Solution

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

Let $\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{(1-x)}+\frac{C}{1+x}$

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

$\Rightarrow 1=A-A x^{2}+B x+B x^{2}+C x-C x^{2}$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$-A+B-C=0$

$B+C=0$

$A=1$

On solving these equations, we obtain

$A=1, B=\frac{1}{2}$, and $C=-\frac{1}{2}$

From equation (1), we obtain $\frac{1}{x(1-x)(1+x)}=\frac{1}{x}+\frac{1}{2(1-x)}-\frac{1}{2(1+x)}$

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

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

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

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

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

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

Solution

$ \begin{aligned} \frac{1}{\sqrt{x+a}+\sqrt{x+b}} & =\frac{1}{\sqrt{x+a}+\sqrt{x+b}} \times \frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}} \\ & =\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} \\ & =\frac{(\sqrt{x+a}-\sqrt{x+b})}{a-b} \end{aligned} $

$\Rightarrow \int \frac{1}{\sqrt{x+a}-\sqrt{x+b}} d x=\frac{1}{a-b} \int(\sqrt{x+a}-\sqrt{x+b}) d x$

$ \begin{aligned} & =\frac{1}{(a-b)}[\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}-\frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}}] \\ & =\frac{2}{3(a-b)}[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}]+C \end{aligned} $

Solution

$ \begin{aligned} & \frac{1}{x \sqrt{a x-x^{2}}} \\ & \text{ Let } x=\frac{a}{t} \Rightarrow d x=-\frac{a}{t^{2}} d t \\ & \Rightarrow \int \frac{1}{x \sqrt{a x-x^{2}}} d x=\int \frac{1}{\frac{a}{t} \sqrt{a \cdot \frac{a}{t}-(\frac{a}{t})^{2}}}(-\frac{a}{t^{2}} d t) \\ &=-\int \frac{1}{a t} \cdot \frac{1}{\sqrt{\frac{1}{t}-\frac{1}{t^{2}}}} d t \\ &=-\frac{1}{a} \int \frac{1}{\sqrt{\frac{t^{2}}{t}-\frac{t^{2}}{t^{2}}} d t} \\ &=-\frac{1}{a} \int \frac{1}{\sqrt{t-1} d t} \\ &=-\frac{1}{a}[2 \sqrt{t-1}]+C \\ &=-\frac{1}{a}[2 \sqrt{\frac{a}{x}-1}]+C \\ &=-\frac{2}{a}(\frac{\sqrt{a-x}}{\sqrt{x}})+C \\ &=-\frac{2}{a}(\sqrt{\frac{a-x}{x}})+C \end{aligned} $

Solution

$\frac{1}{x^{2}(x^{4}+1)^{\frac{3}{4}}}$

Multiplying and dividing by $x^{-3}$, we obtain

$ \begin{aligned} \frac{x^{-3}}{x^{2} \cdot x^{-3}(x^{4}+1)^{\frac{3}{4}}} & =\frac{x^{-3}(x^{4}+1)^{\frac{-3}{4}}}{x^{2} \cdot x^{-3}} \\ & =\frac{(x^{4}+1)^{\frac{-3}{4}}}{x^{5} \cdot(x^{4})^{-\frac{3}{4}}} \\ & =\frac{1}{x^{5}}(\frac{x^{4}+1}{x^{4}})^{-\frac{3}{4}} \\ & =\frac{1}{x^{5}}(1+\frac{1}{x^{4}})^{-\frac{3}{4}} \end{aligned} $

Let $\frac{1}{x^{4}}=t \Rightarrow-\frac{4}{x^{5}} d x=d t \Rightarrow \frac{1}{x^{5}} d x=-\frac{d t}{4}$

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

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

$ \begin{aligned} & =-\frac{1}{4}[\frac{(1+t)^{\frac{1}{4}}}{\frac{1}{4}}]+C \\ & =-\frac{1}{4} \frac{(1+\frac{1}{x^{4}})^{\frac{1}{4}}}{\frac{1}{4}}+C \\ & =-(1+\frac{1}{x^{4}})^{\frac{1}{4}}+C \end{aligned} $

Solution

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

Let $x=t^{6} \Rightarrow d x=6 t^{5} d t$

$\therefore \int \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}} d x=\int \frac{1}{x^{\frac{1}{3}}(1+x^{\frac{1}{6}})} d x$

$=\int \frac{6 t^{5}}{t^{2}(1+t)} d t$

$=6 \int \frac{t^{3}}{(1+t)} d t$

On dividing, we obtain

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

Solution

Let $\frac{5 x}{(x+1)(x^{2}+9)}=\frac{A}{(x+1)}+\frac{B x+C}{(x^{2}+9)}$

$\Rightarrow 5 x=A(x^{2}+9)+(B x+C)(x+1)$

$\Rightarrow 5 x=A x^{2}+9 A+B x^{2}+B x+C x+C$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$A+B=0$

$B+C=5$

$9 A+C=0$

On solving these equations, we obtain

$A=-\frac{1}{2}, B=\frac{1}{2}$, and $C=\frac{9}{2}$

From equation (1), we obtain

$ \begin{aligned} & \frac{5 x}{(x+1)(x^{2}+9)}=\frac{-1}{2(x+1)}+\frac{\frac{x}{2}+\frac{9}{2}}{(x^{2}+9)} \\ & \begin{aligned} \int \frac{5 x}{(x+1)(x^{2}+9)} d x & =\int{\frac{-1}{2(x+1)}+\frac{(x+9)}{2(x^{2}+9)}} d x \\ & =-\frac{1}{2} \log |x+1|+\frac{1}{2} \int \frac{x}{x^{2}+9} d x+\frac{9}{2} \int \frac{1}{x^{2}+9} d x \\ & =-\frac{1}{2} \log |x+1|+\frac{1}{4} \int \frac{2 x}{x^{2}+9} d x+\frac{9}{2} \int \frac{1}{x^{2}+9} d x \\ & =-\frac{1}{2} \log |x+1|+\frac{1}{4} \log |x^{2}+9|+\frac{9}{2} \cdot \frac{1}{3} \tan ^{-1} \frac{x}{3} \\ & =-\frac{1}{2} \log |x+1|+\frac{1}{4} \log (x^{2}+9)+\frac{3}{2} \tan ^{-1} \frac{x}{3}+C \end{aligned} \end{aligned} $

Solution

$ \frac{\sin x}{\sin (x-a)} $

Let $x-a=t d x=d t$

$ \begin{aligned} \int \frac{\sin x}{\sin (x-a)} d x & =\int \frac{\sin (t+a)}{\sin t} d t \\ & =\int \frac{\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} $

Solution

$ \begin{aligned} \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} & =\frac{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 \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x & =\int x^{2} d x=\frac{x^{3}}{3}+C \end{aligned} $

Solution

$\frac{\cos x}{\sqrt{4-\sin ^{2} x}}$

Let $\sin x=t \cos x d x=d t$

$\Rightarrow \int \frac{\cos x}{\sqrt{4-\sin ^{2} x}} d x=\int \frac{d t}{\sqrt{(2)^{2}-(t)^{2}}}$

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

Solution

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

Solution

$\frac{1}{\cos (x+a) \cos (x+b)}$

Multiplying and dividing by $\sin (a-b)$, we obtain

$ \begin{aligned} & \frac{1}{\sin (a-b)}[\frac{\sin (a-b)}{\cos (x+a) \cos (x+b)}] \\ & =\frac{1}{\sin (a-b)}[\frac{\sin [(x+a)-(x+b)]}{\cos (x+a) \cos (x+b)}] \\ & =\frac{1}{\sin (a-b)}[\frac{\sin (x+a) \cdot \cos (x+b)-\cos (x+a) \sin (x+b)}{\cos (x+a) \cos (x+b)}] \\ & =\frac{1}{\sin (a-b)}[\frac{\sin (x+a)}{\cos (x+a)}-\frac{\sin (x+b)}{\cos (x+b)}] \\ & =\frac{1}{\sin (a-b)}[\tan (x+a)-\tan (x+b)] \end{aligned} $

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

Solution

$\frac{x^{3}}{\sqrt{1-x^{8}}}$

Let $x^{4}=t 4 x^{3} d x=d t$ $\Rightarrow \int \frac{x^{3}}{\sqrt{1-x^{8}}} d x=\frac{1}{4} \int \frac{d t}{\sqrt{1-t^{2}}}$

$ \begin{aligned} & =\frac{1}{4} \sin ^{-1} t+C \\ & =\frac{1}{4} \sin ^{-1}(x^{4})+C \end{aligned} $

Solution

$\frac{e^{x}}{(1+e^{x})(2+e^{x})}$

Let $e^{x}=t e^{x} d x=d t$

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

$=\int[\frac{1}{(t+1)}-\frac{1}{(t+2)}] d t$

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

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

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

Solution

$\therefore \frac{1}{(x^{2}+1)(x^{2}+4)}=\frac{A x+B}{(x^{2}+1)}+\frac{C x+D}{(x^{2}+4)}$

$\Rightarrow 1=(A x+B)(x^{2}+4)+(C x+D)(x^{2}+1)$

$\Rightarrow 1=A x^{3}+4 A x+B x^{2}+4 B+C x^{3}+C x+D x^{2}+D$

Equating the coefficients of $x^{3}, x^{2}, x$, and constant term, we obtain

$A+C=0$

$B+D=0$

$4 A+C=0$

$4 B+D=1$

On solving these equations, we obtain

$A=0, B=\frac{1}{3}, C=0$, and $D=-\frac{1}{3}$

From equation (1), we obtain

$ \begin{aligned} & \frac{1}{(x^{2}+1)(x^{2}+4)}=\frac{1}{3(x^{2}+1)}-\frac{1}{3(x^{2}+4)} \\ & \begin{aligned} \int \frac{1}{(x^{2}+1)(x^{2}+4)} d x & =\frac{1}{3} \int \frac{1}{x^{2}+1} d x-\frac{1}{3} \int \frac{1}{x^{2}+4} d x \\ & =\frac{1}{3} \tan ^{-1} x-\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2}+C \\ & =\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C \end{aligned} \end{aligned} $

Solution

$\cos ^{3} x e^{\log \sin x}=\cos ^{3} x \times \sin x$

Let $\cos x=t -\sin x d x=d t$ $\Rightarrow \int \cos ^{3} x e^{\log \sin x} d x=\int \cos ^{3} x \sin x d x$

$ \begin{aligned} & =-\int t^{3} \cdot d t \\ & =-\frac{t^{4}}{4}+C \\ & =-\frac{\cos ^{4} x}{4}+C \end{aligned} $

Solution

$e^{3 \log x}(x^{4}+1)^{-1}=e^{\log x^{3}}(x^{4}+1)^{-1}=\frac{x^{3}}{(x^{4}+1)}$

Let $x^{4}+1=t \Rightarrow 4 x^{3} d x=d t$

$\Rightarrow \int e^{3 \log x}(x^{4}+1)^{-1} d x=\int \frac{x^{3}}{(x^{4}+1)} d x$

$ \begin{aligned} & =\frac{1}{4} \int \frac{d t}{t} \\ & =\frac{1}{4} \log |t|+C \\ & =\frac{1}{4} \log |x^{4}+1|+C \\ & =\frac{1}{4} \log (x^{4}+1)+C \end{aligned} $

Solution

$ f^{\prime}(a x+b)[f(a x+b)]^{n} $

Let $f(a x+b)=t \Rightarrow a f^{\prime}(a x+b) d x=d t$

$\Rightarrow \int f^{\prime}(a x+b)[f(a x+b)]^{n} d x=\frac{1}{a} \int t^{n} d t$

$ \begin{aligned} & =\frac{1}{a}[\frac{t^{n+1}}{n+1}] \\ & =\frac{1}{a(n+1)}(f(a x+b))^{n+1}+C \end{aligned} $

Solution

$ \begin{aligned} \frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}} & =\frac{1}{\sqrt{\sin ^{3} x(\sin x \cos \alpha+\cos x \sin \alpha)}} \\ & =\frac{1}{\sqrt{\sin ^{4} x \cos \alpha+\sin ^{3} x \cos x \sin \alpha}} \\ & =\frac{1}{\sin ^{2} x \sqrt{\cos \alpha+\cot x \sin \alpha}} \\ & =\frac{cosec^{2}x}{\sqrt{\cos \alpha+\cot x \sin \alpha}} \end{aligned} $

Let $\cos \alpha+\cot x \sin \alpha=t \Rightarrow-cosec^{2} x \sin \alpha d x=d t$

$ \begin{aligned} \therefore \int \frac{1}{\sin ^{3} x \sin (x+\alpha)} d x & =\int \frac{cosec^{2} x}{\sqrt{\cos \alpha+\cot x \sin \alpha}} d x \\ & =\frac{-1}{\sin \alpha} \int \frac{d t}{\sqrt{t}} \\ & =\frac{-1}{\sin \alpha}[2 \sqrt{t}]+C \\ & =\frac{-1}{\sin \alpha}[2 \sqrt{\cos \alpha+\cot x \sin \alpha}]+C \\ & =\frac{-2}{\sin \alpha} \sqrt{\cos \alpha+\frac{\cos x \sin \alpha}{\sin x}}+C \\ & =\frac{-2}{\sin \alpha} \sqrt{\frac{\sin x \cos \alpha+\cos x \sin \alpha}{\sin x}}+C \\ & =-\frac{2}{\sin \alpha} \sqrt{\frac{\sin (x+\alpha)}{\sin x}+C} \end{aligned} $

Solution

$ I=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x $

Let $x=\cos ^{2} \theta \Rightarrow d x=-2 \sin \theta \cos \theta d \theta$

$ I=\int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin \theta \cos \theta) d \theta $

$ \begin{aligned} & =-\int \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \sin 2 \theta d \theta \\ & =-\int \tan \frac{\theta}{2} \cdot 2 \sin \theta \cos \theta d \theta \\ & =-2 \int \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}(2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}) \cos \theta d \theta \end{aligned} $

$=-4 \int \sin ^{2} \frac{\theta}{2} \cos \theta d \theta$

$=-4 \int \sin ^{2} \frac{\theta}{2} \cdot(2 \cos ^{2} \frac{\theta}{2}-1) d \theta$

$=-4 \int(2 \sin ^{2} \frac{\theta}{2} \cos ^{2} \frac{\theta}{2}-\sin ^{2} \frac{\theta}{2}) d \theta$

$=-8 \int \sin ^{2} \frac{\theta}{2} \cdot \cos ^{2} \frac{\theta}{2} d \theta+4 \int \sin ^{2} \frac{\theta}{2} d \theta$

$=-2 \int \sin ^{2} \theta d \theta+4 \int \sin ^{2} \frac{\theta}{2} d \theta$

$=-2 \int(\frac{1-\cos 2 \theta}{2}) d \theta+4 \int \frac{1-\cos \theta}{2} d \theta$

$=-2[\frac{\theta}{2}-\frac{\sin 2 \theta}{4}]+4[\frac{\theta}{2}-\frac{\sin \theta}{2}]+C$

$=-\theta+\frac{\sin 2 \theta}{2}+2 \theta-2 \sin \theta+C$

$=\theta+\frac{\sin 2 \theta}{2}-2 \sin \theta+C$

$=\theta+\frac{2 \sin \theta \cos \theta}{2}-2 \sin \theta+C$

$=\theta+\sqrt{1-\cos ^{2} \theta} \cdot \cos \theta-2 \sqrt{1-\cos ^{2} \theta}+C$

$=\cos ^{-1} \sqrt{x}+\sqrt{1-x} \cdot \sqrt{x}-2 \sqrt{1-x}+C$

$=-2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x(1-x)}+C$

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

Solution

$ \begin{aligned} I & =\int(\frac{2+\sin 2 x}{1+\cos 2 x}) e^{x} \\ & =\int(\frac{2+2 \sin x \cos x}{2 \cos ^{2} x}) e^{x} \\ & =\int(\frac{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 $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)}$

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

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

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

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$A+C=1$

$3 A+B+2 C=1$

$2 A+2 B+C=1$

On solving these equations, we obtain

$A=-2, B=1$, and $C=3$

From equation (1), we obtain

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

Solution

$I=\tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$

Let $x=\cos \theta \Rightarrow d x=-\sin \theta d \theta$

$I=\int \tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-\sin \theta d \theta)$

$=-\int \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \sin \theta d \theta$

$=-\int \tan ^{-1} \tan \frac{\theta}{2} \cdot \sin \theta d \theta$

$=-\frac{1}{2} \int \theta \cdot \sin \theta d \theta$

$=-\frac{1}{2}[\theta \cdot(-\cos \theta)-\int 1 \cdot(-\cos \theta) d \theta]$

$=-\frac{1}{2}[-\theta \cos \theta+\sin \theta]$

$=+\frac{1}{2} \theta \cos \theta-\frac{1}{2} \sin \theta$

$=\frac{1}{2} \cos ^{-1} x \cdot x-\frac{1}{2} \sqrt{1-x^{2}}+C$

$=\frac{x}{2} \cos ^{-1} x-\frac{1}{2} \sqrt{1-x^{2}}+C$

$=\frac{1}{2}(x \cos ^{-1} x-\sqrt{1-x^{2}})+C$

Solution

$ \begin{aligned} \frac{\sqrt{x^{2}+1}[\log (x^{2}+1)-2 \log x]}{x^{4}} & =\frac{\sqrt{x^{2}+1}}{x^{4}}[\log (x^{2}+1)-\log x^{2}] \\ & =\frac{\sqrt{x^{2}+1}}{x^{4}}[\log (\frac{x^{2}+1}{x^{2}})] \\ & =\frac{\sqrt{x^{2}+1}}{x^{4}} \log (1+\frac{1}{x^{2}}) \\ & =\frac{1}{x^{3}} \sqrt{\frac{x^{2}+1}{x^{2}}} \log (1+\frac{1}{x^{2}}) \\ & =\frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} \log (1+\frac{1}{x^{2}}) \end{aligned} $

Let $1+\frac{1}{x^{2}}=t \Rightarrow \frac{-2}{x^{3}} d x=d t$

$ \begin{aligned} \therefore I & =\int \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} \log (1+\frac{1}{x^{2}}) d x \\ & =-\frac{1}{2} \int \sqrt{t} \log t d t \\ & =-\frac{1}{2} \int t^{\frac{1}{2}} \cdot \log t d t \end{aligned} $

Integrating by parts, we obtain

$ \begin{aligned} I & =-\frac{1}{2}[\log t \cdot \int t^{\frac{1}{2}} d t-{(\frac{d}{d t} \log t) \int t^{\frac{1}{2}} d t} d t] \\ & =-\frac{1}{2}[\log t \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}}-\int \frac{1}{t} \cdot \frac{t^{\frac{3}{2}}}{\frac{3}{2}} d t] \\ & =-\frac{1}{2}[\frac{2}{3} t^{\frac{3}{2}} \log t-\frac{2}{3} \int t^{\frac{1}{2}} d t] \\ & =-\frac{1}{2}[\frac{2}{3} t^{\frac{3}{2}} \log t-\frac{4}{9} t^{\frac{3}{2}}] \\ & =-\frac{1}{3} t^{\frac{3}{2}} \log t+\frac{2}{9} t^{\frac{3}{2}} \\ & =-\frac{1}{3} t^{\frac{3}{2}}[\log t-\frac{2}{3}] \\ & =-\frac{1}{3}(1+\frac{1}{x^{2}})^{\frac{3}{2}}[\log (1+\frac{1}{x^{2}})-\frac{2}{3}]+C \end{aligned} $

Solution

$ \begin{aligned} I & =\int _{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-\sin x}{1-\cos x}) d x \\ & =\int _{\frac{\pi}{2}}^{\pi} e^{x}(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}}) d x \\ & =\int _{\frac{\pi}{2}}^{\pi} e^{x}(\frac{cosec^{2} \frac{x}{2}}{2}-\cot \frac{x}{2}) d x \end{aligned} $

Let $f(x)=-\cot \frac{x}{2}$

$\Rightarrow f^{\prime}(x)=-(-\frac{1}{2} cosec^{2} \frac{x}{2})=\frac{1}{2} cosec^{2} \frac{x}{2}$

$\therefore I=\int _{\frac{\pi}{2}}^{\pi} e^{x}(f(x)+f^{\prime}(x)] d x$

$=[e^{x} \cdot f(x) d x] _{\frac{\pi}{2}}^{x}$

$=-[e^{x} \cdot \cot \frac{x}{2}] _{\frac{\pi}{2}}^{\pi}$

$=-[e^{\pi} \times \cot \frac{\pi}{2}-e^{\frac{\pi}{2}} \times \cot \frac{\pi}{4}]$

$=-[e^{\pi} \times 0-e^{\frac{\pi}{2}} \times 1]$

$=e^{\frac{\pi}{2}}$

Solution

Let $I=\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$

$ \begin{aligned} & \Rightarrow I=\int_0^{\frac{\pi}{4}} \frac{\frac{(\sin x \cos x)}{\cos ^{4} x}}{\frac{(\cos ^{4} x+\sin ^{4} x)}{\cos ^{4} x}} d x \\ & \Rightarrow I=\int_0^{\frac{\pi}{4}} \frac{\tan x \sec ^{2} x}{1+\tan ^{4} x} d x \end{aligned} $

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

$ \begin{aligned} \therefore I & =\frac{1}{2} \int_0^{1} \frac{d t}{1+t^{2}} \\ & =\frac{1}{2}[\tan ^{-1} t]_0^{1} \\ & =\frac{1}{2}[\tan ^{-1} 1-\tan ^{-1} 0] \\ & =\frac{1}{2}[\frac{\pi}{4}] \\ & =\frac{\pi}{8} \end{aligned} $

$ \text{ When } x=0, t=0 \text{ and when } x=\frac{\pi}{4}, t=1 $

Solution

Let $I=\int_0^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4(1-\cos ^{2} x)} d x$

$\Rightarrow I=\int_0^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4-4 \cos ^{2} x} d x$

$\Rightarrow I=\frac{-1}{3} \int_0^{\frac{\pi}{2}} \frac{4-3 \cos ^{2} x-4}{4-3 \cos ^{2} x} d x$

$\Rightarrow I=\frac{-1}{3} \int_0^{\frac{\pi}{2}} \frac{4-3 \cos ^{2} x}{4-3 \cos ^{2} x} d x+\frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4}{4-3 \cos ^{2} x} d x$

$\Rightarrow I=\frac{-1}{3} \int_0^{\frac{\pi}{2}} 1 d x+\frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4 \sec ^{2} x}{4 \sec ^{2} x-3} d x$

$\Rightarrow I=\frac{-1}{3}[x]_0^{\frac{\pi}{2}}+\frac{1}{3} \int_0^{\frac{\pi}{2}} \frac{4 \sec ^{2} x}{4(1+\tan ^{2} x)-3} d x$

$\Rightarrow I=-\frac{\pi}{6}+\frac{2}{3} \int_0^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x$

Consider, $\int_0^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x$

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

When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\infty$

$\Rightarrow \int_0^{\frac{\pi}{2}} \frac{2 \sec ^{2} x}{1+4 \tan ^{2} x} d x=\int_0^{\infty} \frac{d t}{1+t^{2}}$

$ \begin{aligned} & =[\tan ^{-1} t]_0^{\infty} \\ & =[\tan ^{-1}(\infty)-\tan ^{-1}(0)] \\ & =\frac{\pi}{2} \end{aligned} $

Therefore, from (1),we obtain

$I=-\frac{\pi}{6}+\frac{2}{3}[\frac{\pi}{2}]=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$

Solution

Let $I=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$

$\Rightarrow I=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x+\cos x)}{\sqrt{-(-\sin 2 x)}} d x$

$\Rightarrow I=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{-(-1+1-2 \sin x \cos x)}} d x$

$\Rightarrow I=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x+\cos x)}{\sqrt{1-(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x)}} d x$

$\Rightarrow I=\int _{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x+\cos x) d x}{\sqrt{1-(\sin x-\cos x)^{2}}}$

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

When $x=\frac{\pi}{6}, t=(\frac{1-\sqrt{3}}{2})$ and when $\quad x=\frac{\pi}{3}, t=(\frac{\sqrt{3}-1}{2})$

$I=\int _{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{d t}{\sqrt{1-t^{2}}}$

$\Rightarrow I=\int _{-(\frac{\sqrt{3}-1}{2})}^{.\frac{\sqrt{3}-1}{2})} \frac{d t}{\sqrt{1-t^{2}}}$

As $\frac{1}{\sqrt{1-(-t)^{2}}}=\frac{1}{\sqrt{1-t^{2}}}$, therefore, $\frac{1}{\sqrt{1-t^{2}}}$ 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^{\sqrt{3}-1} \frac{d t}{\sqrt{1-t^{2}}} \\ & =[2 \sin ^{-1} t]_0^{\frac{\sqrt{3}-1}{2}} \\ & =2 \sin ^{-1}(\frac{\sqrt{3}-1}{2}) \end{aligned} $

Solution

$ \begin{aligned} & \text{ Let } I=\int_0^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}} \\ & \begin{aligned} I & =\int_0^{1} \frac{1}{(\sqrt{1+x}-\sqrt{x})} \times \frac{(\sqrt{1+x}+\sqrt{x})}{(\sqrt{1+x}+\sqrt{x})} d x \\ & =\int_0^{1} \frac{\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 \\ & =[\frac{2}{3}(1+x)^{\frac{3}{2}}]_0^{1}+[\frac{2}{3}(x)^{\frac{3}{2}}]_0^{1} \\ & =\frac{2}{3}[(2)^{\frac{3}{2}}-1]+\frac{2}{3}[1] \\ & =\frac{2}{3}(2)^{\frac{3}{2}} \\ & =\frac{2 \cdot 2 \sqrt{2}}{3} \\ & =\frac{4 \sqrt{2}}{3} \end{aligned} \\ & \text{ (1) } \end{aligned} $

Solution

Let $I=\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$

Also, let $\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$

When $x=0, t=-1$ and when $x=\frac{\pi}{4}, t=0$

$\Rightarrow(\sin x-\cos x)^{2}=t^{2}$

$\Rightarrow \sin ^{2} x+\cos ^{2} x-2 \sin x \cos x=t^{2}$

$\Rightarrow 1-\sin 2 x=t^{2}$

$\Rightarrow \sin 2 x=1-t^{2}$

$\therefore I=\int _{-1}^{0} \frac{d t}{9+16(1-t^{2})}$

$=\int _{-1}^{0} \frac{d t}{9+16-16 t^{2}}$

$=\int _{-1}^{0} \frac{d t}{25-16 t^{2}}=\int _{-1}^{0} \frac{d t}{(5)^{2}-(4 t)^{2}}$

$=\frac{1}{4}[\frac{1}{2(5)} \log |\frac{5+4 t}{5-4 t}|] _{-1}^{0}$

$=\frac{1}{40}[\log (1)-\log |\frac{1}{9}|]$

$=\frac{1}{40} \log 9$

Solution

Let $I=\int_0^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x=\int_0^{\frac{\pi}{2}} 2 \sin x \cos x \tan ^{-1}(\sin x) d x$

Also, let $\sin x=t \Rightarrow \cos x d x=d t$

When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=1$

$\Rightarrow I=2 \int_0^{1} t \tan ^{-1}(t) d t$

Consider $\int t \cdot \tan ^{-1} t d t=\tan ^{-1} t \cdot \int t d t-\int{\frac{d}{d t}(\tan ^{-1} t) \int t d t} d t$

$ \begin{aligned} & =\tan ^{-1} t \cdot \frac{t^{2}}{2}-\int \frac{1}{1+t^{2}} \cdot \frac{t^{2}}{2} d t \\ & =\frac{t^{2} \tan ^{-1} t}{2}-\frac{1}{2} \int \frac{t^{2}+1-1}{1+t^{2}} d t \\ & =\frac{t^{2} \tan ^{-1} t}{2}-\frac{1}{2} \int 1 d t+\frac{1}{2} \int \frac{1}{1+t^{2}} d t \\ & =\frac{t^{2} \tan ^{-1} t}{2}-\frac{1}{2} \cdot t+\frac{1}{2} \tan ^{-1} t \end{aligned} $

$ \begin{aligned} \Rightarrow \int_0^{1} t \cdot \tan ^{-1} t d t & =[\frac{t^{2} \cdot \tan ^{-1} t}{2}-\frac{t}{2}+\frac{1}{2} \tan ^{-1} t]_0^{1} \\ & =\frac{1}{2}[\frac{\pi}{4}-1+\frac{\pi}{4}] \\ & =\frac{1}{2}[\frac{\pi}{2}-1]=\frac{\pi}{4}-\frac{1}{2} \end{aligned} $

From equation (1), we obtain

$I=2[\frac{\pi}{4}-\frac{1}{2}]=\frac{\pi}{2}-1$

Solution

Let $I=\int_1^{4}[|x-1|+|x-2|+|x-3|] d x$

$\Rightarrow I=\int^{4}|x-1| d x+\int_1^{4}|x-2| d x+\int^{4}|x-3| d x$

$I=I_1+I_2+I_3$

where, $I_1=\int_1^{4}|x-1| d x, I_2=\int_1^{4}|x-2| d x$, and $I_3=\int_1^{4}|x-3| d x$

$I_1=\int_1^{4}|x-1| d x$

$(x-1) \geq 0$ for $1 \leq x \leq 4$

$\therefore I_1=\int_1^{4}(x-1) d x$

$\Rightarrow I_1=[\frac{x^{2}}{x}-x]_1^{4}$

$\Rightarrow I_1=[8-4-\frac{1}{2}+1]=\frac{9}{2}$

$I_2=\int^{4}|x-2| d x$

$x-2 \geq 0$ for $2 \leq x \leq 4$ and $x-2 \leq 0$ for $1 \leq x \leq 2$

$\therefore I_2=\int^{2}(2-x) d x+\int_2^{4}(x-2) d x$

$\Rightarrow I_2=[2 x-\frac{x^{2}}{2}]_1^{2}+[\frac{x^{2}}{2}-2 x]_2^{4}$

$\Rightarrow I_2=[4-2-2+\frac{1}{2}]+[8-8-2+4]$

$\Rightarrow I_2=\frac{1}{2}+2=\frac{5}{2}$ $I_3=\int^{4}|x-3| d x$

$x-3 \geq 0$ for $3 \leq x \leq 4$ and $x-3 \leq 0$ for $1 \leq x \leq 3$

$\therefore I_3=\int_1^{3}(3-x) d x+\int_3^{4}(x-3) d x$

$\Rightarrow I_3=[3 x-\frac{x^{2}}{2}]_1^{3}+[\frac{x^{2}}{2}-3 x]_3^{4}$

$\Rightarrow I_3=[9-\frac{9}{2}-3+\frac{1}{2}]+[8-12-\frac{9}{2}+9]$

$\Rightarrow I_3=[6-4]+[\frac{1}{2}]=\frac{5}{2}$

From equations (1), (2), (3), and (4), we obtain

$I=\frac{9}{2}+\frac{5}{2}+\frac{5}{2}=\frac{19}{2}$

Solution

Let $I=\int_1^{3} \frac{d x}{x^{2}(x+1)}$

Also, let $\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}$

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

$\Rightarrow 1=A x^{2}+A x+B x+B+C x^{2}$

Equating the coefficients of $x^{2}, x$, and constant term, we obtain

$A+C=0$

$A+B=0$

$B=1$

On solving these equations, we obtain

$A=-1, C=1$, and $B=1$ $\therefore \frac{1}{x^{2}(x+1)}=\frac{-1}{x}+\frac{1}{x^{2}}+\frac{1}{(x+1)}$

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

$ =[-\log x-\frac{1}{x}+\log (x+1)]_1^{3} $

$ =[\log (\frac{x+1}{x})-\frac{1}{x}]_1^{3} $

$=\log (\frac{4}{3})-\frac{1}{3}-\log (\frac{2}{1})+1$

$=\log 4-\log 3-\log 2+\frac{2}{3}$

$=\log 2-\log 3+\frac{2}{3}$

$=\log (\frac{2}{3})+\frac{2}{3}$

Hence, the given result is proved.

Solution

Let $I=\int_0^{1} x e^{x} d x$

Integrating by parts, we obtain

$ \begin{aligned} I & =x \int_0^{1} e^{x} d x-\int_0^{1}{(\frac{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} $

Hence, the given result is proved.

Solution

Let $I=\int _{-1}^{1} x^{17} \cos ^{4} x d x$

Also, let $f(x)=x^{17} \cos ^{4} x$

$\Rightarrow f(-x)=(-x)^{17} \cos ^{4}(-x)=-x^{17} \cos ^{4} x=-f(x)$

Therefore, $f(x)$ is an odd function.

It is known that if $f(x)$ is an odd function, then $\int _{-a}^{a} f(x) d x=0$

$\therefore I=\int _{-1}^{1} x^{17} \cos ^{4} x d x=0$

Hence, the given result is proved.

Solution

Let $I=\int_0^{\frac{\pi}{2}} \sin ^{3} x d x$

$I=\int_0^{\frac{\pi}{2}} \sin ^{2} x \cdot \sin x d x$

$=\int_0^{\frac{\pi}{2}}(1-\cos ^{2} x) \sin x d x$

$=\int_0^{\frac{\pi}{2}} \sin x d x-\int_0^{\frac{\pi}{2}} \cos ^{2} x \cdot \sin x d x$

$=[-\cos x]_0^{\frac{\pi}{2}}+[\frac{\cos ^{3} x}{3}]_0^{\frac{\pi}{2}}$

$=1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}$

Hence, the given result is proved.

Solution

Let $I=\int_0^{\frac{\pi}{4}} 2 \tan ^{3} x d x$

$I=2 \int_0^{\frac{\pi}{4}} \tan ^{2} x \tan x d x=2 \int_0^{\frac{\pi}{4}}(\sec ^{2} x-1) \tan x d x$

$=2 \int_0^{\frac{\pi}{4}} \sec ^{2} x \tan x d x-2 \int_0^{\frac{\pi}{4}} \tan x d x$

$=2[\frac{\tan ^{2} x}{2}]_0^{\frac{\pi}{4}}+2[\log \cos x]_0^{\frac{\pi}{4}}$

$=1+2[\log \cos \frac{\pi}{4}-\log \cos 0]$

$=1+2[\log \frac{1}{\sqrt{2}}-\log 1]$

$=1-\log 2-\log 1=1-\log 2$

Hence, the given result is proved.

Solution

Let $I=\int_0^{1} \sin ^{-1} x d x$

$\Rightarrow I=\int_0^{1} \sin ^{-1} x \cdot 1 \cdot d x$

Integrating by parts, we obtain

$ \begin{aligned} I & =[\sin ^{-1} x \cdot x]_0^{1}-\int_0^{1} \frac{1}{\sqrt{1-x^{2}}} \cdot x d x \\ & =[x \sin ^{-1} x]_0^{1}+\frac{1}{2} \int_0^{1} \frac{(-2 x)}{\sqrt{1-x^{2}}} d x \end{aligned} $

Let $1-x^{2}=t \square-2 x d x=d t$

When $x=0, t=1$ and when $x=1, t=0$

$ \begin{aligned} I & =[x \sin ^{-1} x]_0^{1}+\frac{1}{2} \int_0^{0} \frac{d t}{\sqrt{t}} \\ & =[x \sin ^{-1} x]_0^{1}+\frac{1}{2}[2 \sqrt{t}]_1^{0} \\ & =\sin ^{-1}(1)+[-\sqrt{1}] \\ & =\frac{\pi}{2}-1 \end{aligned} $

Hence, the given result is proved.

Solution

Let $I=\int \frac{d x}{e^{x}+e^{-x}} d x=\int \frac{e^{x}}{e^{2 x}+1} d x$

Also, let $e^{x}=t \Rightarrow e^{x} d x=d t$

$ \begin{aligned} \therefore I & =\int \frac{d t}{1+t^{2}} \\ & =\tan ^{-1} t+C \\ & =\tan ^{-1}(e^{x})+C \end{aligned} $

Hence, the correct Answer is A.

Solution

Let $I=\frac{\cos 2 x}{(\cos x+\sin x)^{2}}$

$ \begin{aligned} I & =\int \frac{\cos ^{2} x-\sin ^{2} x}{(\cos x+\sin x)^{2}} d x \\ & =\int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^{2}} d x \\ & =\int \frac{\cos x-\sin x}{\cos +\sin x} d x \end{aligned} $

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

$ \begin{aligned} \therefore I & =\int \frac{d t}{t} \\ & =\log |t|+C \\ & =\log |\cos x+\sin x|+C \end{aligned} $

Hence, the correct Answer is B.

Solution

Let $I=\int_a^{b} x f(x) d x$

$ \begin{aligned} & I=\int_a^{b}(a+b-x) f(a+b-x) d x \\ & (\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 \\ & \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=(\frac{a+b}{2}) \int_a^{b} f(x) d x \end{aligned} $

Hence, the correct Answer is D.