x}{4}) d x & =x \int e^{x} d x-\int{\dfrac{d}{d x}( x) \int e^{x} d x} d x+{\dfrac{-\cos \dfrac{\pi x}{4}}{\dfrac{\pi}{4}}} \\ & =x e^{x}-\int e^{x} d x-\dfrac{4 cos(\dfrac{\pi x}{4})}{\pi} \\ & =x e^{x}-e^{x}-\dfrac{4 cos(\dfrac{\pi x}{4})}{\pi} \\ & =F(x) \end{aligned} $

कलन के द्वितीय मूल तत्व प्रमेय के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} I & =F(1)-F(0) \\ & =(1 . e^{1}-e^{1}-\dfrac{4}{\pi} \cos \dfrac{\pi}{4})-(0 . e^{0}-e^{0}-\dfrac{4}{\pi} \cos 0) \\ & =e-e-\dfrac{4}{\pi}(\dfrac{1}{\sqrt{2}})+1+\dfrac{4}{\pi} \\ & =1+\dfrac{4}{\pi}-\dfrac{2 \sqrt{2}}{\pi} \end{aligned} $

हल

$ \int \dfrac{d x}{1+x^{2}}=\tan ^{-1} x=F(x) $

कलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} & \int_1^{\sqrt{3}} \dfrac{d x}{1+x^{2}}=F(\sqrt{3})-F(1) \\ & =\tan ^{-1} \sqrt{3}-\tan ^{-1} 1 \\ & =\dfrac{\pi}{3}-\dfrac{\pi}{4} \\ & =\dfrac{\pi}{12} \end{aligned} $

अतः, सही उत्तर D है।

हल

$ \int \dfrac{d x}{4+9 x^{2}}=\int \dfrac{d x}{(2)^{2}+(3 x)^{2}} $

मान लीजिए $3 x=t \Rightarrow 3 d x=d t$

$\therefore \int \dfrac{d x}{(2)^{2}+(3 x)^{2}}=\dfrac{1}{3} \int \dfrac{d t}{(2)^{2}+t^{2}}$

$ \begin{aligned} & =\dfrac{1}{3}[\dfrac{1}{2} \tan ^{-1} \dfrac{t}{2}] \\ & =\dfrac{1}{6} \tan ^{-1}(\dfrac{3 x}{2}) \\ & =F(x) \end{aligned} $

कलन के द्वितीय मूल थेयोरम के अनुसार, हम प्राप्त करते हैं

$ \begin{aligned} \int_0^{\dfrac{2}{3}} \dfrac{d x}{4+9 x^{2}} & =F(\dfrac{2}{3})-F(0) \\ & =\dfrac{1}{6} \tan ^{-1}(\dfrac{3}{2} \cdot \dfrac{2}{3})-\dfrac{1}{6} \tan ^{-1} 0 \\ & =\dfrac{1}{6} \tan ^{-1} 1-0 \\ & =\dfrac{1}{6} \times \dfrac{\pi}{4} \\ & =\dfrac{\pi}{24} \end{aligned} $

अतः, सही उत्तर C है।

हल

$\int_0^{1} \dfrac{x}{x^{2}+1} d x$

मान लीजिए $x^{2}+1=t \Rightarrow 2 x d x=d t$

जब $x=0, t=1$ और जब $x=1, t=2$

$\therefore \int_0^{1} \dfrac{x}{x^{2}+1} d x=\dfrac{1}{2} \int_1^{2} \dfrac{d t}{t}$

$=\dfrac{1}{2}[\log |t|]_1^{2}$

$=\dfrac{1}{2}[\log 2-\log 1]$

$=\dfrac{1}{2} \log 2$

हल

मान लीजिए $I=\int_0^{\dfrac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi=\int_0^{\dfrac{\pi}{2}} \sqrt{\sin \phi} \cos ^{4} \phi \cos \phi d \phi$

इसके अतिरिक्त, मान लीजिए $\sin \phi=t \Rightarrow \cos \phi d \phi=d t$

जब $\phi=0, t=0$ और जब $\phi=\dfrac{\pi}{2}, t=1$

$\therefore I=\displaystyle\int_0^{1} \sqrt{t}(1-t^{2})^{2} d t$

$=\displaystyle\int_0^{1} t^{\dfrac{1}{2}}(1+t^{4}-2 t^{2}) d t$

$=\displaystyle\int_0^{1}[t^{\dfrac{1}{2}}+t^{\dfrac{9}{2}}-2 t^{\dfrac{5}{2}}] d t$

$=\Bigg[\dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}}+\dfrac{t^{\dfrac{11}{2}}}{\dfrac{11}{2}}-\dfrac{2 t^{\dfrac{7}{2}}}{\dfrac{7}{2}}\Bigg]_0^{1}$

$=\dfrac{2}{3}+\dfrac{2}{11}-\dfrac{4}{7}$

$=\dfrac{154+42-132}{231}$

$=\dfrac{64}{231}$

हल

मान लीजिए $I=\int_0^{1} \sin ^{-1}(\dfrac{2 x}{1+x^{2}}) d x$

इसके अतिरिक्त, मान लीजिए $x=\tan \theta \Rightarrow d x=\sec ^{2} \theta d \theta$

जब $x=0, \theta=0$ और जब $x=1, \theta=\dfrac{\pi}{4}$

$ \begin{aligned} I & =\int_0^{\pi/4} \sin ^{-1}(\dfrac{2 \tan \theta}{1+\tan ^{2} \theta}) \sec ^{2} \theta d \theta \\ & =\int_0^{\dfrac{\pi}{4}} \sin ^{-1}(\sin 2 \theta) \sec ^{2} \theta d \theta \\ & =\int_0^{\dfrac{\pi}{4}} 2 \theta \cdot \sec ^{2} \theta d \theta \\ & =2 \int_0^{\pi/4} \theta \cdot \sec ^{2} \theta d \theta

\end{aligned} $

$\theta$ को पहला फलन लेकर और $\sec ^{2} \theta$ को दूसरा फलन लेकर समाकलन द्वारा बाहर लेने पर, हम प्राप्त करते हैं

$ \begin{aligned} I & =2\Bigg[\theta \int \sec ^{2} \theta d \theta-\int{(\dfrac{d}{d x} \theta) \int \sec ^{2} \theta d \theta} d \theta\Bigg]_0^{\dfrac{\pi}{4}} \\ & =2\Bigg[\theta \tan \theta-\int \tan \theta d \theta\Bigg]_0^{\dfrac{\pi}{4}} \\ & =2\Bigg[\theta \tan \theta+\log |\cos \theta|\Bigg]_0^{\dfrac{\pi}{4}} \\ & =2[\dfrac{\pi}{4} \tan \dfrac{\pi}{4}+\log |\cos \dfrac{\pi}{4}|-\log |\cos 0|] \\ & =2[\dfrac{\pi}{4}+\log (\dfrac{1}{\sqrt{2}})-\log 1] \\ & =2[\dfrac{\pi}{4}-\dfrac{1}{2} \log 2] \\ & =\dfrac{\pi}{2}-\log 2 \end{aligned} $

Solution

$\int_0^{2} x \sqrt{x+2} d x$

Let $x+2=t^{2} \Rightarrow d x=2 t d t$

When $x=0, t=\sqrt{2}$ and when $x=2, t=2$

$\therefore \displaystyle\int_0^{2} x \sqrt{x+2} d x=\int _{\sqrt{2}}^{2}(t^{2}-2) \sqrt{t^{2}} 2 t\ d t$

$=2 \displaystyle\int _{\sqrt{2}}^{2}(t^{2}-2)t^{2} d t$

$=2 \displaystyle\int _{\sqrt{2}}^{2}(t^{4}-2 t^{2}) d t$

$=2\Bigg[\dfrac{t^{5}}{5}-\dfrac{2 t^{3}}{3}\Bigg] _{\sqrt{2}}^{2}$

$=2\Bigg[\dfrac{32}{5}-\dfrac{16}{3}-\dfrac{4 \sqrt{2}}{5}+\dfrac{4 \sqrt{2}}{3}\Bigg]$

$=2\Bigg[\dfrac{96-80-12 \sqrt{2}+20 \sqrt{2}}{15}\Bigg]$

$=2\Bigg[\dfrac{16+8 \sqrt{2}}{15}\Bigg]$

$=\dfrac{16(2+\sqrt{2})}{15}$

$=\dfrac{16 \sqrt{2}(\sqrt{2}+1)}{15}$

Solution

$\displaystyle\int_0^{\dfrac{\pi}{2}} \dfrac{\sin x}{1+\cos ^{2} x} d x$

Let $\cos x=t \Rightarrow -\sin x d x=d t$

When $x=0, t=1$ and when $x=\dfrac{\pi}{2}, t=0$

$\Rightarrow \displaystyle\int_0^{\dfrac{\pi}{2}} \dfrac{\sin x}{1+\cos ^{2} x} d x=-\int_0^{1} \dfrac{d t}{1+t^{2}}$

$=-[\tan ^{-1} t]_1^{0}$

$=-[\tan ^{-1} 0-\tan ^{-1} 1]$

$=-[-\dfrac{\pi}{4}]$

$=\dfrac{\pi}{4}$

Solution

$\displaystyle\int_0^{2} \dfrac{d x}{x+4-x^{2}}=\int_0^{2} \dfrac{d x}{-(x^{2}-x-4)}$

$=\displaystyle\int_0^{2} \dfrac{d x}{-(x^{2}-x+\dfrac{1}{4}-\dfrac{1}{4}-4)}$

$=\displaystyle\int_0^{2} \dfrac{d x}{-[(x-\dfrac{1}{2})^{2}-\dfrac{17}{4}]}$

$=\displaystyle\int_0^{2} \dfrac{d x}{(\dfrac{\sqrt{17}}{2})^{2}-(x-\dfrac{1}{2})^{2}}$

Let $\quad x-\dfrac{1}{2}=t \Rightarrow d x=d t$

When $x=0, t=-\dfrac{1}{2}$ and when $x=2, t=\dfrac{3}{2}$

$\therefore \displaystyle\int_0^{2} \dfrac{d x}{(\dfrac{\sqrt{17}}{2})^{2}-(x-\dfrac{1}{2})^{2}}=\displaystyle \int_{-\dfrac{1}{2}}^{\dfrac{3}{2}} \dfrac{d t}{(\dfrac{\sqrt{17}}{2})^{2}-t^{2}}$

$=\Bigg[\dfrac{1}{2(\dfrac{\sqrt{17}}{2})} \log \dfrac{\dfrac{\sqrt{17}}{2}+t}{\dfrac{\sqrt{17}}{2}-t}\Bigg] _{-\dfrac{1}{2}}^{\dfrac{3}{2}}$

$=\dfrac{1}{\sqrt{17}}\Bigg[\log \dfrac{\dfrac{\sqrt{17}}{2}+\dfrac{3}{2}}{\dfrac{\sqrt{17}}{2}-\dfrac{3}{2}}-\dfrac{\log \dfrac{\sqrt{17}}{2}-\dfrac{1}{2}}{\log \dfrac{\sqrt{17}}{2}+\dfrac{1}{2}}\Bigg]$

$=\dfrac{1}{\sqrt{17}}\Bigg[\log \dfrac{\sqrt{17}+3}{\sqrt{17}-3}-\log \dfrac{\sqrt{17}-1}{\sqrt{17}+1}\Bigg]$

$=\dfrac{1}{\sqrt{17}} \log \dfrac{\sqrt{17}+3}{\sqrt{17}-3} \times \dfrac{\sqrt{17}+1}{\sqrt{17}-1}$

$=\dfrac{1}{\sqrt{17}} \log [\dfrac{17+3+4 \sqrt{17}}{17+3-4 \sqrt{17}}]$

$=\dfrac{1}{\sqrt{17}} \log [\dfrac{20+4 \sqrt{17}}{20-4 \sqrt{17}}]$

$=\dfrac{1}{\sqrt{17}} \log (\dfrac{5+\sqrt{17}}{5-\sqrt{17}})$

$=\dfrac{1}{\sqrt{17}} \log [\dfrac{(5+\sqrt{17})(5+\sqrt{17})}{25-17}]$

$=\dfrac{1}{\sqrt{17}} \log [\dfrac{25+17+10 \sqrt{17}}{8}]$

$=\dfrac{1}{\sqrt{17}} \log (\dfrac{42+10 \sqrt{17}}{8})$

$=\dfrac{1}{\sqrt{17}} \log (\dfrac{21+5 \sqrt{17}}{4})$

Solution

$\displaystyle\int _{-1}^{1} \dfrac{d x}{x^{2}+2 x+5}=\int _{-1}^{1} \dfrac{d x}{(x^{2}+2 x+1)+4}=\int _{-1}^{1} \dfrac{d x}{(x+1)^{2}+(2)^{2}}$

Let $x+1=t \Rightarrow d x=d t$

जब $x=-1, t=0$ और जब $x=1, t=2$

$\therefore \displaystyle\int _{-1}^{1} \dfrac{d x}{(x+1)^{2}+(2)^{2}}=\int_0^{2} \dfrac{d t}{t^{2}+2^{2}}$

$=\Bigg[\dfrac{1}{2} \tan ^{-1} \dfrac{t}{2}\Bigg]_0^{2}$

$=\dfrac{1}{2} \tan ^{-1} 1-\dfrac{1}{2} \tan ^{-1} 0$

$=\dfrac{1}{2}(\dfrac{\pi}{4})=\dfrac{\pi}{8}$

Solution

$\int(\dfrac{1}{x}-\dfrac{1}{2 x^{2}}) e^{2 x} d x$

मान लीजिए $2 x=t \Rightarrow 2 d x=d t$

जब $x=1, t=2$ और जब $x=2, t=4$

$ \begin{aligned} \therefore \int_1^{2}(\dfrac{1}{x}-\dfrac{1}{2 x^{2}}) e^{2 x} d x & =\dfrac{1}{2} \int_2^{4}(\dfrac{2}{t}-\dfrac{2}{t^{2}}) e^{t} d t \\ & =\int_2^{4}(\dfrac{1}{t}-\dfrac{1}{t^{2}}) e^{t} d t \end{aligned} $

मान लीजिए $\dfrac{1}{t}=f(t)$

तब, $f^{\prime}(t)=-\dfrac{1}{t^{2}}$

$ \begin{aligned} \Rightarrow \int_2^{4}(\dfrac{1}{t}-\dfrac{1}{t^{2}}) e^{t} d t & =\int_2^{4} e^{t}[f(t)+f^{\prime}(t)] d t \\ & =[e^{t} f(t)]_2^{4} \\ & =[e^{t} \cdot \dfrac{2}{t}]_2^{4} \\ & =[\dfrac{e^{t}}{t}]_2^{4} \\ & =\dfrac{e^{4}}{4}-\dfrac{e^{2}}{2} \\ & =\dfrac{e^{2}(e^{2}-2)}{4} \end{aligned} $

Solution

मान लीजिए $I=\displaystyle\int _{\dfrac{1}{3}}^{1} \dfrac{(x-x^{3})^{\dfrac{1}{3}}}{x^{4}} d x$

इसके अतिरिक्त, मान लीजिए $x=\sin \theta \Rightarrow d x=\cos \theta d \theta$

जब $x=\dfrac{1}{3}, \theta=\sin ^{-1}(\dfrac{1}{3})$ और जब $x=1, \theta=\dfrac{\pi}{2}$

$\Rightarrow I=\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}} \dfrac{(\sin \theta-\sin ^{3} \theta)^{\dfrac{1}{3}}}{\sin ^{4} \theta} \cos \theta d \theta$

$ =\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}} \dfrac{(\sin \theta)^{\dfrac{1}{3}}(1-\sin ^{2} \theta)^{\dfrac{1}{3}}}{\sin ^{4} \theta} \cos \theta d \theta $

$

$=\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}} \dfrac{(\sin \theta)^{\dfrac{1}{3}}(\cos \theta)^{\dfrac{2}{3}}}{\sin ^{4} \theta} \cos \theta d \theta$

$=\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}} \dfrac{(\sin \theta)^{\dfrac{1}{3}}(\cos \theta)^{\dfrac{2}{3}}}{\sin ^{2} \theta \sin ^{2} \theta} \cos \theta d \theta$

$=\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}} \dfrac{(\cos \theta)^{\dfrac{5}{3}}}{(\sin \theta)^{\dfrac{5}{3}}} cosec^{2} \theta d \theta$

$=\int _{\sin ^{-1}(\dfrac{1}{3})}^{\dfrac{\pi}{2}}(\cot \theta)^{\dfrac{5}{3}} cosec^{2} \theta d \theta$

$\cot \theta=t \Rightarrow -cosec 2 \theta d \theta=d t$

$ \begin{aligned} \therefore I & =-\int _{2 \sqrt{2}}^{0}(t)^{\dfrac{5}{3}} d t \\ & =-\Bigg[\dfrac{3}{8}(t)^{\dfrac{8}{3}}\Bigg] _{2 \sqrt{2}}^{0} \\ & =-\dfrac{3}{8}\Bigg[(t)^{\dfrac{8}{3}}\Bigg] _{2 \sqrt{2}}^{0} \\ & =-\dfrac{3}{8}[-(2 \sqrt{2})^{\dfrac{8}{3}}] \\ & =\dfrac{3}{8}[(\sqrt{8})^{\dfrac{8}{3}}] \\ & =\dfrac{3}{8}[(8)^{\dfrac{4}{3}}] \\ & =\dfrac{3}{8}[16] \\ & =3 \times 2 \\ & =6 \end{aligned} $

$ \text{ जब } \theta=\sin ^{-1}(\dfrac{1}{3}), t=2 \sqrt{2} \text{ और जब } \theta=\dfrac{\pi}{2}, t=0 $

इसलिए, सही उत्तर A है।

हल

$f(x)=\int_0^{x} t \sin t d t$

भाग विधि द्वारा समाकलन करने पर, हम प्राप्त करते हैं

$ \begin{aligned} f(x) & =t \int_0^{x} \sin t d t-\int_0^{x}{\dfrac{d}{d t} (t) \int \sin t d t} d t \\ & =[t(-\cos t)]_0^{x}-\int_0^{x}(-\cos t) d t \\ & =[-t \cos t+\sin t]_0^{x} \\ & =-x \cos x+\sin x \\ \Rightarrow f^{\prime}(x) & =-[{x(-\sin x)}+\cos x]+\cos x \\ & =x \sin x-\cos x+\cos x \\ & =x \sin x \end{aligned} $

इसलिए, सही उत्तर B है।

हल

$I=\int_0^{\dfrac{\pi}{2}} \cos ^{2} x d x \qquad…(1) $

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \cos ^{2}(\dfrac{\pi}{2}-x) d x \qquad (\because\int _0^{a} f(x) d x=\int _0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \sin ^{2} x d x \qquad…(2)$

(1) और (2) को जोड़ने पर, हम प्राप्त करते हैं

$2 I=\int_0^{\dfrac{\pi}{2}}(\sin ^{2} x+\cos ^{2} x) d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\dfrac{\pi}{2}}$

$\Rightarrow 2 I=\dfrac{\pi}{2}$

$\Rightarrow I=\dfrac{\pi}{4}$

हल

$\int_0^{\dfrac{\pi}{2}} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

मान लीजिए $I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sqrt{\sin (\dfrac{\pi}{2}-x)}}{\sqrt{\sin (\dfrac{\pi}{2}-x)}+\sqrt{\cos (\dfrac{\pi}{2}-x)}} d x $

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sqrt{\cos(x) }}{\sqrt{\cos(x) }+\sqrt{\sin x}} d x \qquad…(2)$

(1) और (2) को जोड़ने पर, हम प्राप्त करते हैं

$2 I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\dfrac{\pi}{2}}$

$\Rightarrow 2 I=\dfrac{\pi}{2}$

$\Rightarrow I=\dfrac{\pi}{4}$

हल

$(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

मान लीजिए $I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{\dfrac{3}{2}} x}{\sin ^{\dfrac{3}{2}} x+\cos ^{\dfrac{3}{2}} x} d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{\dfrac{3}{2}}(\dfrac{\pi}{2}-x)}{\sin ^{\dfrac{3}{2}}(\dfrac{\pi}{2}-x)+\cos ^{\dfrac{3}{2}}(\dfrac{\pi}{2}-x)} d x$

$ (\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x) $

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos ^{\dfrac{3}{2}} x}{\sin ^{\dfrac{3}{2}} x+\cos ^{\dfrac{3}{2}} x} d x \qquad…(2)$

(1) और (2) को जोड़ने पर हम प्राप्त करते हैं

$2 I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{\dfrac{3}{2}} x+\cos ^{\dfrac{3}{2}} x}{\sin ^{\dfrac{3}{2}} x+\cos ^{\dfrac{3}{2}} x} d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\dfrac{\pi}{2}}$

$\Rightarrow 2 I=\dfrac{\pi}{2}$

$\Rightarrow I=\dfrac{\pi}{4}$

Solution

मान लीजिए $I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos ^{5}(\dfrac{\pi}{2}-x)}{\sin ^{5}(\dfrac{\pi}{2}-x)+\cos ^{5}(\dfrac{\pi}{2}-x)} d x $

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x \qquad…(2)$

(1) और (2) को जोड़ने पर हम प्राप्त करते हैं

$2 I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin ^{5} x+\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} 1 \cdot d x$

$\Rightarrow 2 I=[x]_0^{\dfrac{\pi}{2}}$

$\Rightarrow 2 I=\dfrac{\pi}{2}$

$\Rightarrow I=\dfrac{\pi}{4}$

Solution

मान लीजिए $I=\int _{-5}^{5}|x+2| d x$

देखा जा सकता है कि $(x+2) \leq 0$ अंतराल $[-5,-2]$ पर और $(x+2) \geq 0$ अंतराल $[-2,5]$ पर है।

$ \begin{aligned} \therefore I & =\int _{-5}^{-2}-(x+2) d x+\int _{-2}^{5}(x+2) d x \quad(\because\int_a^{b} f(x)=\int_a^{c} f(x)+\int_c^{b} f(x)) \\ I & =-[\dfrac{x^{2}}{2}+2 x] _{-5}^{-2}+[\dfrac{x^{2}}{2}+2 x] _{-2}^{5} \\ & =-[\dfrac{(-2)^{2}}{2}+2(-2)-\dfrac{(-5)^{2}}{2}-2(-5)]+[\dfrac{(5)^{2}}{2}+2(5)-\dfrac{(-2)^{2}}{2}-2(-2)] \\ `

& =-[2-4-\dfrac{25}{2}+10]+[\dfrac{25}{2}+10-2+4] \\ & =-2+4+\dfrac{25}{2}-10+\dfrac{25}{2}+10-2+4 \\ & =29 \end{aligned} $

Solution

Let $I=\int_2^{8}|x-5| d x$

It can be seen that $(x-5) \leq 0$ on $[2,5]$ and $(x-5) \geq 0$ on $[5,8]$.

$ \begin{aligned} &I=\int_2^{5}-(x-5) d x+\int_5^{8}(x-5) d x & (\because\int_a^{b} f(x)=\int_a^{c} f(x)+\int_c^{b} f(x)) \ & =-[\dfrac{x^{2}}{2}-5 x]_2^{5}+[\dfrac{x^{2}}{2}-5 x]_5^{8} \ & =-[\dfrac{25}{2}-25-2+10]+[32-40-\dfrac{25}{2}+25] \ &=9 \end{aligned} $

Solution

$ \begin{aligned} \text{ Let } & I=\int_0^{1} x(1-x)^{n} d x \\ \therefore I & =\int_0^{1}(1-x)(1-(1-x))^{n} d x \\ & =\int_0^{1}(1-x)(x)^{n} d x \\ & =\int_0^{1}(x^{n}-x^{n+1}) d x \\ & =\bigg[\dfrac{x^{n+1}}{n+1}-\dfrac{x^{n+2}}{n+2}\bigg]_0^{1} \qquad(\because\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x) \\ & =\bigg[\dfrac{1}{n+1}-\dfrac{1}{n+2}\bigg] \\ & =\dfrac{(n+2)-(n+1)}{(n+1)(n+2)} \\ & =\dfrac{1}{(n+1)(n+2)} \end{aligned} $

Solution

Let $I=\int_0^{\dfrac{\pi}{4}} \log (1+\tan x) d x$

$\therefore I=\int_0^{\dfrac{\pi}{4}} \log [1+\tan (\dfrac{\pi}{4}-x)] d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{4}} \log\bigg(1+\dfrac{\tan \dfrac{\pi}{4}-\tan x}{1+\tan \dfrac{\pi}{4} \tan x}\bigg) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{4}} \log ({1+\dfrac{1-\tan x}{1+\tan x}}) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{4}} \log \dfrac{2}{(1+\tan x)} d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{4}} \log 2 d x-\int_0^{\dfrac{\pi}{4}} \log (1+\tan x) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{4}} \log 2 d x-I$

$\Rightarrow 2 I=[x \log 2]_0^{\dfrac{\pi}{4}}$

$\Rightarrow 2 I=\dfrac{\pi}{4} \log 2$

$\Rightarrow I=\dfrac{\pi}{8} \log 2$

हल

यह ज्ञात है कि, $\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x$

मान लीजिए $I=\int_0^{2} x \sqrt{2-x} d x$

$ \begin{aligned} I & =\int_0^{2}(2-x) \sqrt{x} d x \\ & =\int_0^{2}{2 x^{\dfrac{1}{2}}-x^{\dfrac{3}{2}}} d x \\ & =\bigg[2(\dfrac{x^{\dfrac{3}{2}}}{\dfrac{3}{2}})-\dfrac{x^{\dfrac{5}{2}}}{\dfrac{5}{2}}\bigg]_0^{2} \\ & =\bigg[\dfrac{4}{3} x^{\dfrac{3}{2}}-\dfrac{2}{5} x^{\dfrac{5}{2}}\bigg]_0^{2} \\ & =\dfrac{4}{3}(2)^{\dfrac{3}{2}}-\dfrac{2}{5}(2)^{\dfrac{5}{2}} \\ & =\dfrac{4 \times 2 \sqrt{2}}{3}-\dfrac{2}{5} \times 4 \sqrt{2} \\ & =\dfrac{8 \sqrt{2}}{3}-\dfrac{8 \sqrt{2}}{5} \\ & =\dfrac{40 \sqrt{2}-24 \sqrt{2}}{15} \\ & =\dfrac{16 \sqrt{2}}{15} \end{aligned} $

हल

मान लीजिए $I=\int_0^{\dfrac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}{2 \log \sin x-\log (2 \sin x \cos x)} d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}{2 \log \sin x-\log \sin x-\log \cos x-\log 2} d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}{\log \sin x-\log \cos x-\log 2} d x$

यह ज्ञात है कि, $(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}{\log \cos x-\log \sin x-\log 2} d x \qquad…(2)$

(1) और (2) को जोड़ने पर हम प्राप्त करते हैं

$ \begin{aligned} & 2 I=\int_0^{\dfrac{\pi}{2}}(-\log 2-\log 2) d x \\ & \Rightarrow 2 I=-2 \log 2 \int_0^{\dfrac{\pi}{2}} 1 d x \\ & \Rightarrow I=-\log 2[\dfrac{\pi}{2}] \\ & \Rightarrow I=\dfrac{\pi}{2}(-\log 2) \\ & \Rightarrow I=\dfrac{\pi}{2}[\log \dfrac{1}{2}] \\ & \Rightarrow I=\dfrac{\pi}{2} \log \dfrac{1}{2} \end{aligned} $

हल

मान लीजिए $I=\int _{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \sin ^{2} x d x$

क्योंकि $\sin ^{2}(-x)=(\sin (-x))^{2}=(-\sin x)^{2}=\sin ^{2} x$, इसलिए, $\sin ^{2} x$ एक विषम फलन है।

यह ज्ञात है कि यदि $f(x)$ एक विषम फलन है, तो $\int _{-a}^{a} f(x) d x=2 \int_0^{a} f(x) d x$

$ \begin{aligned} I & =2 \int_0^{\dfrac{\pi}{2}} \sin ^{2} x d x \\ & =2 \int_0^{\dfrac{\pi}{2}} \dfrac{1-\cos 2 x}{2} d x \\ & =\int_0^{\dfrac{\pi}{2}}(1-\cos 2 x) d x \\ & =[x-\dfrac{\sin 2 x}{2}]_0^{\dfrac{\pi}{2}} \\ & =\dfrac{\pi}{2} \end{aligned} $

Solution

मान लीजिए $I=\int_0^{\pi} \dfrac{x d x}{1+\sin x} \qquad…(1)$

$\Rightarrow I=\int_0^{\pi} \dfrac{(\pi-x)}{1+\sin (\pi-x)} d x$

$(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\pi} \dfrac{(\pi-x)}{1+\sin x} d x \qquad…(2)$

(1) और (2) को जोड़ने पर हम प्राप्त करते हैं

$ \begin{aligned} & 2 I=\int_0^{\pi} \dfrac{\pi}{1+\sin x} d x \\ & \Rightarrow 2 I=\pi \int_0^{\pi} \dfrac{(1-\sin x)}{(1+\sin x)(1-\sin x)} d x \\ & \Rightarrow 2 I=\pi \int_0^{\pi} \dfrac{1-\sin x}{\cos ^{2} x} d x \\ & \Rightarrow 2 I=\pi \int_0^{\pi}{\sec ^{2} x-\tan x \sec x} d x \\ & \Rightarrow 2 I=\pi[\tan x-\sec x]_0^{\pi} \\ & \Rightarrow 2 I=\pi[2] \\ & \Rightarrow I=\pi \end{aligned} $

Solution

मान लीजिए $I=\int _{\dfrac{-\pi}{2}}^{\dfrac{\pi}{2}} \sin ^{7} x d x$

क्योंकि $\sin ^{7}(-x)=(\sin (-x))^{7}=(-\sin x)^{7}=-\sin ^{7} x$, अतः, $\sin ^{2} x$ एक विषम फलन है।

यह ज्ञात है कि, यदि $f(x)$ एक विषम फलन है, तो $\int _{-a}^{a} f(x) d x=0$

$\therefore I=\int _{\dfrac{-\pi}{2}}^{\dfrac{\pi}{2}} \sin ^{7} x d x=0$

Solution

मान लीजिए $I=\int_0^{2 \pi} \cos ^{5} x d x$

$\cos ^{5}(2 \pi-x)=\cos ^{5} x$

यह ज्ञात है कि,

$ \begin{aligned} \int_0^{2 a} f(x) d x & =2 \int_0^{a} f(x) d x \text{, यदि } f(2 a-x)=f(x) \\ & =0 \text{ यदि } f(2 a-x)=-f(x) \end{aligned} $

$\therefore I=2 \int_0^{\pi} \cos ^{5} x d x$

$\Rightarrow I=2(0)=0 \quad[\because\cos ^{5}(\pi-x)=-\cos ^{5} x]$

Solution

Let $I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin x-\cos x}{1+\sin x \cos x} d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\sin (\dfrac{\pi}{2}-x)-\cos (\dfrac{\pi}{2}-x)}{1+\sin (\dfrac{\pi}{2}-x) \cos (\dfrac{\pi}{2}-x)} d x \qquad (\because\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x) $

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos x-\sin x}{1+\sin x \cos x} d x \qquad…(2)$

Adding (1) and (2), we obtain

$ \begin{aligned} & 2 I=\int_0^{\dfrac{\pi}{2}} \dfrac{0}{1+\sin x \cos x} d x \\ & \Rightarrow I=0 \end{aligned} $

Solution

$$ \begin{align*} & \text{ Let } I=\int_0^{\pi} \log (1+\cos x) d x \qquad…(1)\\ & \Rightarrow I=\int_0^{\pi} \log (1+\cos (\pi-x)) d x \\ & \Rightarrow I=\int_0^{\pi} \log (1-\cos x) d x \qquad(\because\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x) \qquad…(2) \end{align*} $$

Adding (1) and (2), we obtain

$2 I=\int_0^{\pi}{\log (1+\cos x)+\log (1-\cos x)} d x$

$\Rightarrow 2 I=\int_0^{\pi} \log (1-\cos ^{2} x) d x$

$\Rightarrow 2 I=\int_0^{\pi} \log \sin ^{2} x d x$

$\Rightarrow 2 I=2 \int_0^{\pi} \log \sin x d x$

$\Rightarrow I=\int_0^{\pi} \log \sin x d x$

$\sin (\pi-x)=\sin x$

$\therefore I=2 \int_0^{\dfrac{\pi}{2}} \log \sin x d x \qquad…(3)$

$\Rightarrow I=2 \int_0^{\dfrac{\pi}{2}} \log \sin (\dfrac{\pi}{2}-x) d x=2 \int_0^{\dfrac{\pi}{2}} \log \cos x d x \qquad…(4)$

Adding (3) and (4), we obtain

$2 I=2 \int_0^{\dfrac{\pi}{2}}(\log \sin x+\log \cos x) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}(\log \sin x+\log \cos x+\log 2-\log 2) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}}(\log 2 \sin x \cos x-\log 2) d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \log \sin 2 x d x-\int_0^{\dfrac{\pi}{2}} \log 2 d x$

Let $2 x=t \Rightarrow 2 d x=d t$

जब $x=0, t=0$ और जब $x=\dfrac{\pi}{2}, \pi=t$

$\therefore I=\dfrac{1}{2} \int_0^{\pi} \log (\sin t) d t-\dfrac{\pi}{2} \log 2$

$\Rightarrow I=\dfrac{1}{2} I-\dfrac{\pi}{2} \log 2$

$\Rightarrow \dfrac{I}{2}=-\dfrac{\pi}{2} \log 2$

$\Rightarrow I=-\pi \log 2$

Solution

Let $I=\int_0^{a} \dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x \qquad…(1)$

यह ज्ञात है कि, $(\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$I=\int_0^{a} \dfrac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}} d x \qquad…(2)$

(1) और (2) को जोड़ने पर, हम प्राप्त करते हैं

$2 I=\int_0^{a} \dfrac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}} d x$

$\Rightarrow 2 I=\int_0^{a} 1 d x$

$\Rightarrow 2 I=[x]_0^{a}$

$\Rightarrow 2 I=a$

$\Rightarrow I=\dfrac{a}{2}$

Solution

$I=\int_0^{4}|x-1| d x$

यह देखा जा सकता है कि, $(x-1) \leq 0$ जब $0 \leq x \leq 1$ और $(x-1) \geq 0$ जब $1 \leq x \leq 4$

$ \begin{aligned} I & =\int_0^{1}|x-1| d x+\int_1^{4}|x-1| d x \qquad(\because\int_a^{b} f(x)=\int_a^{c} f(x)+\int_c^{b} f(x)) \\ & =\int_0^{1}-(x-1) d x+\int_1^{4}(x-1) d x \\ & =[x-\dfrac{x^{2}}{2}]_0^{1}+[\dfrac{x^{2}}{2}-x]_1^{4} \\ & =1-\dfrac{1}{2}+\dfrac{(4)^{2}}{2}-4-\dfrac{1}{2}+1 \\ & =1-\dfrac{1}{2}+8-4-\dfrac{1}{2}+1 \\ & =5 \end{aligned} $

Solution

$$ \begin{align*} & \text{ मान लीजिए } I=\int_0^{a} f(x) g(x) d x \qquad…(1)\\ & \Rightarrow I=\int_0^{a} f(a-x) g(a-x) d x \\ & \Rightarrow I=\int_0^{a} f(x) g(a-x) d x \qquad…(2) \end{align*} $$

(1) और (2) को जोड़ने पर, हम प्राप्त करते हैं

$ \begin{aligned} & 2 I=\int_0^{a}{f(x) g(x)+f(x) g(a-x)} d x \\ & \Rightarrow 2 I=\int_0^{a} [f(x){g(x)+g(a-x)}] d x \\ & \Rightarrow 2 I=\int_0^{a} f(x) \times 4 d x \quad[\because g(x)+g(a-x)=4] \\ `

& \Rightarrow I=2 \int_0^{a} f(x) d x \end{aligned} $

समाधान

मान लीजिए $I=\int _{\dfrac{-\pi}{2}}^{\dfrac{\pi}{2}}(x^{3}+x \cos x+\tan ^{5} x+1) d x$

$\Rightarrow I=\int _{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} x^{3} d x+\int _{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}}x \cos x+\int _{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} \tan ^{5} x d x+\int _{-\dfrac{\pi}{2}}^{\dfrac{\pi}{2}} 1 \cdot d x$

यह ज्ञात है कि यदि $f(x)$ एक सम फलन है, तो $\int _{-a}^{a} f(x) d x=2 \int_0^{a} f(x) d x$ और

यदि $f(x)$ एक विषम फलन है, तो $\int _{-a}^{a} f(x) d x=0$

$ \begin{aligned} I & =0+0+0+2 \int_0^{\dfrac{\pi}{2}} 1 \cdot d x \\ & =2[x]_0^{\dfrac{\pi}{2}} \\ & =\dfrac{2 \pi}{2} \\ I & =\pi \end{aligned} $

अतः, सही उत्तर C है।

समाधान

मान लीजिए $I=\int_0^{\dfrac{\pi}{2}} \log (\dfrac{4+3 \sin x}{4+3 \cos x}) d x \qquad…(1)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \log [\dfrac{4+3 \sin (\dfrac{\pi}{2}-x)}{4+3 \cos (\dfrac{\pi}{2}-x)}] d x \quad(\because\int_0^{a} f(x) d x=\int_0^{a} f(a-x) d x)$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \log (\dfrac{4+3 \cos x}{4+3 \sin x}) d x \qquad…(2)$

(1) और (2) को जोड़ने पर हमें प्राप्त होता है

$2 I=\int_0^{\dfrac{\pi}{2}}{\log (\dfrac{4+3 \sin x}{4+3 \cos x})+\log (\dfrac{4+3 \cos x}{4+3 \sin x})} d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} \log (\dfrac{4+3 \sin x}{4+3 \cos x} \times \dfrac{4+3 \cos x}{4+3 \sin x}) d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} \log 1 d x$

$\Rightarrow 2 I=\int_0^{\dfrac{\pi}{2}} 0 d x$

$\Rightarrow I=0$

अतः, सही उत्तर C है।

हल

$\dfrac{1}{x-x^{3}}=\dfrac{1}{x(1-x^{2})}=\dfrac{1}{x(1-x)(1+x)}$

मान लीजिए $\dfrac{1}{x(1-x)(1+x)}=\dfrac{A}{x}+\dfrac{B}{(1-x)}+\dfrac{C}{1+x}$

$\Rightarrow 1=A(1-x^{2})+B x(1+x)+C x(1-x)$

$\Rightarrow 1=A-A x^{2}+B x+B x^{2}+C x-C x^{2}$

$ x^{2}, x $, और स्थिरांक के गुणांक के तुलना करने पर, हम प्राप्त करते हैं

$-A+B-C=0$

$B+C=0$

$A=1$

इन समीकरणों को हल करने पर, हम प्राप्त करते हैं

$A=1, B=\dfrac{1}{2}$, और $C=-\dfrac{1}{2}$

समीकरण (1) से, हम प्राप्त करते हैं $\dfrac{1}{x(1-x)(1+x)}=\dfrac{1}{x}+\dfrac{1}{2(1-x)}-\dfrac{1}{2(1+x)}$

$\Rightarrow \int \dfrac{1}{x(1-x)(1+x)} d x=\int \dfrac{1}{x} d x+\dfrac{1}{2} \int \dfrac{1}{1-x} d x-\dfrac{1}{2} \int \dfrac{1}{1+x} d x$

$=\log |x|-\dfrac{1}{2} \log |(1-x)|-\dfrac{1}{2} \log |(1+x)|$

$=\log |x|-\log |(1-x)^{\dfrac{1}{2}}|-\log |(1+x)^{\dfrac{1}{2}}|$

$=\log |\dfrac{x}{(1-x)^{\dfrac{1}{2}}(1+x)^{\dfrac{1}{2}}}|+C$

$=\log |(\dfrac{x^{2}}{1-x^{2}})^{\dfrac{1}{2}}|+C$

$=\dfrac{1}{2} \log |\dfrac{x^{2}}{1-x^{2}}|+C$

हल

$ \begin{aligned} \dfrac{1}{\sqrt{x+a}+\sqrt{x+b}} & =\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \dfrac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}} \\ & =\dfrac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} \\ & =\dfrac{(\sqrt{x+a}-\sqrt{x+b})}{a-b} \end{aligned} $

$\Rightarrow \int \dfrac{1}{\sqrt{x+a}-\sqrt{x+b}} d x=\dfrac{1}{a-b} \int(\sqrt{x+a}-\sqrt{x+b}) d x$

$ \begin{aligned} & =\dfrac{1}{(a-b)}[\dfrac{(x+a)^{\dfrac{3}{2}}}{\dfrac{3}{2}}-\dfrac{(x+b)^{\dfrac{3}{2}}}{\dfrac{3}{2}}] \\ & =\dfrac{2}{3(a-b)}[(x+a)^{\dfrac{3}{2}}-(x+b)^{\dfrac{3}{2}}]+C \end{aligned} $

हल

$ \begin{aligned} & \dfrac{1}{x \sqrt{a x-x^{2}}} \\ & \text{ मान लीजिए } x=\dfrac{a}{t} \Rightarrow d x=-\dfrac{a}{t^{2}} d t \\ & \Rightarrow \int \dfrac{1}{x \sqrt{a x-x^{2}}} d x=\int \dfrac{1}{\dfrac{a}{t} \sqrt{a \cdot \dfrac{a}{t}-(\dfrac{a}{t})^{2}}}(-\dfrac{a}{t^{2}} d t) \\ &=-\int \dfrac{1}{a t} \cdot \dfrac{1}{\sqrt{\dfrac{1}{t}-\dfrac{1}{t^{2}}}} d t \\ &=-\dfrac{1}{a} \int \dfrac{1}{\sqrt{\dfrac{t^{2}}{t}-\dfrac{t^{2}}{t^{2}}} d t} \\ &=-\dfrac{1}{a} \int \dfrac{1}{\sqrt{t-1} d t} \\ &=-\dfrac{1}{a}[2 \sqrt{t-1}]+C \\ &=-\dfrac{1}{a}[2 \sqrt{\dfrac{a}{x}-1}]+C \\ &=-\dfrac{2}{a}(\dfrac{\sqrt{a-x}}{\sqrt{x}})+C \\ &=-\dfrac{2}{a}(\sqrt{\dfrac{a-x}{x}})+C \end{aligned} $

हल

$\dfrac{1}{x^{2}(x^{4}+1)^{\dfrac{3}{4}}}$

$ x^{-3} $ से गुणा और विभाजन करने पर, हम प्राप्त करते हैं

$ \begin{aligned} \dfrac{x^{-3}}{x^{2} \cdot x^{-3}(x^{4}+1)^{\dfrac{3}{4}}} & =\dfrac{x^{-3}(x^{4}+1)^{\dfrac{-3}{4}}}{x^{2} \cdot x^{-3}} \\ & =\dfrac{(x^{4}+1)^{\dfrac{-3}{4}}}{x^{5} \cdot(x^{4})^{-\dfrac{3}{4}}} \\ & =\dfrac{1}{x^{5}}(\dfrac{x^{4}+1}{x^{4}})^{-\dfrac{3}{4}} \\ & =\dfrac{1}{x^{5}}(1+\dfrac{1}{x^{4}})^{-\dfrac{3}{4}} \end{aligned} $

मान लीजिए $\dfrac{1}{x^{4}}=t \Rightarrow-\dfrac{4}{x^{5}} d x=d t \Rightarrow \dfrac{1}{x^{5}} d x=-\dfrac{d t}{4}$

$\therefore \int \dfrac{1}{x^{2}(x^{4}+1)^{\dfrac{3}{4}}} d x=\int \dfrac{1}{x^{5}}(1+\dfrac{1}{x^{4}})^{-\dfrac{3}{4}} d x$

$ =-\dfrac{1}{4} \int(1+t)^{-\dfrac{3}{4}} d t $

$ \begin{aligned} & =-\dfrac{1}{4}[\dfrac{(1+t)^{\dfrac{1}{4}}}{\dfrac{1}{4}}]+C \\ & =-\dfrac{1}{4} \dfrac{(1+\dfrac{1}{x^{4}})^{\dfrac{1}{4}}}{\dfrac{1}{4}}+C \\ & =-(1+\dfrac{1}{x^{4}})^{\dfrac{1}{4}}+C \end{aligned} $

हल

$\dfrac{1}{x^{\dfrac{1}{2}}+x^{\dfrac{1}{3}}}=\dfrac{1}{x^{\dfrac{1}{3}}(1+x^{\dfrac{1}{6}})}$

मान लीजिए $x=t^{6} \Rightarrow d x=6 t^{5} d t$

$\therefore \int \dfrac{1}{x^{\dfrac{1}{2}}+x^{\dfrac{1}{3}}} d x=\int \dfrac{1}{x^{\dfrac{1}{3}}(1+x^{\dfrac{1}{6}})} d x$

$=\int \dfrac{6 t^{5}}{t^{2}(1+t)} d t$

$=6 \int \dfrac{t^{3}}{(1+t)} d t$

भाग देकर हम प्राप्त करते हैं

$ \begin{aligned} \int \dfrac{1}{x^{\dfrac{1}{2}}+x^{\dfrac{1}{3}}} d x & =6 \int{(t^{2}-t+1)-\dfrac{1}{1+t}} d t \\ & =6[(\dfrac{t^{3}}{3})-(\dfrac{t^{2}}{2})+t-\log |1+t|] \\ & =2 x^{\dfrac{1}{2}}-3 x^{\dfrac{1}{3}}+6 x^{\dfrac{1}{6}}-6 \log (1+x^{\dfrac{1}{6}})+C \\ & =2 \sqrt{x}-3 x^{\dfrac{1}{3}}+6 x^{\dfrac{1}{6}}-6 \log (1+x^{\dfrac{1}{6}})+C \end{aligned} $

हल

मान लीजिए $\dfrac{5 x}{(x+1)(x^{2}+9)}=\dfrac{A}{(x+1)}+\dfrac{B x+C}{(x^{2}+9)}$

$\Rightarrow 5 x=A(x^{2}+9)+(B x+C)(x+1)$

$\Rightarrow 5 x=A x^{2}+9 A+B x^{2}+B x+C x+C$

$ x^{2}, x $, और अचर पद के गुणांक के तुलना करने पर हम प्राप्त करते हैं

$A+B=0$

$B+C=5$

$9 A+C=0$

इन समीकरणों को हल करने पर हम प्राप्त करते हैं

$A=-\dfrac{1}{2}, B=\dfrac{1}{2}$, और $C=\dfrac{9}{2}$

समीकरण (1) से हम प्राप्त करते हैं

$ \begin{aligned} & \dfrac{5 x}{(x+1)(x^{2}+9)}=\dfrac{-1}{2(x+1)}+\dfrac{\dfrac{x}{2}+\dfrac{9}{2}}{(x^{2}+9)} \\ & \begin{aligned} \int \dfrac{5 x}{(x+1)(x^{2}+9)} d x & =\int{\dfrac{-1}{2(x+1)}+\dfrac{(x+9)}{2(x^{2}+9)}} d x \\ & =-\dfrac{1}{2} \log |x+1|+\dfrac{1}{2} \int \dfrac{x}{x^{2}+9} d x+\dfrac{9}{2} \int \dfrac{1}{x^{2}+9} d x \\ & =-\dfrac{1}{2} \log |x+1|+\dfrac{1}{4} \int \dfrac{2 x}{x^{2}+9} d x+\dfrac{9}{2} \int \dfrac{1}{x^{2}+9} d x \\ & =-\dfrac{1}{2} \log |x+1|+\dfrac{1}{4} \log |x^{2}+9|+\dfrac{9}{2} \cdot \dfrac{1}{3} \tan ^{-1} \dfrac{x}{3} \\ & =-\dfrac{1}{2} \log |x+1|+\dfrac{1}{4} \log (x^{2}+9)+\dfrac{3}{2} \tan ^{-1} \dfrac{x}{3}+C \end{aligned} \end{aligned} $

हल

$ \dfrac{\sin x}{\sin (x-a)} $

मान लीजिए $x-a=t \Rightarrow d x=d t$

$ \begin{aligned} \int \dfrac{\sin x}{\sin (x-a)} d x & =\int \dfrac{\sin (t+a)}{\sin t} d t \\ & =\int \dfrac{\sin t \cos a+\cos t \sin a}{\sin t} d t \\ & =\int(\cos a+\cot t \sin a) d t \\ & =t \cos a+\sin a \log |\sin t|+C_1 \\ & =(x-a) \cos a+\sin a \log |\sin (x-a)|+C_1 \\ & =x \cos a+\sin a \log |\sin (x-a)|-a \cos a+C_1 \\ & =\sin a \log |\sin (x-a)|+x \cos a+C \end{aligned} $

हल

$ \begin{aligned} \dfrac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} & =\dfrac{e^{4 \log x}(e^{\log x}-1)}{e^{2 \log x}(e^{\log x}-1)} \\ & =e^{2 \log x} \\ & =e^{\log x^{2}} \\ & =x^{2} \\ \therefore \int \dfrac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x & =\int x^{2} d x=\dfrac{x^{3}}{3}+C \end{aligned} $

हल

$\dfrac{\cos x}{\sqrt{4-\sin ^{2} x}}$

मान लीजिए $\sin x=t \Rightarrow \cos x d x=d t$

$\Rightarrow \int \dfrac{\cos x}{\sqrt{4-\sin ^{2} x}} d x=\int \dfrac{d t}{\sqrt{(2)^{2}-(t)^{2}}}$

$ \begin{aligned} & =\sin ^{-1}(\dfrac{t}{2})+C \\ & =\sin ^{-1}(\dfrac{\sin x}{2})+C \end{aligned} $

हल

$ \begin{aligned} & \dfrac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}=\dfrac{(\sin ^{4} x+\cos ^{4} x)(\sin ^{4} x-\cos ^{4} x)}{\sin ^{2} x+\cos ^{2} x-\sin ^{2} x \cos ^{2} x-\sin ^{2} x \cos ^{2} x} \\ &=\dfrac{(\sin ^{4} x+\cos ^{4} x)(\sin ^{2} x+\cos ^{2} x)(\sin ^{2} x-\cos ^{2} x)}{(\sin ^{2} x-\sin ^{2} x \cos ^{2} x)+(\cos ^{2} x-\sin ^{2} x \cos ^{2} x)} \\ &=\dfrac{(\sin ^{4} x+\cos ^{4} x)(\sin ^{2} x-\cos ^{2} x)}{\sin ^{2} x(1-\cos ^{2} x)+\cos ^{2} x(1-\sin ^{2} x)} \\

&=\dfrac{-(\sin ^{4} x+\cos ^{4} x)(\cos ^{2} x-\sin ^{2} x)}{(\sin ^{4} x+\cos ^{4} x)} \\ &=-\cos 2 x \\ & \therefore \int \dfrac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x=\int-\cos 2 x d x=-\dfrac{\sin 2 x}{2}+C \end{aligned} $

Solution

$\dfrac{1}{\cos (x+a) \cos (x+b)}$

$\sin (a-b)$ से गुणा और विभाजन करके, हम प्राप्त करते हैं

$ \begin{aligned} & \dfrac{1}{\sin (a-b)}\bigg[\dfrac{\sin (a-b)}{\cos (x+a) \cos (x+b)}\bigg] \\ & =\dfrac{1}{\sin (a-b)}\bigg[\dfrac{\sin [(x+a)-(x+b)]}{\cos (x+a) \cos (x+b)}\bigg] \\ & =\dfrac{1}{\sin (a-b)}\bigg[\dfrac{\sin (x+a) \cdot \cos (x+b)-\cos (x+a) \sin (x+b)}{\cos (x+a) \cos (x+b)}\bigg] \\ & =\dfrac{1}{\sin (a-b)}\bigg[\dfrac{\sin (x+a)}{\cos (x+a)}-\dfrac{\sin (x+b)}{\cos (x+b)}\bigg] \\ & =\dfrac{1}{\sin (a-b)}[\tan (x+a)-\tan (x+b)] \end{aligned} $

$ \begin{aligned} \int \dfrac{1}{\cos (x+a) \cos (x+b)} d x & =\dfrac{1}{\sin (a-b)} \int[\tan (x+a)-\tan (x+b)] d x \\ & =\dfrac{1}{\sin (a-b)}[-\log |\cos (x+a)|+\log |\cos (x+b)|]+C \\ & =\dfrac{1}{\sin (a-b)} \log |\dfrac{\cos (x+b)}{\cos (x+a)}|+C \end{aligned} $

Solution

$\dfrac{x^{3}}{\sqrt{1-x^{8}}}$

मान लीजिए $x^{4}=t \Rightarrow 4 x^{3} d x=d t$

$\Rightarrow \int \dfrac{x^{3}}{\sqrt{1-x^{8}}} d x=\dfrac{1}{4} \int \dfrac{d t}{\sqrt{1-t^{2}}}$

$ \begin{aligned} & =\dfrac{1}{4} \sin ^{-1} t+C \\ & =\dfrac{1}{4} \sin ^{-1}(x^{4})+C \end{aligned} $

Solution

$\dfrac{e^{x}}{(1+e^{x})(2+e^{x})}$

मान लीजिए $e^{x}=t \Rightarrow e^{x} d x=d t$

$\Rightarrow \int \dfrac{e^{x}}{(1+e^{x})(2+e^{x})} d x=\int \dfrac{d t}{(t+1)(t+2)}$

$=\int[\dfrac{1}{(t+1)}-\dfrac{1}{(t+2)}] d t$

$=\log |t+1|-\log |t+2|+C$

$=\log |\dfrac{t+1}{t+2}|+C$

$=\log |\dfrac{1+e^{x}}{2+e^{x}}|+C$

हल

$\therefore \dfrac{1}{(x^{2}+1)(x^{2}+4)}=\dfrac{A x+B}{(x^{2}+1)}+\dfrac{C x+D}{(x^{2}+4)} \qquad…(1)$

$\Rightarrow 1=(A x+B)(x^{2}+4)+(C x+D)(x^{2}+1)$

$\Rightarrow 1=A x^{3}+4 A x+B x^{2}+4 B+C x^{3}+C x+D x^{2}+D$

$ x^{3}, x^{2}, x $ और स्थिरांक के गुणांक के तुलना करने पर, हम प्राप्त करते हैं

$ A+C=0 $

$ B+D=0 $

$ 4 A+C=0 $

$ 4 B+D=1 $

इन समीकरणों को हल करने पर, हम प्राप्त करते हैं

$ A=0, B=\dfrac{1}{3}, C=0 $, और $ D=-\dfrac{1}{3} $

समीकरण (1) से, हम प्राप्त करते हैं

$ \begin{aligned} & \dfrac{1}{(x^{2}+1)(x^{2}+4)}=\dfrac{1}{3(x^{2}+1)}-\dfrac{1}{3(x^{2}+4)} \\ & \begin{aligned} \int \dfrac{1}{(x^{2}+1)(x^{2}+4)} d x & =\dfrac{1}{3} \int \dfrac{1}{x^{2}+1} d x-\dfrac{1}{3} \int \dfrac{1}{x^{2}+4} d x \\ & =\dfrac{1}{3} \tan ^{-1} x-\dfrac{1}{3} \cdot \dfrac{1}{2} \tan ^{-1} \dfrac{x}{2}+C \\ & =\dfrac{1}{3} \tan ^{-1} x-\dfrac{1}{6} \tan ^{-1} \dfrac{x}{2}+C \end{aligned} \end{aligned} $

हल

$\cos ^{3} x e^{\log \sin x}=\cos ^{3} x \times \sin x$

मान लीजिए $\cos x=t -\sin x d x=d t$ $\Rightarrow \int \cos ^{3} x e^{\log \sin x} d x=\int \cos ^{3} x \sin x d x$

$ \begin{aligned} & =-\int t^{3} \cdot d t \\ & =-\dfrac{t^{4}}{4}+C \\ & =-\dfrac{\cos ^{4} x}{4}+C \end{aligned} $

हल

$e^{3 \log x}(x^{4}+1)^{-1}=e^{\log x^{3}}(x^{4}+1)^{-1}=\dfrac{x^{3}}{(x^{4}+1)}$

मान लीजिए $x^{4}+1=t \Rightarrow 4 x^{3} d x=d t$

$\Rightarrow \int e^{3 \log x}(x^{4}+1)^{-1} d x=\int \dfrac{x^{3}}{(x^{4}+1)} d x$

$ \begin{aligned} & =\dfrac{1}{4} \int \dfrac{d t}{t} \\ & =\dfrac{1}{4} \log |t|+C \\ & =\dfrac{1}{4} \log |x^{4}+1|+C \\ & =\dfrac{1}{4} \log (x^{4}+1)+C \end{aligned} $

हल

$ f^{\prime}(a x+b)[f(a x+b)]^{n} `

$

मान लीजिए $f(a x+b)=t \Rightarrow a f^{\prime}(a x+b) d x=d t$

$\Rightarrow \int f^{\prime}(a x+b)[f(a x+b)]^{n} d x=\dfrac{1}{a} \int t^{n} d t$

$ \begin{aligned} & =\dfrac{1}{a}[\dfrac{t^{n+1}}{n+1}] \\ & =\dfrac{1}{a(n+1)}(f(a x+b))^{n+1}+C \end{aligned} $

Solution

$ \begin{aligned} \dfrac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}} & =\dfrac{1}{\sqrt{\sin ^{3} x(\sin x \cos \alpha+\cos x \sin \alpha)}} \\ & =\dfrac{1}{\sqrt{\sin ^{4} x \cos \alpha+\sin ^{3} x \cos x \sin \alpha}} \\ & =\dfrac{1}{\sin ^{2} x \sqrt{\cos \alpha+\cot x \sin \alpha}} \\ & =\dfrac{cosec^{2}x}{\sqrt{\cos \alpha+\cot x \sin \alpha}} \end{aligned} $

मान लीजिए $\cos \alpha+\cot x \sin \alpha=t \Rightarrow-cosec^{2} x \sin \alpha d x=d t$

$ \begin{aligned} \therefore \int \dfrac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}} d x & =\int \dfrac{cosec^{2} x}{\sqrt{\cos \alpha+\cot x \sin \alpha}} d x \\ & =\dfrac{-1}{\sin \alpha} \int \dfrac{d t}{\sqrt{t}} \\ & =\dfrac{-1}{\sin \alpha}[2 \sqrt{t}]+C \\ & =\dfrac{-1}{\sin \alpha}[2 \sqrt{\cos \alpha+\cot x \sin \alpha}]+C \\ & =\dfrac{-2}{\sin \alpha} \sqrt{\cos \alpha+\dfrac{\cos x \sin \alpha}{\sin x}}+C \\ & =\dfrac{-2}{\sin \alpha} \sqrt{\dfrac{\sin x \cos \alpha+\cos x \sin \alpha}{\sin x}}+C \\ & =-\dfrac{2}{\sin \alpha} \sqrt{\dfrac{\sin (x+\alpha)}{\sin x}+C} \end{aligned} $

Solution

$ I=\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}} d x $

मान लीजिए $x=\cos ^{2} \theta \Rightarrow d x=-2 \sin \theta \cos \theta d \theta$

$ I=\int \sqrt{\dfrac{1-\cos \theta}{1+\cos \theta}}(-2 \sin \theta \cos \theta) d \theta $

$ \begin{aligned} & =-\int \sqrt{\dfrac{2 \sin ^{2} \dfrac{\theta}{2}}{2 \cos ^{2} \dfrac{\theta}{2}}} \sin 2 \theta d \theta \\ & =-\int \tan \dfrac{\theta}{2} \cdot 2 \sin \theta \cos \theta d \theta \\ & =-2 \int \dfrac{\sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}}(2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2}) \cos \theta d \theta

\end{aligned} $

$=-4 \int \sin ^{2} \dfrac{\theta}{2} \cos \theta d \theta$

$=-4 \int \sin ^{2} \dfrac{\theta}{2} \cdot(2 \cos ^{2} \dfrac{\theta}{2}-1) d \theta$

$=-4 \int(2 \sin ^{2} \dfrac{\theta}{2} \cos ^{2} \dfrac{\theta}{2}-\sin ^{2} \dfrac{\theta}{2}) d \theta$

$=-8 \int \sin ^{2} \dfrac{\theta}{2} \cdot \cos ^{2} \dfrac{\theta}{2} d \theta+4 \int \sin ^{2} \dfrac{\theta}{2} d \theta$

$=-2 \int \sin ^{2} \theta d \theta+4 \int \sin ^{2} \dfrac{\theta}{2} d \theta$

$=-2 \int(\dfrac{1-\cos 2 \theta}{2}) d \theta+4 \int \dfrac{1-\cos \theta}{2} d \theta$

$=-2[\dfrac{\theta}{2}-\dfrac{\sin 2 \theta}{4}]+4[\dfrac{\theta}{2}-\dfrac{\sin \theta}{2}]+C$

$=-\theta+\dfrac{\sin 2 \theta}{2}+2 \theta-2 \sin \theta+C$

$=\theta+\dfrac{\sin 2 \theta}{2}-2 \sin \theta+C$

$=\theta+\dfrac{2 \sin \theta \cos \theta}{2}-2 \sin \theta+C$

$=\theta+\sqrt{1-\cos ^{2} \theta} \cdot \cos \theta-2 \sqrt{1-\cos ^{2} \theta}+C$

$=\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}-2 \sqrt{1-x}+C$

$=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}+C$

$=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x-x^{2}}+C$

Solution

$ \begin{aligned} I & =\int(\dfrac{2+\sin 2 x}{1+\cos 2 x}) e^{x} \\ & =\int(\dfrac{2+2 \sin x \cos x}{2 \cos ^{2} x}) e^{x} \\ & =\int(\dfrac{1+\sin x \cos x}{\cos ^{2} x}) e^{x} \\ & =\int(\sec ^{2} x+\tan x) e^{x} \end{aligned} $

Let $f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^{2} x$

$ \begin{aligned} \therefore I & =\int(f(x)+f^{\prime}(x)] e^{x} d x \\ & =e^{x} f(x)+C \\ & =e^{x} \tan x+C \end{aligned} $

Solution

Let $\dfrac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\dfrac{A}{(x+1)}+\dfrac{B}{(x+1)^{2}}+\dfrac{C}{(x+2)} \qquad…(1)$

$\Rightarrow x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C(x^{2}+2 x+1)$

$\Rightarrow x^{2}+x+1=A(x^{2}+3 x+2)+B(x+2)+C(x^{2}+2 x+1)$

$\Rightarrow x^{2}+x+1=(A+C) x^{2}+(3 A+B+2 C) x+(2 A+2 B+C)$

समीकरणों के गुणांक $x^{2}, x$, और अचर पद के बराबर करके, हम प्राप्त करते हैं

$A+C=1$

$3 A+B+2 C=1$

$2 A+2 B+C=1$

इन समीकरणों को हल करने पर, हम प्राप्त करते हैं

$A=-2, B=1$, और $C=3$

समीकरण (1) से, हम प्राप्त करते हैं

$ \begin{aligned} \dfrac{x^{2}+x+1}{(x+1)^{2}(x+2)} & =\dfrac{-2}{(x+1)}+\dfrac{3}{(x+2)}+\dfrac{1}{(x+1)^{2}} \\ \int \dfrac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x & =-2 \int \dfrac{1}{x+1} d x+3 \int \dfrac{1}{(x+2)} d x+\int \dfrac{1}{(x+1)^{2}} d x \\ & =-2 \log |x+1|+3 \log |x+2|-\dfrac{1}{(x+1)}+C \end{aligned} $

Solution

$I=\tan ^{-1} \sqrt{\dfrac{1-x}{1+x}} d x$

मान लीजिए $x=\cos \theta \Rightarrow d x=-\sin \theta d \theta$

$I=\int \tan ^{-1} \sqrt{\dfrac{1-\cos \theta}{1+\cos \theta}}(-\sin \theta d \theta)$

$=-\int \tan ^{-1} \sqrt{\dfrac{2 \sin ^{2} \dfrac{\theta}{2}}{2 \cos ^{2} \dfrac{\theta}{2}}} \sin \theta d \theta$

$=-\int \tan ^{-1} \tan \dfrac{\theta}{2} \cdot \sin \theta d \theta$

$=-\dfrac{1}{2} \int \theta \cdot \sin \theta d \theta$

$=-\dfrac{1}{2}[\theta \cdot(-\cos \theta)-\int 1 \cdot(-\cos \theta) d \theta]$

$=-\dfrac{1}{2}[-\theta \cos \theta+\sin \theta]$

$=+\dfrac{1}{2} \theta \cos \theta-\dfrac{1}{2} \sin \theta$

$=\dfrac{1}{2} \cos ^{-1} x \cdot x-\dfrac{1}{2} \sqrt{1-x^{2}}+C$

$=\dfrac{x}{2} \cos ^{-1} x-\dfrac{1}{2} \sqrt{1-x^{2}}+C$

$=\dfrac{1}{2}(x \cos ^{-1} x-\sqrt{1-x^{2}})+C$

Solution

$ \begin{aligned} \dfrac{\sqrt{x^{2}+1}[\log (x^{2}+1)-2 \log x]}{x^{4}} & =\dfrac{\sqrt{x^{2}+1}}{x^{4}}[\log (x^{2}+1)-\log x^{2}] \\ & =\dfrac{\sqrt{x^{2}+1}}{x^{4}}[\log (\dfrac{x^{2}+1}{x^{2}})] \\ & =\dfrac{\sqrt{x^{2}+1}}{x^{4}} \log (1+\dfrac{1}{x^{2}}) \\ & =\dfrac{1}{x^{3}} \sqrt{\dfrac{x^{2}+1}{x^{2}}} \log (1+\dfrac{1}{x^{2}}) \\ & =\dfrac{1}{x^{3}} \sqrt{1+\dfrac{1}{x^{2}}} \log (1+\dfrac{1}{x^{2}}) \end{aligned} $

$

मान लीजिए $1+\dfrac{1}{x^{2}}=t \Rightarrow \dfrac{-2}{x^{3}} d x=d t$

$ \begin{aligned} \therefore I & =\int \dfrac{1}{x^{3}} \sqrt{1+\dfrac{1}{x^{2}}} \log (1+\dfrac{1}{x^{2}}) d x \\ & =-\dfrac{1}{2} \int \sqrt{t} \log t d t \\ & =-\dfrac{1}{2} \int t^{\dfrac{1}{2}} \cdot \log t d t \end{aligned} $

भाग द्वारा समाकलन करने पर, हम प्राप्त करते हैं

$ \begin{aligned} I & =-\dfrac{1}{2}[\log t \cdot \int t^{\dfrac{1}{2}} d t-{(\dfrac{d}{d t} \log t) \int t^{\dfrac{1}{2}} d t} d t] \\ & =-\dfrac{1}{2}[\log t \cdot \dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}}-\int \dfrac{1}{t} \cdot \dfrac{t^{\dfrac{3}{2}}}{\dfrac{3}{2}} d t] \\ & =-\dfrac{1}{2}[\dfrac{2}{3} t^{\dfrac{3}{2}} \log t-\dfrac{2}{3} \int t^{\dfrac{1}{2}} d t] \\ & =-\dfrac{1}{2}[\dfrac{2}{3} t^{\dfrac{3}{2}} \log t-\dfrac{4}{9} t^{\dfrac{3}{2}}] \\ & =-\dfrac{1}{3} t^{\dfrac{3}{2}} \log t+\dfrac{2}{9} t^{\dfrac{3}{2}} \\ & =-\dfrac{1}{3} t^{\dfrac{3}{2}}[\log t-\dfrac{2}{3}] \\ & =-\dfrac{1}{3}(1+\dfrac{1}{x^{2}})^{\dfrac{3}{2}}[\log (1+\dfrac{1}{x^{2}})-\dfrac{2}{3}]+C \end{aligned} $

हल

$ \begin{aligned} I & =\int _{\dfrac{\pi}{2}}^{\pi} e^{x}(\dfrac{1-\sin x}{1-\cos x}) d x \\ & =\int _{\dfrac{\pi}{2}}^{\pi} e^{x}(\dfrac{1-2 \sin \dfrac{x}{2} \cos \dfrac{x}{2}}{2 \sin ^{2} \dfrac{x}{2}}) d x \\ & =\int _{\dfrac{\pi}{2}}^{\pi} e^{x}(\dfrac{cosec^{2} \dfrac{x}{2}}{2}-\cot \dfrac{x}{2}) d x \end{aligned} $

मान लीजिए $f(x)=-\cot \dfrac{x}{2}$

$\Rightarrow f^{\prime}(x)=-(-\dfrac{1}{2} cosec^{2} \dfrac{x}{2})=\dfrac{1}{2} cosec^{2} \dfrac{x}{2}$

$\therefore I=\int _{\dfrac{\pi}{2}}^{\pi} e^{x}(f(x)+f^{\prime}(x)] d x$

$=\bigg[e^{x} \cdot f(x) d x\bigg] _{\dfrac{\pi}{2}}^{x}$

$=-\bigg[e^{x} \cdot \cot \dfrac{x}{2}\bigg] _{\dfrac{\pi}{2}}^{\pi}$

$=-[e^{\pi} \times \cot \dfrac{\pi}{2}-e^{\dfrac{\pi}{2}} \times \cot \dfrac{\pi}{4}]$

$=-[e^{\pi} \times 0-e^{\dfrac{\pi}{2}} \times 1]$

$=e^{\dfrac{\pi}{2}}$

Solution

Let $I=\int_0^{\dfrac{\pi}{4}} \dfrac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$

$ \begin{aligned} & \Rightarrow I=\int_0^{\dfrac{\pi}{4}} \dfrac{\dfrac{(\sin x \cos x)}{\cos ^{4} x}}{\dfrac{(\cos ^{4} x+\sin ^{4} x)}{\cos ^{4} x}} d x \\ & \Rightarrow I=\int_0^{\dfrac{\pi}{4}} \dfrac{\tan x \sec ^{2} x}{1+\tan ^{4} x} d x \end{aligned} $

Let $\tan ^{2} x=t \Rightarrow 2 \tan x \sec ^{2} x d x=d t$

$ \text{ When } x=0, t=0 \text{ and when } x=\dfrac{\pi}{4}, t=1 $

$ \begin{aligned} \therefore I & =\dfrac{1}{2} \int_0^{1} \dfrac{d t}{1+t^{2}} \\ & =\dfrac{1}{2}[\tan ^{-1} t]_0^{1} \\ & =\dfrac{1}{2}[\tan ^{-1} 1-\tan ^{-1} 0] \\ & =\dfrac{1}{2}[\dfrac{\pi}{4}] \\ & =\dfrac{\pi}{8} \end{aligned} $

Solution

Let $I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos ^{2} x}{\cos ^{2} x+4(1-\cos ^{2} x)} d x$

$\Rightarrow I=\int_0^{\dfrac{\pi}{2}} \dfrac{\cos ^{2} x}{\cos ^{2} x+4-4 \cos ^{2} x} d x$

$\Rightarrow I=\dfrac{-1}{3} \int_0^{\dfrac{\pi}{2}} \dfrac{4-3 \cos ^{2} x-4}{4-3 \cos ^{2} x} d x$

$\Rightarrow I=\dfrac{-1}{3} \int_0^{\dfrac{\pi}{2}} \dfrac{4-3 \cos ^{2} x}{4-3 \cos ^{2} x} d x+\dfrac{1}{3} \int_0^{\dfrac{\pi}{2}} \dfrac{4}{4-3 \cos ^{2} x} d x$

$\Rightarrow I=\dfrac{-1}{3} \int_0^{\dfrac{\pi}{2}} 1 d x+\dfrac{1}{3} \int_0^{\dfrac{\pi}{2}} \dfrac{4 \sec ^{2} x}{4 \sec ^{2} x-3} d x$

$\Rightarrow I=\dfrac{-1}{3}[x]_0^{\dfrac{\pi}{2}}+\dfrac{1}{3} \int_0^{\dfrac{\pi}{2}} \dfrac{4 \sec ^{2} x}{4(1+\tan ^{2} x)-3} d x$

$\Rightarrow I=-\dfrac{\pi}{6}+\dfrac{2}{3} \int_0^{\dfrac{\pi}{2}} \dfrac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x \qquad…(1)$

Consider, $\int_0^{\dfrac{\pi}{2}} \dfrac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x$

Let $2 \tan x=t \Rightarrow 2 \sec ^{2} x d x=d t$

When $x=0, t=0$ and when $x=\dfrac{\pi}{2}, t=\infty$

$\Rightarrow \int_0^{\dfrac{\pi}{2}} \dfrac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x=\int_0^{\infty} \dfrac{d t}{1+t^{2}}$

$ \begin{aligned} & =[\tan ^{-1} t]_0^{\infty} \\ & =[\tan ^{-1}(\infty)-\tan ^{-1}(0)] \\ & =\dfrac{\pi}{2} \end{aligned} $

Therefore, from (1),we obtain

$I=-\dfrac{\pi}{6}+\dfrac{2}{3}[\dfrac{\pi}{2}]=\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{6}$

Solution

Let $I=\int _{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$

$\Rightarrow I=\int _{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{(\sin x+\cos x)}{\sqrt{-(-\sin 2 x)}} d x$

$\Rightarrow I=\int _{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{\sin x+\cos x}{\sqrt{-(-1+1-2 \sin x \cos x)}} d x$

$\Rightarrow I=\int _{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{(\sin x+\cos x)}{\sqrt{1-(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x)}} d x$

$\Rightarrow I=\int _{\dfrac{\pi}{6}}^{\dfrac{\pi}{3}} \dfrac{(\sin x+\cos x) d x}{\sqrt{1-(\sin x-\cos x)^{2}}}$

Let $(\sin x-\cos x)=t \Rightarrow(\sin x+\cos x) d x=d t$

When $x=\dfrac{\pi}{6}, t=(\dfrac{1-\sqrt{3}}{2})$ and when $\quad x=\dfrac{\pi}{3}, t=(\dfrac{\sqrt{3}-1}{2})$

$I=\int _{\dfrac{1-\sqrt{3}}{2}}^{\dfrac{\sqrt{3}-1}{2}} \dfrac{d t}{\sqrt{1-t^{2}}}$

$\Rightarrow I=\int _{-(\dfrac{\sqrt{3}-1}{2})}^{(\dfrac{\sqrt{3}-1}{2})} \dfrac{d t}{\sqrt{1-t^{2}}}$

As $\dfrac{1}{\sqrt{1-(-t)^{2}}}=\dfrac{1}{\sqrt{1-t^{2}}}$, therefore, $\dfrac{1}{\sqrt{1-t^{2}}}$ is an even function.

It is known that if $f(x)$ is an even function, then $\int _{-a}^{a} f(x) d x=2 \int_0^{a} f(x) d x$

$ \begin{aligned} \Rightarrow I & =2 \int_0^{\dfrac{\sqrt{3}-1}{2}} \dfrac{d t}{\sqrt{1-t^{2}}} \\ & =\bigg[2 \sin ^{-1} t\bigg]_0^{\dfrac{{\sqrt{3}-1}}{2}} \\

& =2 \sin ^{-1}(\dfrac{\sqrt{3}-1}{2}) \end{aligned} $

Solution

$ \begin{aligned} & \text{ मान लीजिए } I=\int_0^{1} \dfrac{d x}{\sqrt{1+x}-\sqrt{x}} \\ & \begin{aligned} I & =\int_0^{1} \dfrac{1}{(\sqrt{1+x}-\sqrt{x})} \times \dfrac{(\sqrt{1+x}+\sqrt{x})}{(\sqrt{1+x}+\sqrt{x})} d x \\ & =\int_0^{1} \dfrac{\sqrt{1+x}+\sqrt{x}}{1+x-x} d x \\ & =\int_0^{1} \sqrt{1+x} d x+\int_0^{1} \sqrt{x} d x \\ & =[\dfrac{2}{3}(1+x)^{\dfrac{3}{2}}]_0^{1}+[\dfrac{2}{3}(x)^{\dfrac{3}{2}}]_0^{1} \\ & =\dfrac{2}{3}[(2)^{\dfrac{3}{2}}-1]+\dfrac{2}{3}[1] \\ & =\dfrac{2}{3}(2)^{\dfrac{3}{2}} \\ & =\dfrac{2 \cdot 2 \sqrt{2}}{3} \\ & =\dfrac{4 \sqrt{2}}{3} \end{aligned} \\ \end{aligned} $

Solution

मान लीजिए $I=\int_0^{\dfrac{\pi}{4}} \dfrac{\sin x+\cos x}{9+16 \sin 2 x} d x$

इसके अतिरिक्त, मान लीजिए $\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$

जब $x=0, t=-1$ और जब $x=\dfrac{\pi}{4}, t=0$

$\Rightarrow(\sin x-\cos x)^{2}=t^{2}$

$\Rightarrow \sin ^{2} x+\cos ^{2} x-2 \sin x \cos x=t^{2}$

$\Rightarrow 1-\sin 2 x=t^{2}$

$\Rightarrow \sin 2 x=1-t^{2}$

$\therefore I=\int _{-1}^{0} \dfrac{d t}{9+16(1-t^{2})}$

$=\int _{-1}^{0} \dfrac{d t}{9+16-16 t^{2}}$

$=\int _{-1}^{0} \dfrac{d t}{25-16 t^{2}}=\int _{-1}^{0} \dfrac{d t}{(5)^{2}-(4 t)^{2}}$

$=\dfrac{1}{4}[\dfrac{1}{2(5)} \log |\dfrac{5+4 t}{5-4 t}|] _{-1}^{0}$

$=\dfrac{1}{40}[\log (1)-\log |\dfrac{1}{9}|]$

$=\dfrac{1}{40} \log 9$

Solution

मान लीजिए $I=\int_0^{\dfrac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x=\int_0^{\dfrac{\pi}{2}} 2 \sin x \cos x \tan ^{-1}(\sin x) d x$

इसके अतिरिक्त, मान लीजिए $\sin x=t \Rightarrow \cos x d x=d t$

जब $x=0, t=0$ और जब $x=\dfrac{\pi}{2}, t=1$

$\Rightarrow I=2 \int_0^{1} t \tan ^{-1}(t) d t \qquad…(1)$

Consider $\int t \cdot \tan ^{-1} t d t=\tan ^{-1} t \cdot \int t d t-\int{\dfrac{d}{d t}(\tan ^{-1} t) \int t d t} d t$

$ \begin{aligned} & =\tan ^{-1} t \cdot \dfrac{t^{2}}{2}-\int \dfrac{1}{1+t^{2}} \cdot \dfrac{t^{2}}{2} d t \\ & =\dfrac{t^{2} \tan ^{-1} t}{2}-\dfrac{1}{2} \int \dfrac{t^{2}+1-1}{1+t^{2}} d t \\ & =\dfrac{t^{2} \tan ^{-1} t}{2}-\dfrac{1}{2} \int 1 d t+\dfrac{1}{2} \int \dfrac{1}{1+t^{2}} d t \\ & =\dfrac{t^{2} \tan ^{-1} t}{2}-\dfrac{1}{2} \cdot t+\dfrac{1}{2} \tan ^{-1} t \end{aligned} $

$ \begin{aligned} \Rightarrow \int_0^{1} t \cdot \tan ^{-1} t d t & =\bigg[\dfrac{t^{2} \cdot \tan ^{-1} t}{2}-\dfrac{t}{2}+\dfrac{1}{2} \tan ^{-1} t\bigg]_0^{1} \\ & =\dfrac{1}{2}[\dfrac{\pi}{4}-1+\dfrac{\pi}{4}] \\ & =\dfrac{1}{2}[\dfrac{\pi}{2}-1]=\dfrac{\pi}{4}-\dfrac{1}{2} \end{aligned} $

समीकरण (1) से, हम प्राप्त करते हैं

$I=2[\dfrac{\pi}{4}-\dfrac{1}{2}]=\dfrac{\pi}{2}-1$

Solution

मान लीजिए $I=\int_1^{4}[|x-1|+|x-2|+|x-3|] d x$

$\Rightarrow I=\int_1^{4}|x-1| d x+\int_1^{4}|x-2| d x+\int_1^{4}|x-3| d x$

$I=I_1+I_2+I_3 \qquad…(1)$

जहाँ, $I_1=\int_1^{4}|x-1| d x, I_2=\int_1^{4}|x-2| d x$, और $I_3=\int_1^{4}|x-3| d x$

$I_1=\int_1^{4}|x-1| d x$

$(x-1) \geq 0$ जब $1 \leq x \leq 4$

$\therefore I_1=\int_1^{4}(x-1) d x$

$\Rightarrow I_1=[\dfrac{x^{2}}{x}-x]_1^{4}$

$\Rightarrow I_1=[8-4-\dfrac{1}{2}+1]=\dfrac{9}{2}\qquad…(2)$

$I_2=\int_1^{4}|x-2| d x$

$x-2 \geq 0$ जब $2 \leq x \leq 4$ और $x-2 \leq 0$ जब $1 \leq x \leq 2$

$\therefore I_2=\int_1^{2}(2-x) d x+\int_2^{4}(x-2) d x$

$\Rightarrow I_2=[2 x-\dfrac{x^{2}}{2}]_1^{2}+[\dfrac{x^{2}}{2}-2 x]_2^{4}$

$\Rightarrow I_2=[4-2-2+\dfrac{1}{2}]+[8-8-2+4]$

$\Rightarrow I_2=\dfrac{1}{2}+2=\dfrac{5}{2} \qquad…(3)$

$I_3=\int_1^{4}|x-3| d x$

$x-3 \geq 0$ जब $3 \leq x \leq 4$ और $x-3 \leq 0$ जब $1 \leq x \leq 3$

$\therefore I_3=\int_1^{3}(3-x) d x+\int_3^{4}(x-3) d x$

$\Rightarrow I_3=[3 x-\dfrac{x^{2}}{2}]_1^{3}+[\dfrac{x^{2}}{2}-3 x]_3^{4}$

$\Rightarrow I_3=[9-\dfrac{9}{2}-3+\dfrac{1}{2}]+[8-12-\dfrac{9}{2}+9]$

$\Rightarrow I_3=[6-4]+[\dfrac{1}{2}]=\dfrac{5}{2} \qquad…(4)$

समीकरण (1), (2), (3) और (4) से, हम प्राप्त करते हैं

$I=\dfrac{9}{2}+\dfrac{5}{2}+\dfrac{5}{2}=\dfrac{19}{2}$

हल

मान लीजिए $I=\int_1^{3} \dfrac{d x}{x^{2}(x+1)}$

इसके अतिरिक्त, मान लीजिए $\dfrac{1}{x^{2}(x+1)}=\dfrac{A}{x}+\dfrac{B}{x^{2}}+\dfrac{C}{x+1}$

$\Rightarrow 1=A x(x+1)+B(x+1)+C(x^{2})$

$\Rightarrow 1=A x^{2}+A x+B x+B+C x^{2}$

$x^{2}, x$ और अचर पद के गुणांक के तुलना करने पर, हम प्राप्त करते हैं

$A+C=0$

$A+B=0$

$B=1$

इन समीकरणों को हल करने पर, हम प्राप्त करते हैं

$A=-1, C=1$, और $B=1$

$\therefore \dfrac{1}{x^{2}(x+1)}=\dfrac{-1}{x}+\dfrac{1}{x^{2}}+\dfrac{1}{(x+1)}$

$\Rightarrow I=\int_1^{3}{-\dfrac{1}{x}+\dfrac{1}{x^{2}}+\dfrac{1}{(x+1)}} d x$

$ =[-\log x-\dfrac{1}{x}+\log (x+1)]_1^{3} $

$ =[\log (\dfrac{x+1}{x})-\dfrac{1}{x}]_1^{3} $

$=\log (\dfrac{4}{3})-\dfrac{1}{3}-\log (\dfrac{2}{1})+1$

$=\log 4-\log 3-\log 2+\dfrac{2}{3}$

$=\log 2-\log 3+\dfrac{2}{3}$

$=\log (\dfrac{2}{3})+\dfrac{2}{3}$

इसलिए, दिया गया परिणाम सिद्ध हो गया है।

हल

मान लीजिए $I=\int_0^{1} x e^{x} d x$

भाग विधि द्वारा समाकलन करने पर, हम प्राप्त करते हैं

$ \begin{aligned} I & =x \int_0^{1} e^{x} d x-\int_0^{1}{(\dfrac{d}{d x}(x)) \int e^{x} d x} d x \\ & =[x e^{x}]_0^{1}-\int_0^{1} e^{x} d x \\ & =[x e^{x}]_0^{1}-[e^{x}]_0^{1} \\ & =e-e+1 \\ & =1 \end{aligned} $

इसलिए, दिया गया परिणाम सिद्ध हो गया है।

हल

मान लीजिए $I=\int _{-1}^{1} x^{17} \cos ^{4} x d x$

इसके अतिरिक्त, मान लीजिए $f(x)=x^{17} \cos ^{4} x$

$\Rightarrow f(-x)=(-x)^{17} \cos ^{4}(-x)=-x^{17} \cos ^{4} x=-f(x)$

इसलिए, $f(x)$ एक विषम फलन है।

यह ज्ञात है कि यदि $f(x)$ एक विषम फलन है, तो $\int _{-a}^{a} f(x) d x=0$

$\therefore I=\int _{-1}^{1} x^{17} \cos ^{4} x d x=0$

इसलिए, दिए गए परिणाम की साबित कर दिया गया है।

हल

मान लीजिए $I=\int_0^{\dfrac{\pi}{2}} \sin ^{3} x d x$

$I=\int_0^{\dfrac{\pi}{2}} \sin ^{2} x \cdot \sin x d x$

$=\int_0^{\dfrac{\pi}{2}}(1-\cos ^{2} x) \sin x d x$

$=\int_0^{\dfrac{\pi}{2}} \sin x d x-\int_0^{\dfrac{\pi}{2}} \cos ^{2} x \cdot \sin x d x$

$=[-\cos x]_0^{\dfrac{\pi}{2}}+[\dfrac{\cos ^{3} x}{3}]_0^{\dfrac{\pi}{2}}$

$=1+\dfrac{1}{3}[-1]=1-\dfrac{1}{3}=\dfrac{2}{3}$

इसलिए, दिए गए परिणाम की साबित कर दिया गया है।

हल

मान लीजिए $I=\int_0^{\dfrac{\pi}{4}} 2 \tan ^{3} x d x$

$I=2 \int_0^{\dfrac{\pi}{4}} \tan ^{2} x \tan x d x=2 \int_0^{\dfrac{\pi}{4}}(\sec ^{2} x-1) \tan x d x$

$=2 \int_0^{\dfrac{\pi}{4}} \sec ^{2} x \tan x d x-2 \int_0^{\dfrac{\pi}{4}} \tan x d x$

$=2[\dfrac{\tan ^{2} x}{2}]_0^{\dfrac{\pi}{4}}+2[\log \cos x]_0^{\dfrac{\pi}{4}}$

$=1+2[\log \cos \dfrac{\pi}{4}-\log \cos 0]$

$=1+2[\log \dfrac{1}{\sqrt{2}}-\log 1]$

$=1-\log 2-\log 1=1-\log 2$

इसलिए, दिए गए परिणाम की साबित कर दिया गया है।

हल

मान लीजिए $I=\int_0^{1} \sin ^{-1} x d x$

$\Rightarrow I=\int_0^{1} \sin ^{-1} x \cdot 1 \cdot d x$

भाग विधि द्वारा समाकलन करने पर, हम प्राप्त करते हैं

$ \begin{aligned} I & =[\sin ^{-1} x \cdot x]_0^{1}-\int_0^{1} \dfrac{1}{\sqrt{1-x^{2}}} \cdot x d x \\ & =[x \sin ^{-1} x]_0^{1}+\dfrac{1}{2} \int_0^{1} \dfrac{(-2 x)}{\sqrt{1-x^{2}}} d x \end{aligned} $

मान लीजिए $1-x^{2}=t \Rightarrow -2 x d x=d t$

जब $x=0, t=1$ और जब $x=1, t=0$

$ \begin{aligned} I & =[x \sin ^{-1} x]_0^{1}+\dfrac{1}{2} \int_0^{0} \dfrac{d t}{\sqrt{t}} \\ & =[x \sin ^{-1} x]_0^{1}+\dfrac{1}{2}[2 \sqrt{t}]_1^{0} \\ & =\sin ^{-1}(1)+[-\sqrt{1}] \\ & =\dfrac{\pi}{2}-1 \end{aligned} $

$

अतः, दिए गए परिणाम की साबित कर दिया गया है।

हल

मान लीजिए $I=\int \dfrac{d x}{e^{x}+e^{-x}} d x=\int \dfrac{e^{x}}{e^{2 x}+1} d x$

इसके अतिरिक्त, मान लीजिए $e^{x}=t \Rightarrow e^{x} d x=d t$

$ \begin{aligned} \therefore I & =\int \dfrac{d t}{1+t^{2}} \\ & =\tan ^{-1} t+C \\ & =\tan ^{-1}(e^{x})+C \end{aligned} $

अतः, सही उत्तर A है।

हल

मान लीजिए $I=\dfrac{\cos 2 x}{(\cos x+\sin x)^{2}}$

$ \begin{aligned} I & =\int \dfrac{\cos ^{2} x-\sin ^{2} x}{(\cos x+\sin x)^{2}} d x \\ & =\int \dfrac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^{2}} d x \\ & =\int \dfrac{\cos x-\sin x}{\cos x +\sin x} d x \end{aligned} $

मान लीजिए $\cos x+\sin x=t \Rightarrow(\cos x-\sin x) d x=d t$

$ \begin{aligned} \therefore I & =\int \dfrac{d t}{t} \\ & =\log |t|+C \\ & =\log |\cos x+\sin x|+C \end{aligned} $

अतः, सही उत्तर B है।

हल

मान लीजिए $I=\int_a^{b} x f(x) d x$

$ \begin{aligned} & I=\int_a^{b}(a+b-x) f(a+b-x) d x\qquad (\because\int_a^{b} f(x) d x=\int_a^{b} f(a+b-x) d x) \\ & \Rightarrow I=\int_a^{b}(a+b-x) f(x) d x \\

$$ \begin{aligned} & \Rightarrow I=(a+b) \int_a^{b} f(x) d x \quad-I \\ & \Rightarrow I+I=(a+b) \int_a^{b} f(x) d x \\ & \Rightarrow 2 I=(a+b) \int_a^{b} f(x) d x \\ & \Rightarrow I=(\dfrac{a+b}{2}) \int_a^{b} f(x) d x \end{aligned} $$

इसलिए, सही उत्तर D है।