\displaystyle \textbf{Question 1: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/4}\sqrt{1+\sin 2x}\,dx \qquad \text{(ii)}\ \int_{0}^{\pi/4}\sqrt{1-\sin 2x}\,dx \hspace{2.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{}\ \text{(i) Let }I=\int_{0}^{\pi/4}\sqrt{1+\sin 2x}\,dx.
\displaystyle I=\int_{0}^{\pi/4}\sqrt{\sin^{2}x+\cos^{2}x+2\sin x\cos x}\,dx
\displaystyle I=\int_{0}^{\pi/4}\sqrt{(\sin x+\cos x)^{2}}\,dx
\displaystyle I=\int_{0}^{\pi/4}|\cos x+\sin x|\,dx
\displaystyle =\int_{0}^{\pi/4}(\cos x+\sin x)\,dx
\displaystyle =[\sin x-\cos x]_{0}^{\pi/4}
\displaystyle =(\sin\frac{\pi}{4}-\cos\frac{\pi}{4})-(\sin 0-\cos 0)
\displaystyle =\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)-(0-1)=1

\displaystyle \text{(ii) Let }I=\int_{0}^{\pi/4}\sqrt{1-\sin 2x}\,dx.
\displaystyle I=\int_{0}^{\pi/4}\sqrt{\sin^{2}x+\cos^{2}x-2\sin x\cos x}\,dx
\displaystyle I=\int_{0}^{\pi/4}\sqrt{(\cos x-\sin x)^{2}}\,dx
\displaystyle I=\int_{0}^{\pi/4}|\cos x-\sin x|\,dx
\displaystyle \because 0<x<\frac{\pi}{4}\Rightarrow \cos x>\sin x\Rightarrow \cos x-\sin x>0
\displaystyle \Rightarrow |\cos x-\sin x|=\cos x-\sin x
\displaystyle I=\int_{0}^{\pi/4}(\cos x-\sin x)\,dx
\displaystyle =[\sin x+\cos x]_{0}^{\pi/4}
\displaystyle =\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(0+1)=\sqrt{2}-1

\displaystyle \textbf{Question 2: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{1}^{2}\frac{5x^{2}}{x^{2}+4x+3}\,dx \qquad \text{(ii)}\ \int_{1}^{3}\frac{1}{x^{2}(x+1)}\,dx \hspace{2.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }I=\int_{1}^{2}\frac{5x^{2}}{x^{2}+4x+3}\,dx.
\displaystyle I=5\int_{1}^{2}\frac{x^{2}}{x^{2}+4x+3}\,dx =5\int_{1}^{2}\left(1-\frac{4x+3}{x^{2}+4x+3}\right)\,dx
\displaystyle =5\int_{1}^{2}1\,dx-5\int_{1}^{2}\frac{4x+3}{x^{2}+4x+3}\,dx
\displaystyle =5\int_{1}^{2}1\,dx-5\int_{1}^{2}\frac{2(2x+4)-5}{x^{2}+4x+3}\,dx
\displaystyle =5\int_{1}^{2}1\,dx-10\int_{1}^{2}\frac{2x+4}{x^{2}+4x+3}\,dx +25\int_{1}^{2}\frac{1}{x^{2}+4x+3}\,dx
\displaystyle =5[x]_{1}^{2}-10[\log(x^{2}+4x+3)]_{1}^{2} +\frac{25}{2}\left[\log\frac{x+2-1}{x+2+1}\right]_{1}^{2}
\displaystyle =5(2-1)-10(\log 15-\log 8) +\frac{25}{2}\left(\log\frac{3}{5}-\log\frac{2}{4}\right)
\displaystyle =5-10\log\frac{15}{8}+\frac{25}{2}\log\frac{6}{5}

\displaystyle \text{(ii) Let }\frac{1}{x^{2}(x+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}}
\displaystyle \text{Then }1=A x^{2}+(Bx+C)(x+1)
\displaystyle \text{Putting }x=0,x=-1\text{ respectively, we get: }C=1\text{ and }A=1
\displaystyle \text{Equating coefficients of }x^{2}\text{ on both sides, we get: }0=A+B \Rightarrow B=-1
\displaystyle \text{Substituting values of }A,B,C\text{ we obtain}
\displaystyle \frac{1}{x^{2}(x+1)}=\frac{1}{x+1}-\frac{1}{x}+\frac{1}{x^{2}}
\displaystyle \int_{1}^{3}\frac{1}{x^{2}(x+1)}\,dx =\int_{1}^{3}\left(\frac{1}{x+1}-\frac{1}{x}+\frac{1}{x^{2}}\right)dx
\displaystyle =[\log|x+1|-\log|x|-\frac{1}{x}]_{1}^{3}
\displaystyle =(\log 4-\log 3-\frac{1}{3})-(\log 2-\log 1-1)
\displaystyle =\log\frac{4}{3}-\frac{1}{3}+1=\log\frac{2}{3}+\frac{2}{3}

\displaystyle \textbf{Question 3: } \text{Evaluate:}\qquad \int_{\pi/4}^{\pi/2}\cos 2x\log\sin x\,dx \qquad [\text{CBSE 2003}]
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int_{\pi/4}^{\pi/2}\cos 2x\log\sin x\,dx.
\displaystyle I=\left[\frac{1}{2}(\log\sin x)\sin 2x\right]_{\pi/4}^{\pi/2} -\int_{\pi/4}^{\pi/2}\frac{1}{2}\cot x\sin 2x\,dx
\displaystyle \Rightarrow I=\left[0-\frac{1}{2}\log\frac{1}{\sqrt{2}}\right] -\int_{\pi/4}^{\pi/2}\cos^{2}x\,dx
\displaystyle \Rightarrow I=\frac{1}{4}\log 2-\frac{1}{2} \int_{\pi/4}^{\pi/2}(1+\cos 2x)\,dx
\displaystyle \Rightarrow I=\frac{1}{4}\log 2-\frac{1}{2} \left[x+\frac{1}{2}\sin 2x\right]_{\pi/4}^{\pi/2}
\displaystyle =\frac{1}{4}\log 2-\frac{1}{2} \left(\frac{\pi}{2}+0\right)-\left(\frac{\pi}{4}+\frac{1}{2}\right)
\displaystyle =\frac{1}{4}\log 2-\frac{\pi}{8}-\frac{1}{4}

\displaystyle \textbf{Question 4: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/2}\frac{\cos\theta}{(1+\sin\theta)(2+\sin\theta)}\,d\theta \hspace{4.0cm}[\text{CBSE 2004}]
\displaystyle \text{(ii)}\ \int_{0}^{1/2}\frac{1}{(1+x^{2})\sqrt{1-x^{2}}}\,dx \hspace{5.0cm}[\text{CBSE 2015}]
\displaystyle \text{(iii)}\ \int_{0}^{\pi/4}\frac{1}{\cos^{3}x\sqrt{2\sin 2x}}\,dx \hspace{5.0cm}[\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }\sin\theta=t.\text{ Then }d(\sin\theta)=dt \Rightarrow \cos\theta\,d\theta=dt
\displaystyle \text{Also, }\theta=0\Rightarrow t=0 \text{ and }\theta=\frac{\pi}{2}\Rightarrow t=1
\displaystyle I=\int_{0}^{\pi/2}\frac{\cos\theta}{(1+\sin\theta)(2+\sin\theta)}\,d\theta =\int_{0}^{1}\frac{1}{(1+t)(2+t)}\,dt
\displaystyle =\int_{0}^{1}\left(\frac{1}{1+t}-\frac{1}{2+t}\right)dt \qquad [\text{By using partial fractions}]
\displaystyle I=[\log(1+t)-\log(2+t)]_{0}^{1}
\displaystyle =( \log 2-\log 1)-(\log 3-\log 2) =\log 2-\log 3+\log 2=2\log 2-\log 3=\log\frac{4}{3}

\displaystyle \text{(ii) Let }I=\int_{0}^{1/2}\frac{1}{(1+x^{2})\sqrt{1-x^{2}}}\,dx. \text{ Let }x=\sin\theta
\displaystyle \text{Then }dx=\cos\theta\,d\theta \text{ and }x=0\Rightarrow\theta=0,\ x=\frac{1}{2}\Rightarrow\theta=\frac{\pi}{6}
\displaystyle I=\int_{0}^{\pi/6}\frac{\cos\theta}{(1+\sin^{2}\theta)\sqrt{1-\sin^{2}\theta}}\,d\theta =\int_{0}^{\pi/6}\frac{1}{1+\sin^{2}\theta}\,d\theta
\displaystyle I=\int_{0}^{\pi/6}\frac{\sec^{2}\theta}{\sec^{2}\theta+\tan^{2}\theta}\,d\theta \qquad [\text{Dividing N and D by }\cos^{2}\theta]
\displaystyle =\int_{0}^{\pi/6}\frac{\sec^{2}\theta}{1+2\tan^{2}\theta}\,d\theta
\displaystyle \text{Let }\tan\theta=t.\text{ Then }\sec^{2}\theta\,d\theta=dt
\displaystyle \theta=0\Rightarrow t=0,\ \theta=\frac{\pi}{6}\Rightarrow t=\frac{1}{\sqrt{3}}
\displaystyle I=\int_{0}^{1/\sqrt{3}}\frac{1}{1+2t^{2}}\,dt =\frac{1}{2}\int_{0}^{1/\sqrt{3}}\frac{1}{(1/\sqrt{2})^{2}+t^{2}}\,dt
\displaystyle =\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}t)\Big|_{0}^{1/\sqrt{3}} =\frac{1}{\sqrt{2}}\tan^{-1}\sqrt{\frac{2}{3}}

\displaystyle \text{(iii) Let }I=\int_{0}^{\pi/4}\frac{1}{\cos^{3}x\sqrt{2\sin 2x}}\,dx
\displaystyle I=\int_{0}^{\pi/4}\frac{1}{\cos^{3}x\sqrt{4\sin x\cos x}}\,dx =\frac{1}{2}\int_{0}^{\pi/4}\sin^{-1/2}x\cos^{-7/2}x\,dx
\displaystyle \text{Divide numerator and denominator by }\cos^{4}x
\displaystyle I=\frac{1}{2}\int_{0}^{\pi/4}\frac{\sec^{4}x}{\sqrt{\tan x}}\,dx =\frac{1}{2}\int_{0}^{\pi/4}\frac{1+\tan^{2}x}{\sqrt{\tan x}}\sec^{2}x\,dx
\displaystyle \text{Let }\tan x=t.\text{ Then }dt=\sec^{2}x\,dx
\displaystyle x=0\Rightarrow t=0,\ x=\frac{\pi}{4}\Rightarrow t=1
\displaystyle I=\frac{1}{2}\int_{0}^{1}\frac{1+t^{2}}{\sqrt{t}}\,dt =\frac{1}{2}\int_{0}^{1}(t^{-1/2}+t^{3/2})\,dt
\displaystyle =\frac{1}{2}\left[2t^{1/2}+\frac{2}{5}t^{5/2}\right]_{0}^{1} =\frac{1}{2}\times\frac{12}{5}=\frac{6}{5}

\displaystyle \textbf{Question 5: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{1/\sqrt{2}}\frac{\sin^{-1}x}{(1-x^{2})^{3/2}}\,dx \qquad [\text{CBSE 2007}]
\displaystyle \text{(ii)}\ \int_{0}^{1}\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)dx \qquad [\text{CBSE 2002}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }\sin^{-1}x=\theta,\ x=\sin\theta. \text{ Then }dx=\cos\theta\,d\theta
\displaystyle x=0\Rightarrow\theta=0,\ x=\frac{1}{\sqrt{2}}\Rightarrow\theta=\frac{\pi}{4}
\displaystyle I=\int_{0}^{\pi/4}\frac{\theta}{\cos^{3}\theta}\cos\theta\,d\theta =\int_{0}^{\pi/4}\theta\sec^{2}\theta\,d\theta
\displaystyle I=[\theta\tan\theta]_{0}^{\pi/4}-\int_{0}^{\pi/4}\tan\theta\,d\theta
\displaystyle =\frac{\pi}{4}-\frac{1}{2}\log 2

\displaystyle \text{(ii) Let }I=\int_{0}^{1}\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)dx
\displaystyle \because \sin^{-1}\frac{2x}{1+x^{2}}=2\tan^{-1}x
\displaystyle I=2\int_{0}^{1}\tan^{-1}x\,dx
\displaystyle =2\left[x\tan^{-1}x-\frac{1}{2}\log(1+x^{2})\right]_{0}^{1}
\displaystyle =2\left(\frac{\pi}{4}-\frac{1}{2}\log 2\right) =\frac{\pi}{2}-\log 2

\displaystyle \textbf{Question 6: } \text{Evaluate:}\ \int_{0}^{\pi/4}\tan^{3}x\,dx \qquad [\text{CBSE 2004}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi/4}\tan^{3}x\,dx
\displaystyle I=\int_{0}^{\pi/4}(\sec^{2}x-1)\tan x\,dx
\displaystyle \text{Let }\tan x=t,\ dt=\sec^{2}x\,dx
\displaystyle x=0\Rightarrow t=0,\ x=\frac{\pi}{4}\Rightarrow t=1
\displaystyle I=\int_{0}^{1}t^{2}\,dt-\int_{0}^{1}\tan x\,dx
\displaystyle =\frac{1}{2}-\log\sqrt{2}+1 =\frac{1}{2}-\log 2

\displaystyle \textbf{Question 7: } \text{Evaluate:}
\displaystyle \text{}\ \int_{0}^{\pi}\frac{1}{5+4\cos x}\,dx \qquad [\text{CBSE 2005}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi}\frac{1}{5+4\cos x}\,dx
\displaystyle =\int_{0}^{\pi}\frac{1+\tan^{2}\frac{x}{2}} {5(1+\tan^{2}\frac{x}{2})+4(1-\tan^{2}\frac{x}{2})}\,dx
\displaystyle =\int_{0}^{\pi}\frac{\sec^{2}\frac{x}{2}} {9+\tan^{2}\frac{x}{2}}\,dx
\displaystyle \text{Let }\tan\frac{x}{2}=t,\ dx=\frac{2dt}{\sec^{2}\frac{x}{2}}
\displaystyle I=2\int_{0}^{\infty}\frac{dt}{9+t^{2}} =\frac{2}{3}\tan^{-1}\infty-\tan^{-1}0=\frac{\pi}{3}

\displaystyle \textbf{Question 8: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/2}\frac{\cos x}{1+\cos x+\sin x}\,dx \qquad [\text{CBSE 2014}]
\displaystyle \text{(ii)}\ \int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx \qquad [\text{CBSE 2010}]
\displaystyle \text{(iii)}\ \int_{0}^{\pi/2}\sqrt{\tan x+\cot x}\,dx \qquad [\text{CBSE 2002, 2003}]
\displaystyle \text{Answer:}

\displaystyle \text{(i) We have }I=\int_{0}^{\pi/2}\frac{\cos x}{1+\cos x+\sin x}\,dx
\displaystyle =\int_{0}^{\pi/2}\frac{\cos x}{(1+\cos x)+\sin x}\,dx
\displaystyle =\int_{0}^{\pi/2}\frac{\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}} {2\cos^{2}\frac{x}{2}+2\sin^{2}\frac{x}{2}}\,dx
\displaystyle =\int_{0}^{\pi/2}\frac{1-\tan^{2}\frac{x}{2}}{2+2\tan^{2}\frac{x}{2}}\,dx
\displaystyle \text{[Dividing numerator and denominator by }\cos^{2}\frac{x}{2}]
\displaystyle =\frac{1}{2}\int_{0}^{\pi/2}\frac{(1-\tan\frac{x}{2})(1+\tan\frac{x}{2})} {1+\tan^{2}\frac{x}{2}}\,dx
\displaystyle =\frac{1}{2}\int_{0}^{\pi/2}(1-\tan\frac{x}{2})\,dx
\displaystyle =\frac{1}{2}\left[x+2\log\cos\frac{x}{2}\right]_{0}^{\pi/2}
\displaystyle =\frac{1}{2}\left(\frac{\pi}{2}+2\log\frac{1}{\sqrt{2}}\right) =\frac{1}{2}\left(\frac{\pi}{2}-\log 2\right)

\displaystyle \text{(ii) Let }I=\int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx
\displaystyle \sin 2x=1-(\sin x-\cos x)^{2}
\displaystyle I=\int_{0}^{\pi/4}\frac{\sin x+\cos x} {25-16(\sin x-\cos x)^{2}}\,dx
\displaystyle \text{Let }\sin x-\cos x=t,\ dt=(\cos x+\sin x)dx
\displaystyle x=0\Rightarrow t=-1,\ x=\frac{\pi}{4}\Rightarrow t=0
\displaystyle I=\int_{-1}^{0}\frac{dt}{25-16t^{2}} =\frac{1}{40}\left[\log\frac{5+4t}{5-4t}\right]_{-1}^{0}
\displaystyle =\frac{1}{40}(\log 1-\log\frac{9}{1}) =-\frac{1}{40}\log 9

\displaystyle \text{(iii) Let }I=\int_{0}^{\pi/2}\sqrt{\tan x+\cot x}\,dx
\displaystyle =\sqrt{2}\int_{0}^{\pi/2}\frac{\sin x+\cos x} {\sqrt{\sin x\cos x}}\,dx
\displaystyle \text{Let }\sin x-\cos x=t,\ dt=(\cos x+\sin x)dx
\displaystyle x=0\Rightarrow t=-1,\ x=\frac{\pi}{2}\Rightarrow t=1
\displaystyle I=\sqrt{2}\int_{-1}^{1}\frac{dt}{\sqrt{1-t^{2}}} =\sqrt{2}\left[\sin^{-1}t\right]_{-1}^{1}=\sqrt{2}\pi

\displaystyle \textbf{Question 9: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/2}\frac{\sin 2x}{\sin^{4}x+\cos^{4}x}\,dx \qquad [\text{CBSE 2003C}]
\displaystyle \text{(ii)}\ \int_{0}^{\pi/4}\frac{\sin 2x}{\cos^{4}x+\sin^{4}x}\,dx \qquad [\text{CBSE 2013}]
\displaystyle \text{Answer:}

\displaystyle \text{(i) Let }I=\int_{0}^{\pi/2}\frac{\sin 2x}{\sin^{4}x+\cos^{4}x}\,dx =\int_{0}^{\pi/2}\frac{2\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle \text{Dividing numerator and denominator by }\cos^{4}x\text{ we obtain}
\displaystyle I=\int_{0}^{\pi/2}\frac{2\tan x\sec^{2}x}{\tan^{4}x+1}\,dx
\displaystyle \text{Let }t=\tan^{2}x.\text{ Then }dt=d(\tan^{2}x)=2\tan x\sec^{2}x\,dx
\displaystyle \text{Also, }x=0\Rightarrow t=\tan^{2}0=0,\ x=\frac{\pi}{2} \Rightarrow t=\tan^{2}\frac{\pi}{2}=\infty
\displaystyle I=\int_{0}^{\infty}\frac{dt}{t^{2}+1} =\left[\tan^{-1}t\right]_{0}^{\infty} =\tan^{-1}\infty-\tan^{-1}0=\frac{\pi}{2}

\displaystyle \text{(ii) Let }I=\int_{0}^{\pi/4}\frac{\sin 2x}{\cos^{4}x+\sin^{4}x}\,dx =\int_{0}^{\pi/4}\frac{2\sin x\cos x}{\cos^{4}x+\sin^{4}x}\,dx
\displaystyle \text{Dividing numerator and denominator by }\cos^{4}x\text{ we get}
\displaystyle I=\int_{0}^{\pi/4}\frac{2\tan x\sec^{2}x}{1+\tan^{4}x}\,dx
\displaystyle \text{Let }t=\tan^{2}x.\text{ Then }dt=2\tan x\sec^{2}x\,dx
\displaystyle \text{Also, }x=0\Rightarrow t=\tan^{2}0=0,\ x=\frac{\pi}{4} \Rightarrow t=\tan^{2}\frac{\pi}{4}=1
\displaystyle I=\int_{0}^{1}\frac{dt}{1+t^{2}} =\left[\tan^{-1}t\right]_{0}^{1} =\tan^{-1}1-\tan^{-1}0=\frac{\pi}{4}

\displaystyle \textbf{Question 10: } \text{Evaluate:}\ \int_{0}^{\pi/2} \frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x}\,dx\qquad [\text{CBSE 2012}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi/2} \frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x}\,dx
\displaystyle I=\int_{0}^{\pi/2} \frac{\cos^{2}x}{\cos^{2}x+4(1-\cos^{2}x)}\,dx
\displaystyle =\int_{0}^{\pi/2}\frac{\cos^{2}x}{4-3\cos^{2}x}\,dx
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2}\frac{3\cos^{2}x}{4-3\cos^{2}x}\,dx
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2} \frac{(4-3\cos^{2}x)-4}{4-3\cos^{2}x}\,dx
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2} \left(1-\frac{4}{4-3\cos^{2}x}\right)dx
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2}1\,dx +\frac{4}{3}\int_{0}^{\pi/2}\frac{1}{4-3\cos^{2}x}\,dx
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2}1\,dx +\frac{4}{3}\int_{0}^{\pi/2} \frac{\sec^{2}x}{4(1+\tan^{2}x)-3}\,dx
\displaystyle \text{[Dividing N and D by }\cos^{2}x]
\displaystyle =-\frac{1}{3}\int_{0}^{\pi/2}1\,dx +\frac{4}{3}\int_{0}^{\pi/2}\frac{\sec^{2}x}{1+4\tan^{2}x}\,dx
\displaystyle =-\frac{1}{3}\left[x\right]_{0}^{\pi/2} +\frac{4}{3}\int_{0}^{\infty}\frac{dt}{1+4t^{2}},\ \text{where }t=\tan x
\displaystyle =-\frac{\pi}{6}+\frac{4}{3}\times\frac{1}{2} \left[\tan^{-1}(2t)\right]_{0}^{\infty}
\displaystyle =-\frac{\pi}{6}+\frac{2}{3} \left(\frac{\pi}{2}-0\right)=\frac{\pi}{6}

\displaystyle \textbf{Question 11: } \text{Evaluate:}
\displaystyle \text{}\ \int_{-1}^{2}|x^{3}-x|\,dx \qquad [\text{CBSE 2012, 2013, 2016}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-1}^{2}|x^{3}-x|\,dx\text{ and }f(x)=x^{3}-x
\displaystyle \text{Clearly, }f(x)=x^{3}-x=x(x-1)(x+1)
\displaystyle \text{The signs of }f(x)\text{ for different values of }x
\displaystyle \text{We observe that }f(x)>0\text{ for }x\in(-1,0)\cup(1,2) \text{ and }f(x)<0\text{ for }x\in(0,1)
\displaystyle |f(x)|=\begin{cases} x^{3}-x, & x\in(-1,0)\cup(1,2)\\ -(x^{3}-x), & x\in(0,1) \end{cases}
\displaystyle \Rightarrow |x^{3}-x|= \begin{cases} x^{3}-x, & x\in(-1,0)\cup(1,2)\\ -(x^{3}-x), & x\in(0,1) \end{cases}
\displaystyle I=\int_{-1}^{0}|x^{3}-x|\,dx+\int_{0}^{1}|x^{3}-x|\,dx +\int_{1}^{2}|x^{3}-x|\,dx\qquad [\text{Using additive property}]
\displaystyle I=\int_{-1}^{0}(x^{3}-x)\,dx-\int_{0}^{1}(x^{3}-x)\,dx +\int_{1}^{2}(x^{3}-x)\,dx
\displaystyle I=\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0} -\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{0}^{1} +\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{1}^{2}
\displaystyle I=-\left(\frac{1}{4}-\frac{1}{2}\right) -\left(\frac{1}{4}-\frac{1}{2}\right) +\left(\frac{16}{4}-\frac{4}{2}\right)-\left(\frac{1}{4}-\frac{1}{2}\right)
\displaystyle I=\frac{3}{4}+(4-2)=\frac{11}{4}

\displaystyle \textbf{Question 12: } \text{Evaluate:}
\displaystyle \text{}\ \int_{\pi/6}^{\pi/3}\frac{1}{1+\sqrt{\cot x}}\,dx \qquad [\text{CBSE 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{\pi/6}^{\pi/3}\frac{1}{1+\sqrt{\cot x}}\,dx =\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx
\displaystyle \text{Then,}
\displaystyle I=\int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)}+ \sqrt{\cos(\frac{\pi}{2}-x)}}\,dx
\displaystyle \Rightarrow I=\int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx =\int_{\pi/6}^{\pi/3}1\,dx
\displaystyle =\left[x\right]_{\pi/6}^{\pi/3} =\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}
\displaystyle I=\frac{\pi}{12}

\displaystyle \textbf{Question 13 } \text{Prove that:}\ \int_{0}^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx=\frac{\pi}{4} \qquad [\text{CBSE 2002C, 2007}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi/2} \frac{\sin x}{\sin x+\cos x}\,dx
\displaystyle \text{Then,}
\displaystyle I=\int_{0}^{\pi/2} \frac{\sin(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x)+\cos(\frac{\pi}{2}-x)}\,dx \qquad [\text{Using }\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \frac{\cos x}{\cos x+\sin x}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi/2}\frac{\sin x}{\sin x+\cos x}\,dx +\int_{0}^{\pi/2}\frac{\cos x}{\sin x+\cos x}\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2} \frac{\sin x+\cos x}{\sin x+\cos x}\,dx =\int_{0}^{\pi/2}1\,dx
\displaystyle =\left[x\right]_{0}^{\pi/2} =\frac{\pi}{2}-0=\frac{\pi}{2}
\displaystyle \Rightarrow I=\frac{\pi}{4}

\displaystyle \textbf{Question 14: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/2}\log\tan x\,dx\qquad [\text{CBSE 2007}]
\displaystyle \text{(ii)}\ \int_{0}^{\pi/4}\log(1+\tan x)\,dx \qquad [\text{CBSE 2002C,03,04,11,13}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }I=\int_{0}^{\pi/2}\log\tan x\,dx
\displaystyle \text{Then,}
\displaystyle I=\int_{0}^{\pi/2}\log\tan\left(\frac{\pi}{2}-x\right)dx \qquad [\text{Using }\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi/2}\log\cot x\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi/2}(\log\tan x+\log\cot x)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2}\log(\tan x\cot x)\,dx =\int_{0}^{\pi/2}\log 1\,dx=\int_{0}^{\pi/2}0\,dx=0
\displaystyle \Rightarrow I=0

\displaystyle \text{(ii) Let }I=\int_{0}^{\pi/4}\log(1+\tan x)\,dx
\displaystyle \text{Then,}
\displaystyle I=\int_{0}^{\pi/4}\log\!\left(1+\tan\left(\frac{\pi}{4}-x\right)\right)dx \qquad [\text{Using }\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi/4} \log\!\left(1+\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}\right)dx
\displaystyle =\int_{0}^{\pi/4}\log\!\left(1+\frac{1-\tan x}{1+\tan x}\right)dx
\displaystyle =\int_{0}^{\pi/4}\log\!\left(\frac{2}{1+\tan x}\right)dx
\displaystyle =\int_{0}^{\pi/4}\log 2\,dx-\int_{0}^{\pi/4}\log(1+\tan x)\,dx
\displaystyle \Rightarrow I=(\log 2)[x]_{0}^{\pi/4}-I
\displaystyle \Rightarrow 2I=\frac{\pi}{4}\log 2\Rightarrow I=\frac{\pi}{8}\log 2

\displaystyle \textbf{Question 15: } \text{Evaluate:}
\displaystyle \text{}\ \int_{0}^{\pi/2}(2\log\sin x-\log\sin 2x)\,dx \qquad [\text{CBSE 2009}]
\displaystyle \text{Answer:}
\displaystyle \text{We have }I=\int_{0}^{\pi/2}(2\log\sin x-\log\sin 2x)\,dx
\displaystyle =\int_{0}^{\pi/2}\{2\log\sin x-\log(2\sin x\cos x)\}dx
\displaystyle =\int_{0}^{\pi/2}\{2\log\sin x-\log 2-\log\sin x-\log\cos x\}dx
\displaystyle =\int_{0}^{\pi/2}\log\sin x\,dx-\int_{0}^{\pi/2}\log 2\,dx -\int_{0}^{\pi/2}\log\cos x\,dx
\displaystyle =\int_{0}^{\pi/2}\log\sin x\,dx-(\log 2)\int_{0}^{\pi/2}1\,dx -\int_{0}^{\pi/2}\log\cos\left(\frac{\pi}{2}-x\right)dx
\displaystyle =\int_{0}^{\pi/2}\log\sin x\,dx-(\log 2)[x]_{0}^{\pi/2} -\int_{0}^{\pi/2}\log\sin x\,dx
\displaystyle =-(\log 2)\frac{\pi}{2}=-\frac{\pi}{2}\log 2

\displaystyle \textbf{Question 16: } \text{Evaluate:}\ \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}\,dx\qquad [\text{CBSE 2009}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}\,dx
\displaystyle \text{Then,}
\displaystyle I=\int_{0}^{\pi} \frac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)}+e^{-\cos(\pi-x)}}\,dx \qquad [\text{Using }\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi}1\,dx=\pi\Rightarrow I=\frac{\pi}{2}

\displaystyle \textbf{Question 17: } \text{Prove that:}\ \int_{0}^{2a}f(x)\,dx =\int_{0}^{2a}f(2a-x)\,dx\qquad [\text{CBSE 2002C}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{2a}f(x)\,dx
\displaystyle \text{Let }2a-x=t.\text{ Then }d(2a-x)=dt\Rightarrow -dx=dt \Rightarrow dx=-dt
\displaystyle \text{Also, }x=0\Rightarrow t=2a-0=2a\text{ and }x=a\Rightarrow t=2a-a=a
\displaystyle I=\int_{2a}^{a}f(2a-t)(-dt)=-\int_{2a}^{a}f(2a-t)\,dt
\displaystyle \Rightarrow I=\int_{0}^{2a}f(2a-t)\,dt
\displaystyle \Rightarrow I=\int_{0}^{2a}f(2a-x)\,dx
\displaystyle \text{Hence, }\int_{0}^{2a}f(x)\,dx=\int_{0}^{2a}f(2a-x)\,dx

\displaystyle \textbf{Question 18: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi/2}\frac{\sin^{2}x}{\sin x+\cos x}\,dx \qquad [\text{CBSE 2002, 2003, 2016}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }I=\int_{0}^{\pi/2} \frac{\sin^{2}x}{\sin x+\cos x}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \frac{\sin^{2}(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x)+\cos(\frac{\pi}{2}-x)}\,dx \\ \qquad [\text{Using }\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \frac{\cos^{2}x}{\cos x+\sin x}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi/2} \frac{\sin^{2}x}{\sin x+\cos x}\,dx +\int_{0}^{\pi/2}\frac{\cos^{2}x}{\sin x+\cos x}\,dx
\displaystyle =\int_{0}^{\pi/2}\frac{1}{\sin x+\cos x}\,dx
\displaystyle 2I=\int_{0}^{\pi/2} \frac{1} {\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}} +\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}}\,dx
\displaystyle =\int_{0}^{\pi/2} \frac{1+\tan^{2}\frac{x}{2}} {2\tan\frac{x}{2}+1-\tan^{2}\frac{x}{2}}\,dx
\displaystyle =\int_{0}^{\pi/2} \frac{\sec^{2}\frac{x}{2}} {2\tan\frac{x}{2}+1-\tan^{2}\frac{x}{2}}\,dx
\displaystyle \text{Let }\tan\frac{x}{2}=t. \text{ Then }d\!\left(\tan\frac{x}{2}\right)=dt \Rightarrow \sec^{2}\frac{x}{2}\frac{1}{2}dx=dt \Rightarrow \sec^{2}\frac{x}{2}dx=2dt
\displaystyle \text{Also, }x=0\Rightarrow t=\tan 0=0, \ x=\frac{\pi}{2}\Rightarrow t=\tan\frac{\pi}{4}=1
\displaystyle 2I=\int_{0}^{1}\frac{2dt}{2t+1-t^{2}} =2\int_{0}^{1}\frac{dt}{(\sqrt{2})^{2}-(t-1)^{2}}
\displaystyle =2\times\frac{1}{2\sqrt{2}} \left[\log\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}\right]_{0}^{1}
\displaystyle =\frac{1}{\sqrt{2}} \left\{\log\frac{\sqrt{2}}{\sqrt{2}} -\log\frac{\sqrt{2}-1}{\sqrt{2}+1}\right\}
\displaystyle =-\frac{1}{\sqrt{2}} \log\frac{(\sqrt{2}-1)^{2}}{(\sqrt{2}+1)(\sqrt{2}-1)} =-\frac{2}{\sqrt{2}}\log(\sqrt{2}-1)
\displaystyle I=-\frac{1}{\sqrt{2}}\log(\sqrt{2}-1)

\displaystyle \textbf{Question 19: } \text{Evaluate:}\ \int_{0}^{1}\cot^{-1}(1-x+x^{2})\,dx \qquad [\text{CBSE 2008}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{1}\cot^{-1}(1-x+x^{2})\,dx
\displaystyle I=\int_{0}^{1}\tan^{-1}\!\left(\frac{1}{1-x+x^{2}}\right)dx \qquad [\because\ \cot^{-1}x=\tan^{-1}\frac{1}{x},\ x>0]
\displaystyle I=\int_{0}^{1}\tan^{-1}\!\left(\frac{1}{1-x(1-x)}\right)dx
\displaystyle I=\int_{0}^{1}\tan^{-1}\!\left(\frac{x+(1-x)}{1-x(1-x)}\right)dx
\displaystyle I=\int_{0}^{1}\{\tan^{-1}x+\tan^{-1}(1-x)\}\,dx
\displaystyle I=\int_{0}^{1}\tan^{-1}x\,dx+\int_{0}^{1}\tan^{-1}(1-x)\,dx
\displaystyle I=\int_{0}^{1}\tan^{-1}x\,dx+\int_{0}^{1}\tan^{-1}\{1-(1-x)\}\,dx \\ \qquad [\because\ \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle I=\int_{0}^{1}\tan^{-1}x\,dx+\int_{0}^{1}\tan^{-1}x\,dx
\displaystyle I=2\int_{0}^{1}\tan^{-1}x\,dx
\displaystyle I=2\int_{0}^{1}\tan^{-1}x\cdot 1\,dx
\displaystyle I=2\left[x\tan^{-1}x\right]_{0}^{1} -2\int_{0}^{1}\frac{x}{1+x^{2}}\,dx
\displaystyle I=2\left[x\tan^{-1}x\right]_{0}^{1} -\int_{0}^{1}\frac{2x}{1+x^{2}}\,dx
\displaystyle I=2\left[x\tan^{-1}x\right]_{0}^{1} -\left[\log(1+x^{2})\right]_{0}^{1}
\displaystyle I=2\left(\frac{\pi}{4}-0\right)-(\log 2-\log 1) =\frac{\pi}{2}-\log 2

\displaystyle \textbf{Question 20: } \text{Evaluate:}\ \int_{0}^{2\pi}\frac{1}{1+e^{\sin x}}\,dx \qquad [\text{CBSE 2013}]

\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{2\pi}\frac{1}{1+e^{\sin x}}\,dx
\displaystyle I=\int_{0}^{\pi}\left\{\frac{1}{1+e^{\sin x}} +\frac{1}{1+e^{\sin(2\pi-x)}}\right\}dx \\ \qquad [\text{Using }\int_{0}^{2a}f(x)\,dx=\int_{0}^{a}\{f(x)+f(2a-x)\}dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi}\left\{\frac{1}{1+e^{\sin x}} +\frac{1}{1+e^{-\sin x}}\right\}dx
\displaystyle \Rightarrow I=\int_{0}^{\pi}\left\{\frac{1}{1+e^{\sin x}} +\frac{e^{\sin x}}{1+e^{\sin x}}\right\}dx=\int_{0}^{\pi}1\,dx
\displaystyle =\left[x\right]_{0}^{\pi}=\pi

\displaystyle \textbf{Question 21: } \text{Evaluate:}
\displaystyle \text{(i)}\ \int_{0}^{\pi}\frac{x}{1+\sin x}\,dx \qquad [\text{CBSE 2001, 2004, 2010, 2012}]
\displaystyle \text{(ii)}\ \int_{0}^{\pi}\frac{x\tan x}{\sec x+\tan x}\,dx \qquad [\text{CBSE 2008, 2010, 2014}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }I=\int_{0}^{\pi}\frac{x}{1+\sin x}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin(\pi-x)}\,dx \\ \qquad [\because\ \int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin x}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi}\frac{x+\pi-x}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi}\frac{1}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi/2} \left\{\frac{1}{1+\sin x}+\frac{1}{1+\sin(\pi-x)}\right\}dx \\ \qquad [\because\ \int_{0}^{2a}f(x)\,dx=\int_{0}^{a}\{f(x)+f(2a-x)\}dx]
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2}\frac{1}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} \frac{1-\sin x}{1-\sin^{2}x}\,dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} (\sec^{2}x-\tan x\sec x)\,dx
\displaystyle \Rightarrow 2I=2\pi[\tan x-\sec x]_{0}^{\pi/2} =2\pi\left[\frac{\sin x-1}{\cos x}\right]_{0}^{\pi/2}
\displaystyle \Rightarrow 2I=2\pi\left[\frac{\sin^{2}x-1} {\cos x(\sin x+1)}\right]_{0}^{\pi/2} =2\pi\left[\frac{-\cos x}{1+\sin x}\right]_{0}^{\pi/2} =2\pi(0+1)=2\pi
\displaystyle \Rightarrow I=\pi

\displaystyle \text{(ii) Let }I=\int_{0}^{\pi} \frac{x\tan x}{\sec x+\tan x}\,dx=\int_{0}^{\pi}\frac{x\sin x}{1+\sin x}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi} \frac{(\pi-x)\sin(\pi-x)}{1+\sin(\pi-x)}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi} \frac{(\pi-x)\sin x}{1+\sin x}\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi}\frac{\pi\sin x}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi/2} \left\{\frac{\sin x}{1+\sin x} +\frac{\sin(\pi-x)}{1+\sin(\pi-x)}\right\}dx \\ \qquad [\because\ \int_{0}^{2a}f(x)\,dx=\int_{0}^{a}\{f(x)+f(2a-x)\}dx]
\displaystyle \Rightarrow 2I=\pi\int_{0}^{\pi/2} \left(\frac{\sin x}{1+\sin x}+\frac{\sin x}{1+\sin x}\right)dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} \frac{\sin x}{1+\sin x}\,dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} \frac{\sin x(1-\sin x)}{1-\sin^{2}x}\,dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} \frac{\sin x-\sin^{2}x}{\cos^{2}x}\,dx
\displaystyle \Rightarrow 2I=2\pi\int_{0}^{\pi/2} (\tan x\sec x-\tan^{2}x)\,dx
\displaystyle \Rightarrow I=\pi\int_{0}^{\pi/2} (\tan x\sec x-\tan^{2}x)\,dx
\displaystyle \Rightarrow I=\pi\int_{0}^{\pi/2} (\tan x\sec x-(\sec^{2}x-1))\,dx
\displaystyle \Rightarrow I=\pi\int_{0}^{\pi/2} (\sec x\tan x-\sec^{2}x+1)\,dx
\displaystyle \Rightarrow I=\pi[\sec x-\tan x+x]_{0}^{\pi/2}
\displaystyle \Rightarrow I=\pi\left[\frac{1-\sin x}{\cos x}+x\right]_{0}^{\pi/2} \\ =\pi\left[\frac{\cos x}{1+\sin x}+x\right]_{0}^{\pi/2} \\ =\pi\left(0+\frac{\pi}{2}-(1+0)\right) \\ =\frac{\pi}{2}(\pi-2)

\displaystyle \textbf{Question 22: } \text{Evaluate:}\ \int_{0}^{\pi/2} \frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx \qquad [\text{CBSE 2010, 2011, 2014}]

\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi/2} \frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle I=\int_{0}^{\pi/2} \frac{(\frac{\pi}{2}-x)\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2}-x)} {\sin^{4}(\frac{\pi}{2}-x)+\cos^{4}(\frac{\pi}{2}-x)}\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \frac{(\frac{\pi}{2}-x)\sin x\cos x}{\cos^{4}x+\sin^{4}x}\,dx
\displaystyle \Rightarrow I=\frac{\pi}{2}\int_{0}^{\pi/2} \frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx -\int_{0}^{\pi/2} \frac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle \Rightarrow I=\frac{\pi}{2}\int_{0}^{\pi/2} \frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx-I
\displaystyle \Rightarrow 2I=\frac{\pi}{2}\int_{0}^{\pi/2} \frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx
\displaystyle \Rightarrow 2I=\frac{\pi}{2}\int_{0}^{\pi/2} \frac{\tan x\sec^{2}x}{1+\tan^{4}x}\,dx \qquad [\text{Dividing numerator and denominator by }\cos^{4}x]
\displaystyle \Rightarrow 2I=\frac{\pi}{4}\int_{0}^{\pi/2} \frac{2\tan x\sec^{2}x}{1+(\tan^{2}x)^{2}}\,dx
\displaystyle \text{Let }t=\tan^{2}x.\text{ Then }dt=2\tan x\sec^{2}x\,dx
\displaystyle \text{Also, }x=0\Rightarrow t=\tan^{2}0=0,\ x=\frac{\pi}{2} \Rightarrow t=\tan^{2}\frac{\pi}{2}=\infty
\displaystyle 2I=\frac{\pi}{4}\int_{0}^{\infty}\frac{dt}{1+t^{2}}
\displaystyle \Rightarrow 2I=\frac{\pi}{4}[\tan^{-1}t]_{0}^{\infty} =\frac{\pi}{4}\left(\frac{\pi}{2}-0\right)
\displaystyle \Rightarrow I=\frac{\pi^{2}}{16}

\displaystyle \textbf{Question 23: } \text{Evaluate:}\ \int_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{x}}\,dx\qquad [\text{CBSE 2015}]

\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-\pi/2}^{\pi/2} \frac{\cos x}{1+e^{x}}\,dx
\displaystyle I=\int_{-\pi/2}^{\pi/2} \left\{\frac{\cos x}{1+e^{x}}+\frac{\cos(-x)}{1+e^{-x}}\right\}dx \\ \qquad [\because\ \int_{-a}^{a}f(x)\,dx=\int_{0}^{a}\{f(x)+f(-x)\}dx]
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \left\{\frac{\cos x}{1+e^{x}}+\frac{\cos x}{1+e^{-x}}\right\}dx
\displaystyle \Rightarrow I=\int_{0}^{\pi/2} \left\{\frac{1}{1+e^{x}}+\frac{e^{x}}{1+e^{x}}\right\}\cos x\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi/2}\cos x\,dx =[\sin x]_{0}^{\pi/2}=1

\displaystyle \textbf{Question 24: } \text{Evaluate:}
\displaystyle \text{}\ \int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}\,dx \qquad [\text{CBSE 2002, 2008}]

\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}\,dx
\displaystyle I=\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}\times\frac{a-x}{a-x}}\,dx =\int_{-a}^{a}\frac{a-x}{\sqrt{a^{2}-x^{2}}}\,dx
\displaystyle \Rightarrow I=\int_{-a}^{a}\frac{a}{\sqrt{a^{2}-x^{2}}}\,dx -\int_{-a}^{a}\frac{x}{\sqrt{a^{2}-x^{2}}}\,dx=aI_{1}-I_{2}
\displaystyle \text{where }I_{1}=\int_{-a}^{a}\frac{1}{\sqrt{a^{2}-x^{2}}}\,dx \text{ and }I_{2}=\int_{-a}^{a}\frac{x}{\sqrt{a^{2}-x^{2}}}\,dx
\displaystyle \text{Let }f(x)=\frac{1}{\sqrt{a^{2}-x^{2}}} \text{ and }g(x)=\frac{x}{\sqrt{a^{2}-x^{2}}}
\displaystyle f(-x)=\frac{1}{\sqrt{a^{2}-(-x)^{2}}} =\frac{1}{\sqrt{a^{2}-x^{2}}}=f(x)
\displaystyle g(-x)=\frac{-x}{\sqrt{a^{2}-(-x)^{2}}} =-\frac{x}{\sqrt{a^{2}-x^{2}}}=-g(x)
\displaystyle \Rightarrow f(x)\text{ is even and }g(x)\text{ is odd}
\displaystyle \Rightarrow I_{1}=2\int_{0}^{a}\frac{1}{\sqrt{a^{2}-x^{2}}}\,dx \text{ and }I_{2}=0
\displaystyle I_{1}=2\left[\sin^{-1}\frac{x}{a}\right]_{0}^{a} \text{ and }I_{2}=0
\displaystyle I_{1}=2\left(\sin^{-1}1-\sin^{-1}0\right) =2\left(\frac{\pi}{2}-0\right)=\pi
\displaystyle \text{Substituting values of }I_{1}\text{ and }I_{2}\text{ in (i), we get}
\displaystyle I=a\pi-0=a\pi

\displaystyle \textbf{Question 25: } \text{Evaluate:}\ \int_{-\pi}^{\pi}(\cos ax-\sin bx)^{2}\,dx \qquad [\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-\pi}^{\pi}(\cos ax-\sin bx)^{2}\,dx
\displaystyle I=\int_{-\pi}^{\pi}(\cos^{2}ax+\sin^{2}bx-2\cos ax\sin bx)\,dx
\displaystyle \Rightarrow I=\int_{-\pi}^{\pi}\cos^{2}ax\,dx +\int_{-\pi}^{\pi}\sin^{2}bx\,dx-2\int_{-\pi}^{\pi}\cos ax\sin bx\,dx
\displaystyle \Rightarrow I=2\int_{0}^{\pi}\cos^{2}ax\,dx +2\int_{0}^{\pi}\sin^{2}bx\,dx-2\times 0 \qquad [\because\ \cos^{2}ax,\sin^{2}bx\text{ even; }\cos ax\sin bx\text{ odd}]
\displaystyle \Rightarrow I=\int_{0}^{\pi}(1+\cos 2ax)\,dx +\int_{0}^{\pi}(1-\cos 2bx)\,dx
\displaystyle \Rightarrow I=\left[x+\frac{1}{2a}\sin 2ax\right]_{0}^{\pi} +\left[x-\frac{1}{2b}\sin 2bx\right]_{0}^{\pi}
\displaystyle \Rightarrow I=\left(\pi+\frac{1}{2a}\sin 2a\pi-0\right) +\left(\pi-\frac{1}{2b}\sin 2b\pi-0\right)
\displaystyle \Rightarrow I=2\pi+\frac{1}{2a}\sin 2a\pi-\frac{1}{2b}\sin 2b\pi
\displaystyle I= \begin{cases} 2\pi,& a,b\text{ integers}\\[4pt] 2\pi+\frac{1}{2a}\sin 2a\pi-\frac{1}{2b}\sin 2b\pi,& a,b\text{ not integers} \end{cases}

\displaystyle \textbf{Question 26: } \text{Prove that:}\ \int_{0}^{\pi/2}\log\sin x\,dx=\int_{0}^{\pi/2}\log\cos x\,dx =-\frac{\pi}{2}\log 2\qquad [\text{CBSE 2008}]
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi/2}\log\sin x\,dx
\displaystyle \text{Then, }I=\int_{0}^{\pi/2}\log\sin\left(\frac{\pi}{2}-x\right)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\pi/2}\log\cos x\,dx
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle 2I=\int_{0}^{\pi/2}\log\sin x\,dx +\int_{0}^{\pi/2}\log\cos x\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2} (\log\sin x+\log\cos x)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2}\log(\sin x\cos x)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2} \log\left(\frac{2\sin x\cos x}{2}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2}\log\left(\frac{\sin 2x}{2}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2}\log\sin 2x\,dx -\int_{0}^{\pi/2}\log 2\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{\pi/2}\log\sin 2x\,dx -\frac{\pi}{2}\log 2
\displaystyle \text{Let }I_{1}=\int_{0}^{\pi/2}\log\sin 2x\,dx
\displaystyle \text{Putting }2x=t,\text{ we get } I_{1}=\int_{0}^{\pi}\log\sin t\,\frac{dt}{2}
\displaystyle \Rightarrow I_{1}=\frac{1}{2}\int_{0}^{\pi}\log\sin t\,dt
\displaystyle \Rightarrow I_{1}=\frac{1}{2}\times 2\int_{0}^{\pi/2}\log\sin t\,dt \qquad [\text{Using property X}]
\displaystyle \Rightarrow I_{1}=\int_{0}^{\pi/2}\log\sin x\,dx=I \qquad [\text{Using Property I}]
\displaystyle \text{Putting }I_{1}=I\text{ in (iii), we get }2I=I-\frac{\pi}{2}\log 2 \Rightarrow I=-\frac{\pi}{2}\log 2
\displaystyle \text{Hence, }\int_{0}^{\pi/2}\log\sin x\,dx =\int_{0}^{\pi/2}\log\cos x\,dx=-\frac{\pi}{2}\log 2

\displaystyle \textbf{Question 27: } \text{Evaluate the following integrals as limit of sums:}
\displaystyle \text{(i)}\ \int_{0}^{2}(x^{2}+3)\,dx\qquad [\text{CBSE 2001C}]
\displaystyle \text{(ii)}\ \int_{1}^{3}(2x^{2}+5)\,dx\qquad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have,}
\displaystyle \int_{a}^{b}f(x)\,dx=\lim_{h\to 0}h \left[f(a)+f(a+h)+f(a+2h)+\cdots+f\{a+(n-1)h\}\right], \text{ where }h=\frac{b-a}{n}
\displaystyle \text{Here, }a=0,\ b=2,\ f(x)=x^{2}+3\text{ and }h=\frac{2-0}{n}=\frac{2}{n}
\displaystyle I=\int_{0}^{2}(x^{2}+3)\,dx
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[f(0)+f(0+h)+f(0+2h)+\cdots+f\{0+(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[f(0)+f(h)+f(2h)+\cdots+f\{(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[(0+3)+(h^{2}+3)+(2^{2}h^{2}+3)+\cdots+\{(n-1)^{2}h^{2}+3\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0} \left[3n+h^{2}(1^{2}+2^{2}+\cdots+(n-1)^{2})\right]
\displaystyle \Rightarrow I=\lim_{h\to 0} \left\{3n+h^{2}\frac{n(n-1)(2n-1)}{6}\right\}
\displaystyle \Rightarrow I=\lim_{n\to\infty}\frac{2}{n} \left\{3n+\frac{4}{n^{2}}\times\frac{n(n-1)(2n-1)}{6}\right\}
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left\{6+\frac{8}{6}\frac{(n-1)(2n-1)}{n^{2}}\right\}
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left\{6+\frac{8}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right\}
\displaystyle \Rightarrow I=6+\frac{8}{6}(1-0)(2-0) =6+\frac{8}{3}=\frac{26}{3}

\displaystyle \text{(ii) We have,}
\displaystyle \int_{a}^{b}f(x)\,dx=\lim_{h\to 0}h \left[f(a)+f(a+h)+f(a+2h)+\cdots+f\{a+(n-1)h\}\right], \text{ where }h=\frac{b-a}{n}
\displaystyle \text{Here, }a=1,\ b=3,\ f(x)=2x^{2}+5\text{ and }h=\frac{3-1}{n}=\frac{2}{n}
\displaystyle I=\int_{1}^{3}(2x^{2}+5)\,dx
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[f(1)+f(1+h)+f(1+2h)+\cdots+f\{1+(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[2(1)^{2}+5+2(1+h)^{2}+5+\cdots+2\{1+(n-1)h\}^{2}+5\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left\{2\left[1^{2}+(1+h)^{2}+(1+2h)^{2}+\cdots+\{1+(n-1)h\}^{2}\right]+5n\right\}
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left\{2\left[n+2h(1+2+\cdots+(n-1)) +h^{2}(1^{2}+2^{2}+\cdots+(n-1)^{2})\right]+5n\right\}
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left\{2\left[n+2h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6}\right]+5n\right\}
\displaystyle \Rightarrow I=\lim_{n\to\infty}\frac{2}{n} \left\{7n+2\times\frac{2}{n}n(n-1) +2\times\frac{4}{n^{2}}\frac{n(n-1)(2n-1)}{6}\right\}
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[14+8\left(\frac{n-1}{n}\right) +\frac{8}{3}\frac{(n-1)(2n-1)}{n^{2}}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[14+8\left(1-\frac{1}{n}\right) +\frac{8}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle \Rightarrow I=14+8(1-0)+\frac{8}{3}(1-0)(2-0) =14+8+\frac{16}{3}=\frac{82}{3}

\displaystyle \textbf{Question 28: } \text{Evaluate the following integrals as limit of sums:}
\displaystyle \text{(i)}\ \int_{1}^{3}(x^{2}+x)\,dx\qquad [\text{CBSE 2000C}]
\displaystyle \text{(ii)}\ \int_{1}^{3}(x^{2}+5x)\,dx\qquad [\text{CBSE 2010}]
\displaystyle \text{Answer:}
\displaystyle \text{(i) We have,}
\displaystyle \int_{a}^{b}f(x)\,dx=\lim_{h\to 0}h \left[f(a)+f(a+h)+f(a+2h)+\cdots+f\{a+(n-1)h\}\right], \text{ where }h=\frac{b-a}{n}
\displaystyle \text{Here, }a=1,\ b=3,\ f(x)=x^{2}+x\text{ and }h=\frac{3-1}{n} =\frac{2}{n}
\displaystyle I=\int_{1}^{3}(x^{2}+x)\,dx
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[f(1)+f(1+h)+f(1+2h)+\cdots+f\{1+(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[\{1^{2}+1\}+\{(1+h)^{2}+(1+h)\}+\{(1+2h)^{2}+(1+2h)\} +\cdots+\{(1+(n-1)h)^{2}+(1+(n-1)h)\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[\{1^{2}+(1+h)^{2}+(1+2h)^{2}+\cdots+\{1+(n-1)h\}^{2}\} +\{1+(1+h)+(1+2h)+\cdots+(1+(n-1)h)\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left\{n+2h(1+2+\cdots+(n-1)) +h^{2}(1^{2}+2^{2}+\cdots+(n-1)^{2}) +n+h(1+2+\cdots+(n-1))\right\}
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[n+2h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6} +n+h\frac{n(n-1)}{2}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[2n+3h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty}\frac{2}{n} \left[2n+\frac{6}{n}\frac{n(n-1)}{2} +\frac{4}{n^{2}}\frac{n(n-1)(2n-1)}{6}\right] \qquad [\because\ h=\frac{2}{n}]
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[4+6\left(\frac{n-1}{n}\right) +\frac{4}{3}\frac{(n-1)(2n-1)}{n^{2}}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[4+6\left(1-\frac{1}{n}\right) +\frac{4}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle \Rightarrow I=4+6(1-0)+\frac{4}{3}(1-0)(2-0) =4+6+\frac{8}{3}=\frac{38}{3}

\displaystyle \text{(ii) We have,}
\displaystyle \int_{a}^{b}f(x)\,dx=\lim_{h\to 0}h \left[f(a)+f(a+h)+f(a+2h)+\cdots+f\{a+(n-1)h\}\right], \text{ where }h=\frac{b-a}{n}
\displaystyle \text{Here, }a=1,\ b=3,\ f(x)=x^{2}+5x\text{ and }h=\frac{3-1}{n} =\frac{2}{n}
\displaystyle I=\int_{1}^{3}(x^{2}+5x)\,dx
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[f(1)+f(1+h)+f(1+2h)+\cdots+f\{1+(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[\{1^{2}+5(1)\}+\{(1+h)^{2}+5(1+h)\} +\{(1+2h)^{2}+5(1+2h)\} +\cdots+\{(1+(n-1)h)^{2}+5(1+(n-1)h)\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[\{1^{2}+(1+h)^{2}+(1+2h)^{2}+\cdots+\{1+(n-1)h\}^{2}\} +5\{1+(1+h)+(1+2h)+\cdots+(1+(n-1)h)\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left\{n+2h(1+2+\cdots+(n-1)) +h^{2}(1^{2}+2^{2}+\cdots+(n-1)^{2}) +5n+5h(1+2+\cdots+(n-1))\right\}
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[6n+7h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty}\frac{2}{n} \left[6n+\frac{14}{n}\frac{n(n-1)}{2} +\frac{4}{n^{2}}\frac{n(n-1)(2n-1)}{6}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[12+14\left(\frac{n-1}{n}\right) +\frac{8}{3}\frac{(n-1)(2n-1)}{n^{2}}\right]
\displaystyle \Rightarrow I=\lim_{n\to\infty} \left[12+14\left(1-\frac{1}{n}\right) +\frac{8}{3}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right)\right]
\displaystyle \Rightarrow I=12+14(1-0)+\frac{8}{3}(1-0)(2-0) =12+14+\frac{8}{3}=\frac{86}{3}

\displaystyle \textbf{Question 29: } \text{Evaluate the following integrals as a limit of sums.}
\displaystyle \text{}\ \int_{1}^{3}(e^{2-3x}+x^{2}+1)\,dx \qquad [\text{CBSE 2015}]
\displaystyle \text{Answer:}
\displaystyle \text{Here, }a=1,\ b=3\text{ and }f(x)=e^{2-3x}+x^{2}+1. \text{ Therefore, }h=\frac{3-1}{n}=\frac{2}{n}\text{ and }nh=2
\displaystyle \text{Substituting these values in } \\ \int_{a}^{b}f(x)\,dx=\lim_{h\to 0}h \left[f(a)+f(a+h)+f(a+2h)+\cdots+f\{a+(n-1)h\}\right],\text{ we obtain}
\displaystyle I=\int_{1}^{3}(e^{2-3x}+x^{2}+1)\,dx
\displaystyle =\lim_{h\to 0}h\left[f(1)+f(1+h)+f(1+2h)+\cdots+f\{1+(n-1)h\}\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \Big[e^{2-3\cdot1}+1^{2}+1+e^{2-3(1+h)} \\ +(1+h)^{2}+1 +e^{2-3(1+2h)}+(1+2h)^{2}+1+\cdots
\displaystyle \qquad\qquad +e^{2-3\{1+(n-1)h\}}+\{1+(n-1)h\}^{2}+1\Big]
\displaystyle \Rightarrow I=\lim_{h\to 0}h \left[\{e^{-1}+e^{-1-3h}+e^{-1-3(2h)}+\cdots+e^{-1-3(n-1)h}\}\right.
\displaystyle \qquad\qquad \left.+\{1^{2}+(1+h)^{2}+(1+2h)^{2}+\cdots+(1+(n-1)h)^{2}\}+n\right]
\displaystyle \Rightarrow I=\lim_{h\to 0}h\Bigg[ e^{-1}\frac{1-e^{-3nh}}{1-e^{-3h}}  \\ +\left\{n+2h(1+2+\cdots+(n-1)) +h^{2}(1^{2}+2^{2}+\cdots+(n-1)^{2})\right\}+n\Bigg]
\displaystyle \Rightarrow I=\lim_{h\to 0}h\Bigg[ e^{-1}\frac{1-e^{-3nh}}{1-e^{-3h}} +2n+2h\frac{n(n-1)}{2} +h^{2}\frac{n(n-1)(2n-1)}{6}\Bigg]
\displaystyle \Rightarrow I=\lim_{h\to 0}\Bigg[ h e^{-1}\frac{1-e^{-3nh}}{1-e^{-3h}} +2nh+h^{2}n(n-1) +\frac{1}{6}h^{3}n(n-1)(2n-1)\Bigg]
\displaystyle \Rightarrow I=\lim_{h\to 0}\Bigg[ e^{-1}\frac{e^{-6}-1}{-3} +4+2(2-h)+\frac{2}{6}(2-h)(4-h)\Bigg] \qquad [\because\ nh=2]
\displaystyle \Rightarrow I=\frac{e^{-1}}{3}(1-e^{-6}) +4+4+\frac{4}{3} =-\frac{1}{3}(e^{-7}-e^{-1})+\frac{32}{3}

\displaystyle \textbf{Question 31. }\int_0^4(e^{2x}+x)\,dx\text{ is equal to} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{15+e^8}{2} \hspace{2cm} \text{(b) }\frac{16-e^8}{2}
\displaystyle \text{(c) }\frac{e^8-15}{2} \hspace{2cm} \text{(d) }\frac{-e^8-15}{2}
\displaystyle \text{Answer:}
\displaystyle \text{(a) }\int_0^4(e^{2x}+x)\,dx=\int_0^4e^{2x}\,dx+\int_0^4x\,dx
\displaystyle \text{Putting }2x=t\Rightarrow 2\,dx=dt\Rightarrow dx=\frac{dt}{2}
\displaystyle \text{Lower limit: When }x=0\text{, then }t=0
\displaystyle \text{Upper limit: When }x=4\text{, then }t=8
\displaystyle =\int_0^8e^t\frac{dt}{2}+\left(\frac{x^2}{2}\right)_0^4
\displaystyle =\frac{1}{2}\int_0^8e^t\,dt+\frac{1}{2}(x^2)_0^4
\displaystyle =\frac{1}{2}(e^t)_0^8+\frac{1}{2}(16-0)
\displaystyle =\frac{1}{2}(e^8-e^0)+\frac{1}{2}(16)
\displaystyle =\frac{1}{2}(e^8-1)+8
\displaystyle =\frac{e^8-1+16}{2}=\frac{e^8+15}{2}
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\displaystyle \textbf{Question 32. }\text{If }\int_0^a 3x^2\,dx=8\text{, then the value of }'a'\text{ is} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) 2} \hspace{2cm} \text{(b) 4}
\displaystyle \text{(c) 8} \hspace{2cm} \text{(d) 10}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Given, }\int_0^a 3x^2\,dx=8\Rightarrow 3\int_0^a x^2\,dx=8
\displaystyle \Rightarrow 3\left(\frac{x^3}{3}\right)_0^a=8\Rightarrow 3\left(\frac{a^3}{3}-0\right)=8
\displaystyle \Rightarrow a^3=8\Rightarrow a=2
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\displaystyle \textbf{Question 33. }\int_{-1}^1\frac{|x-2|}{x-2}\,dx,\ x\neq 2\text{ is equal to} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) 1} \hspace{2cm} \text{(b) }-1
\displaystyle \text{(c) 2} \hspace{2cm} \text{(d) }-2
\displaystyle \text{Answer:}
\displaystyle \text{(d) }\int_{-1}^1\frac{|x-2|}{x-2}\,dx=\int_{-1}^1\frac{-(x-2)}{x-2}\,dx
\displaystyle =\int_{-1}^1(-1)\,dx=-(x)_{-1}^1=-(1+1)=-2
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\displaystyle \textbf{Question 34. }\int_0^{\pi/6}\sec^2\left(x-\frac{\pi}{6}\right)dx\text{ is equal to} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }\frac{1}{\sqrt{3}} \hspace{2cm} \text{(b) }-\frac{1}{\sqrt{3}}
\displaystyle \text{(c) }\sqrt{3} \hspace{2cm} \text{(d) }-\sqrt{3}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let }I=\int_0^{\pi/6}\sec^2\left(x-\frac{\pi}{6}\right)dx
\displaystyle =\left[\tan\left(x-\frac{\pi}{6}\right)\right]_0^{\pi/6}\qquad\left[\because\int\sec^2x\,dx=\tan x+C\right]
\displaystyle =\tan\left(\frac{\pi}{6}-\frac{\pi}{6}\right)-\tan\left(0-\frac{\pi}{6}\right)
\displaystyle =0+\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}
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\displaystyle \textbf{Question 35. }\text{For any integer }n\text{, the value of }\int_0^\pi e^{\sin^2x}\cos^3(2n+1)x\,dx\text{ is} \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{(a) }-1 \hspace{2cm} \text{(b) 0}
\displaystyle \text{(c) 1} \hspace{2cm} \text{(d) 2}
\displaystyle \text{Answer:}
\displaystyle \text{(b) Let }I=\int_0^\pi e^{\sin^2x}\cos^3(2n+1)x\,dx\qquad\ldots\text{(i)}
\displaystyle =\int_0^\pi e^{\sin^2(\pi-x)}\cos^3(2n+1)(\pi-x)\,dx
\displaystyle \left[\because\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\right]
\displaystyle =\int_0^\pi e^{\sin^2x}[\cos\{(2n+1)\pi-(2n+1)x\}]^3\,dx
\displaystyle =\int_0^\pi e^{\sin^2x}[-\cos(2n+1)x]^3\,dx
\displaystyle \Rightarrow I=-\int_0^\pi e^{\sin^2x}\cos^3(2n+1)x\,dx\qquad\ldots\text{(ii)}
\displaystyle \Rightarrow 2I=0\Rightarrow I=0\qquad\text{[Added Eqs. (i) and (ii)]}
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\displaystyle \textbf{Question 36. }\text{Assertion (A) }\int_2^8\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx=3 \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Reason (R) }\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx
\displaystyle \text{(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).}
\displaystyle \text{(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).}
\displaystyle \text{(c) (A) is correct but (R) is incorrect.}
\displaystyle \text{(d) Both (A) and (R) are incorrect.}
\displaystyle \text{Answer:}
\displaystyle \text{(a) Let }I=\int_2^8\frac{\sqrt{10-x}}{\sqrt{x}+\sqrt{10-x}}\,dx\qquad\ldots\text{(i)}
\displaystyle =\int_2^8\frac{\sqrt{10-(10-x)}}{\sqrt{10-(10-x)}+\sqrt{10-(10-x)}}\,dx
\displaystyle =\int_2^8\frac{\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}\,dx\qquad\ldots\text{(ii)}
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle 2I=\int_2^8\frac{\sqrt{10-x}+\sqrt{x}}{\sqrt{10-x}+\sqrt{x}}\,dx=\int_2^8 1\,dx
\displaystyle \Rightarrow 2I=[x]_2^8=8-2=6\Rightarrow I=3
\displaystyle \text{Thus, Assertion and Reason both are correct and Reason is the correct explanation of Assertion.}
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\displaystyle \left[\because\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\right]
\displaystyle =\int_0^\pi e^{\sin^2x}[\cos\{(2n+1)\pi-(2n+1)x\}]^3\,dx
\displaystyle =\int_0^\pi e^{\sin^2x}[-\cos(2n+1)x]^3\,dx
\displaystyle \Rightarrow I=-\int_0^\pi e^{\sin^2x}\cos^3(2n+1)x\,dx\qquad\ldots\text{(ii)}
\displaystyle \Rightarrow 2I=0\Rightarrow I=0\qquad\text{[Added Eqs. (i) and (ii)]}
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\displaystyle \textbf{Question 37. }\text{Evaluate }\int_{-1}^1\log_e\left(\frac{2-x}{2+x}\right)dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }f(x)=\log_e\left(\frac{2-x}{2+x}\right)
\displaystyle \therefore\ f(-x)=\log_e\left(\frac{2+x}{2-x}\right)=-\log_e\left(\frac{2-x}{2+x}\right)=-f(x)
\displaystyle \text{So, }f(x)\text{ is an odd function.}
\displaystyle \therefore\ \int_{-1}^1\log_e\left(\frac{2-x}{2+x}\right)dx=0
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\displaystyle \textbf{Question 34. }\text{Evaluate }\int_0^{\pi/2}e^x\sin x\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi/2}e^x\sin x\,dx
\displaystyle \text{Here, }\int e^x\sin x\,dx
\displaystyle =\left[\sin x\int e^x\,dx-\int\left(\frac{d}{dx}(\sin x)\int e^x\,dx\right)dx\right]\qquad\text{[using integration by parts]}
\displaystyle =[\sin x\cdot e^x-\int(\cos x\cdot e^x)\,dx]
\displaystyle =\sin x\cdot e^x-\left[\cos x\int e^x\,dx-\int\left(\frac{d}{dx}(\cos x)\int e^x\,dx\right)dx\right]
\displaystyle =\sin x\cdot e^x-\left[\cos x\cdot e^x+\int\sin x\cdot e^x\,dx\right]
\displaystyle =e^x(\sin x-\cos x)-\int\sin x\cdot e^x\,dx
\displaystyle \Rightarrow 2\int e^x\sin x\,dx=e^x(\sin x-\cos x)
\displaystyle \Rightarrow\int e^x\sin x\,dx=\frac{1}{2}e^x(\sin x-\cos x)
\displaystyle \therefore\ I=\int_0^{\pi/2}e^x\sin x\,dx=\frac{1}{2}[e^x(\sin x-\cos x)]_0^{\pi/2}
\displaystyle =\frac{1}{2}\left[e^{\pi/2}\left(\sin\frac{\pi}{2}-\cos\frac{\pi}{2}\right)-e^0(\sin 0-\cos 0)\right]
\displaystyle =\frac{1}{2}(e^{\pi/2}+1)
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\displaystyle \textbf{Question 38. }\text{Evaluate }\int_{1/3}^1\frac{(x-x^3)^{1/3}}{x^4}\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{1/3}^1\frac{(x-x^3)^{1/3}}{x^4}\,dx
\displaystyle =\int_{1/3}^1\frac{(x^3)^{1/3}\left(\dfrac{1}{x^2}-1\right)^{1/3}}{x^4}\,dx
\displaystyle =\int_{1/3}^1\frac{x\left(\dfrac{1}{x^2}-1\right)^{1/3}}{x^4}\,dx
\displaystyle =\int_{1/3}^1\frac{\left(\dfrac{1}{x^2}-1\right)^{1/3}}{x^3}\,dx
\displaystyle \text{On putting }\frac{1}{x^2}=t
\displaystyle \Rightarrow\frac{-2}{x^3}\,dx=dt\Rightarrow\frac{dx}{x^3}=-\frac{dt}{2}
\displaystyle \text{Lower limit: When }x=\frac{1}{3}\text{, then }t=9
\displaystyle \text{Upper limit: When }x=1\text{, then }t=1
\displaystyle \therefore\ I=\int_9^1\frac{(t-1)^{1/3}}{(-2)}\,dt=-\frac{1}{2}\int_9^1(t-1)^{1/3}\,dt
\displaystyle =-\frac{1}{2}\left[\frac{(t-1)^{1/3+1}}{\dfrac{1}{3}+1}\right]_9^1=-\frac{1}{2}\times\frac{3}{4}[(t-1)^{4/3}]_9^1
\displaystyle =-\frac{3}{8}[(1-1)^{4/3}-(9-1)^{4/3}]
\displaystyle =-\frac{3}{8}[0-8^{4/3}]=-\frac{3}{8}(-16)=6
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\displaystyle \textbf{Question 39. }\text{Evaluate }\int_1^3\{|(x-1)|+|(x-2)|\}\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_1^3\{|(x-1)|+|(x-2)|\}\,dx
\displaystyle \text{For }1\leq x<3,\ |x-1|=x-1
\displaystyle 1<x<2,\ |x-2|=2-x
\displaystyle 2\leq x<3,\ |x-2|=x-2
\displaystyle \therefore\ I=\int_1^2(x-1)\,dx+\int_1^2(2-x)\,dx+\int_2^3(x-2)\,dx
\displaystyle =\left(\frac{x^2}{2}-x\right)_1^2+\left(2x-\frac{x^2}{2}\right)_1^2+\left(\frac{x^2}{2}-2x\right)_2^3
\displaystyle \Rightarrow I=\left(\frac{9}{2}-3\right)-\left(\frac{1}{2}-1\right)+\left(4-\frac{4}{2}\right)-\left(2-\frac{1}{2}\right)+\left(\frac{9}{2}-6\right)-\left(\frac{4}{2}-4\right)
\displaystyle =(4-2)+\left(2-\frac{3}{2}\right)+\left(\frac{5}{2}-2\right)=2+\frac{1}{2}+\frac{1}{2}=3
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\displaystyle \textbf{Question 40. }\text{Evaluate }\int_{\pi/4}^{\pi/2}e^{2x}\left(\frac{1-\sin 2x}{1-\cos 2x}\right)dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{\pi/4}^{\pi/2}e^{2x}\left(\frac{1-\sin 2x}{1-\cos 2x}\right)dx
\displaystyle \text{Now, }\int e^{2x}\left(\frac{1-\sin 2x}{1-\cos 2x}\right)dx=\int e^{2x}\left(\frac{1-2\sin x\cos x}{2\sin^2x}\right)dx
\displaystyle =\frac{1}{2}\int e^{2x}\left(\frac{1}{\sin^2x}-\frac{2\sin x\cos x}{\sin^2x}\right)dx
\displaystyle =\frac{1}{2}\int e^{2x}(\text{cosec}^2x-2\cot x)\,dx
\displaystyle =\frac{1}{2}\int e^{2x}\text{cosec}^2x\,dx-\int e^{2x}\cot x\,dx
\displaystyle =\frac{1}{2}\left[e^{2x}\int\text{cosec}^2x\,dx-\int\left(\frac{d}{dx}(e^{2x})\int\text{cosec}^2x\,dx\right)dx\right]-\int e^{2x}\cot x\,dx
\displaystyle =\frac{1}{2}[e^{2x}(-\cot x)-\int 2e^{2x}(-\cot x)\,dx]-\int e^{2x}\cot x\,dx
\displaystyle =\frac{1}{2}e^{2x}(-\cot x)+\int e^{2x}\cot x\,dx-\int e^{2x}\cot x\,dx
\displaystyle =-\frac{1}{2}e^{2x}\cot x
\displaystyle \therefore\ I=-\frac{1}{2}(e^{2x}\cot x)_{\pi/4}^{\pi/2}
\displaystyle =-\frac{1}{2}\left[e^\pi\cot\frac{\pi}{2}-e^{\pi/2}\cot\frac{\pi}{4}\right]
\displaystyle =-\frac{1}{2}(0-e^{\pi/2})=\frac{1}{2}e^{\pi/2}
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\displaystyle \textbf{Question 41. }\text{Evaluate }\int_1^4\frac{1}{\sqrt{2x+1}-\sqrt{2x-1}}\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_1^4\frac{1}{\sqrt{2x+1}-\sqrt{2x-1}}\,dx
\displaystyle =\int_1^4\frac{\sqrt{2x+1}+\sqrt{2x-1}}{(2x+1)-(2x-1)}\,dx
\displaystyle =\frac{1}{2}\int_1^4\{(2x+1)^{1/2}+(2x-1)^{1/2}\}\,dx
\displaystyle =\frac{1}{2}\left[\int_1^4(2x+1)^{1/2}\,dx+\int_1^4(2x-1)^{1/2}\,dx\right]
\displaystyle =\frac{1}{2}\left[\frac{1}{2}\cdot\frac{(2x+1)^{3/2}}{3/2}+\frac{1}{2}\cdot\frac{(2x-1)^{3/2}}{3/2}\right]_1^4
\displaystyle =\frac{1}{2}\times\frac{2}{2}\times\frac{2}{3}[\{(2x+1)^{3/2}\}_1^4+\{(2x-1)^{3/2}\}_1^4]
\displaystyle =\frac{1}{6}[(9)^{3/2}-(3)^{3/2}+(7)^{3/2}-(1)^{3/2}]
\displaystyle =\frac{1}{6}[27-3^{3/2}+7^{3/2}-1]=\frac{1}{6}(26-3^{3/2}+7^{3/2})
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\displaystyle \textbf{Question 42. }\text{Evaluate }\int_1^3\frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}}\,dx. \hspace{1.2cm} \text{[CBSE 2023]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_1^3\frac{\sqrt{4-x}}{\sqrt{x}+\sqrt{4-x}}\,dx\qquad\ldots\text{(i)}
\displaystyle =\int_1^3\frac{\sqrt{4-(4-x)}}{\sqrt{4-(4-x)}+\sqrt{4-(4-x)}}\,dx\qquad\left[\because\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx\right]
\displaystyle =\int_1^3\frac{\sqrt{x}}{\sqrt{4-x}+\sqrt{x}}\,dx\qquad\ldots\text{(ii)}
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle 2I=\int_1^3\frac{\sqrt{4-x}+\sqrt{x}}{\sqrt{x}+\sqrt{4-x}}\,dx=\int_1^3 1\,dx=[x]_1^3=3-1=2
\displaystyle \Rightarrow I=1
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\displaystyle \textbf{Question 43. }\text{Evaluate }\int_0^{\pi/2}2\sin x\cos x\tan^{-1}(\sin x)\,dx. \hspace{1.2cm} \text{[CBSE 2023; CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi/2}2\sin x\cos x\tan^{-1}(\sin x)\,dx
\displaystyle \text{Put }\sin x=t\Rightarrow\cos x\,dx=dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=\sin 0=0
\displaystyle \text{Upper limit: When }x=\frac{\pi}{2}\text{, then }t=\sin\frac{\pi}{2}=1
\displaystyle \therefore\ I=2\int_0^1 t\times\tan^{-1}t\,dt
\displaystyle \text{Applying integration by parts, taking }\tan^{-1}t\text{ as Ist function and }t\text{ as IInd function, we get}
\displaystyle I=2\left[\frac{t^2}{2}\times\tan^{-1}t\right]_0^1-2\int_0^1\frac{1}{1+t^2}\times\frac{t^2}{2}\,dt
\displaystyle \left[\because\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}\right]
\displaystyle \Rightarrow I=2\left[\frac{t^2}{2}\times\tan^{-1}t\right]_0^1-\int_0^1\frac{t^2}{1+t^2}\,dt
\displaystyle \Rightarrow I=2\times\frac{1}{2}\times\tan^{-1}(1)-\int_0^1\frac{1+t^2-1}{1+t^2}\,dt
\displaystyle \Rightarrow I=1\times\frac{\pi}{4}-\int_0^1\left(1-\frac{1}{1+t^2}\right)dt
\displaystyle \Rightarrow I=\frac{\pi}{4}-[t-\tan^{-1}t]_0^1=\frac{\pi}{4}-1+\tan^{-1}(1)
\displaystyle =\frac{\pi}{4}-1+\frac{\pi}{4}=\frac{2\pi}{4}-1
\displaystyle \therefore\ I=\left(\frac{\pi}{2}-1\right)
\\

\displaystyle \textbf{Question 44. }\text{Evaluate }\int_{-2}^1\sqrt{5-4x-x^2}\,dx. \hspace{1.2cm} \text{[CBSE 2022 (Term II)]}
\displaystyle \text{Answer:}
\displaystyle I=\int_{-2}^1\sqrt{5-4x-x^2}\,dx
\displaystyle =\int_{-2}^1\sqrt{5-(x^2+4x+4)+4}\,dx
\displaystyle =\int_{-2}^1\sqrt{9-(x+2)^2}\,dx
\displaystyle \text{Let }x+2=t\Rightarrow dx=dt
\displaystyle \text{Lower limit: When }x=-2\text{, then }t=0
\displaystyle \text{Upper limit: When }x=1\text{, then }t=3
\displaystyle \therefore\ I=\int_0^3\sqrt{3^2-t^2}\,dt
\displaystyle =\left[\frac{1}{2}t\sqrt{3^2-t^2}+\frac{1}{2}\cdot 3^2\sin^{-1}\left(\frac{t}{3}\right)\right]_0^3
\displaystyle =0+\frac{9}{2}\sin^{-1}1-0=\frac{9}{2}\cdot\frac{\pi}{2}=\frac{9\pi}{4}
\\

\displaystyle \textbf{Question 45. }\text{If }[x]\text{ denotes the greatest integer function, then find }\int_0^{3/2}[x^2]\,dx. \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{3/2}[x^2]\,dx=\int_0^1 0\,dx+\int_1^{\sqrt{2}}1\,dx+\int_{\sqrt{2}}^{3/2}2\,dx
\displaystyle =0+[x]_1^{\sqrt{2}}+[2x]_{\sqrt{2}}^{3/2}
\displaystyle =\sqrt{2}-1+3-2\sqrt{2}
\displaystyle \Rightarrow I=2-\sqrt{2}
\\

\displaystyle \textbf{Question 46. }\text{Evaluate }\int_0^{2\pi}|\sin x|\,dx. \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{2\pi}|\sin x|\,dx
\displaystyle =\int_0^\pi\sin x\,dx-\int_\pi^{2\pi}\sin x\,dx
\displaystyle =[-\cos x]_0^\pi+[\cos x]_\pi^{2\pi}
\displaystyle =-[\cos\pi-\cos 0]+[\cos 2\pi-\cos\pi]
\displaystyle =-[-1-1]+[1+1]=2+2=4
\\

\displaystyle \textbf{Question 47. }\text{If }\int_0^a\frac{dx}{1+4x^2}=\frac{\pi}{8}\text{, then find the value of }a. \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{We have, }\int_0^a\frac{dx}{1+4x^2}=\frac{\pi}{8}
\displaystyle \Rightarrow\frac{1}{2}[\tan^{-1}2x]_0^a=\frac{\pi}{8}\Rightarrow\tan^{-1}2a=\frac{\pi}{4}
\displaystyle \Rightarrow 2a=\tan\frac{\pi}{4}\Rightarrow 2a=1\Rightarrow a=\frac{1}{2}
\\

\displaystyle \textbf{Question 48. }\text{Find the value of }\int_1^4|x-5|\,dx. \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_1^4|x-5|\,dx
\displaystyle =\int_1^4-(x-5)\,dx\qquad[\because|x-5|=-(x-5),\ x<5]
\displaystyle =-\left[\frac{x^2}{2}-5x\right]_1^4=-\left[\left(\frac{16}{2}-20\right)-\left(\frac{1}{2}-5\right)\right]
\displaystyle =-\left[-12+\frac{9}{2}\right]=\frac{15}{2}
\\

\displaystyle \textbf{Question 49. }\text{Evaluate }\int_1^2\left[\frac{1}{x}-\frac{1}{2x^2}\right]e^{2x}\,dx. \hspace{1.2cm} \text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_1^2\left[\frac{1}{x}-\frac{1}{2x^2}\right]e^{2x}\,dx
\displaystyle \text{Put }2x=t\Rightarrow x=\frac{1}{2}t\Rightarrow dx=\frac{1}{2}\,dt
\displaystyle \text{Lower limit: When }x=1\text{, then }t=2
\displaystyle \text{Upper limit: When }x=2\text{, then }t=4
\displaystyle \therefore\ I=\frac{1}{2}\int_2^4\left[\frac{2}{t}-\frac{2}{t^2}\right]e^t\,dt
\displaystyle =\int_2^4\left(\frac{1}{t}+\frac{-1}{t^2}\right)e^t\,dt
\displaystyle \left[\because\int e^x\{f(x)+f'(x)\}\,dx=e^x f(x)+C\right]
\displaystyle =\left[\frac{e^t}{t}\right]_2^4=\frac{e^4}{4}-\frac{e^2}{2}=\frac{e^2}{4}(e^2-2)
\\

\displaystyle \textbf{Question 50. }\text{Evaluate }\int_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx. \hspace{0.2cm} \text{[CBSE 2020, 2017; CBSE 2013, 2012 2011C, 2009C, 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx\qquad\ldots\text{(i)}
\displaystyle \Rightarrow I=\int_0^\pi\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\,dx\qquad\left[\because\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\right]
\displaystyle =\int_0^\pi\frac{(\pi-x)\sin x\,dx}{1+\cos^2x}\qquad\ldots\text{(ii)}
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle 2I=\pi\int_0^\pi\frac{\sin x\,dx}{1+\cos^2x}\Rightarrow I=\frac{\pi}{2}\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx
\displaystyle \text{Using }\int_0^{2a}f(x)\,dx=2\int_0^a f(x)\,dx\text{, if }f(2a-x)=f(x)
\displaystyle \therefore\ I=\frac{\pi}{2}\times 2\int_0^{\pi/2}\frac{\sin x}{1+\cos^2x}\,dx
\displaystyle \Rightarrow I=\pi\int_0^{\pi/2}\frac{\sin x}{1+\cos^2x}\,dx
\displaystyle \text{Put }\cos x=t\Rightarrow-\sin x\,dx=dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=1
\displaystyle \text{Upper limit: When }x=\frac{\pi}{2}\text{, then }t=0
\displaystyle \therefore\ I=\pi\int_1^0\frac{-dt}{1+t^2}\Rightarrow I=-\pi[\tan^{-1}t]_1^0
\displaystyle \Rightarrow I=-\pi[\tan^{-1}0-\tan^{-1}1]
\displaystyle \Rightarrow I=-\pi\left[0-\frac{\pi}{4}\right]=\frac{\pi^2}{4}
\\

\displaystyle \textbf{Question 51. }\text{Prove that }\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\text{, hence evaluate }\int_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx. \hspace{0.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{To prove }\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx
\displaystyle \text{Consider, RHS}=\int_0^a f(a-x)\,dx
\displaystyle \text{Putting }t=a-x\text{, then }dt=-dx
\displaystyle \text{Lower limit: When }x=0\text{, then }t=a
\displaystyle \text{Upper limit: When }x=a\text{, then }t=0
\displaystyle \text{Now, RHS}=-\int_a^0 f(t)\,dt=\int_0^a f(t)\,dt
\displaystyle =\int_0^a f(x)\,dx=\text{LHS}\qquad\textbf{Hence proved.}
\displaystyle \text{Now, let }I=\int_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx\qquad\ldots\text{(i)}
\displaystyle =\int_0^\pi\frac{(\pi-x)\sin(\pi-x)\,dx}{1+\cos^2(\pi-x)}\qquad\left[\because\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\right]
\displaystyle =\int_0^\pi\frac{(\pi-x)\sin x\,dx}{1+\cos^2x}
\displaystyle =\pi\int_0^\pi\frac{\sin x\,dx}{1+\cos^2x}-\int_0^\pi\frac{x\sin x\,dx}{1+\cos^2x}
\displaystyle I=\pi\int_0^\pi\frac{\sin x\,dx}{1+\cos^2x}-I\qquad\text{[from Eq. (i)]}
\displaystyle \Rightarrow 2I=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx
\displaystyle \Rightarrow I=\frac{\pi}{2}\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx
\displaystyle \text{Put }\cos x=t\Rightarrow-\sin x\,dx=dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=1
\displaystyle \text{Upper limit: When }x=\pi\text{, then }t=-1
\displaystyle \text{Now, }I=-\frac{\pi}{2}\int_1^{-1}\frac{dt}{1+t^2}=\frac{\pi}{2}\int_{-1}^1\frac{dt}{1+t^2}
\displaystyle =\frac{\pi}{2}[\tan^{-1}t]_{-1}^1=\frac{\pi}{2}[\tan^{-1}(1)-\tan^{-1}(-1)]
\displaystyle =\frac{\pi}{2}\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right]=\frac{\pi}{2}\left[\frac{\pi}{2}\right]=\frac{\pi^2}{4}
\\

\displaystyle \textbf{Question 52. }\text{Prove that }\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\text{ and hence evaluate }\int_0^{\pi/2}\frac{x}{\sin x+\cos x}\,dx. \hspace{0.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{To prove }\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx
\displaystyle \text{Put }a-x=t\Rightarrow dx=-dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=a
\displaystyle \text{Upper limit: When }x=a\text{, then }t=0
\displaystyle \therefore\ \int_0^a f(a-x)\,dx=-\int_a^0 f(t)\,dt=\int_0^a f(t)\,dt=\int_0^a f(x)\,dx=\text{LHS}\qquad\textbf{Hence proved.}
\displaystyle \text{Now, let }I=\int_0^{\pi/2}\frac{x}{\sin x+\cos x}\,dx\qquad\ldots\text{(i)}
\displaystyle \Rightarrow I=\int_0^{\pi/2}\frac{\left(\dfrac{\pi}{2}-x\right)}{\sin\left\{\left(\dfrac{\pi}{2}\right)-x\right\}+\cos\left\{\left(\dfrac{\pi}{2}\right)-x\right\}}\,dx\qquad\left[\because\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\right]
\displaystyle \Rightarrow I=\int_0^{\pi/2}\frac{\left(\dfrac{\pi}{2}-x\right)}{(\cos x+\sin x)}\,dx\qquad\ldots\text{(ii)}
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle 2I=\frac{\pi}{2}\int_0^{\pi/2}\frac{dx}{\sin x+\cos x}
\displaystyle \Rightarrow I=\frac{\pi}{4}\int_0^{\pi/2}\frac{dx}{\sin x+\cos x}
\displaystyle =\frac{\pi}{4}\int_0^{\pi/2}\frac{dx}{\dfrac{2\tan\left(\dfrac{x}{2}\right)}{1+\tan^2\left(\dfrac{x}{2}\right)}+\dfrac{1-\tan^2\left(\dfrac{x}{2}\right)}{1+\tan^2\left(\dfrac{x}{2}\right)}}
\displaystyle =\frac{\pi}{4}\int_0^{\pi/2}\frac{\sec^2\left(\dfrac{x}{2}\right)}{1-\tan^2\left(\dfrac{x}{2}\right)+2\tan\left(\dfrac{x}{2}\right)}\,dx
\displaystyle \text{[dividing numerator and denominator by }\cos^2\left(\dfrac{x}{2}\right)\text{]}
\displaystyle =\frac{\pi}{4}\int_0^{\pi/2}\frac{\sec^2\left(\dfrac{x}{2}\right)}{1+3\tan^2\left(\dfrac{x}{2}\right)}\,dx
\displaystyle \text{[dividing numerator and denominator by }\cos^2x\text{]}
\displaystyle \text{Put }\tan x=t\Rightarrow\sec^2x\,dx=dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=0
\displaystyle \text{Upper limit: When }x=\frac{\pi}{4}\text{, then }t=1
\displaystyle \frac{\pi}{2}\int_0^1\frac{1}{1+3t}\,dt=\frac{\pi}{2}\times\frac{1}{3}\int_0^1\frac{1}{\left(\dfrac{1}{\sqrt{3}}\right)^2+t^2}\,dt
\displaystyle =\frac{\pi}{6}(\sqrt{3})[\tan^{-1}\sqrt{3}\,t]_0^1
\displaystyle \left[\because\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\right]
\displaystyle =\frac{\pi}{6}(\sqrt{3})(\tan^{-1}\sqrt{3}-\tan^{-1}0)=\frac{\pi}{6}(\sqrt{3})\left(\frac{\pi}{3}-0\right)
\displaystyle =\frac{\sqrt{3}\pi^2}{18}
\\

\displaystyle \textbf{Question 53. }\text{Evaluate }\int_{-\pi}^\pi(1-x^2)\sin x\cos^2x\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-\pi}^\pi(1-x^2)\sin x\cos^2x\,dx
\displaystyle \text{Again, let }f(x)=(1-x^2)\sin x\cos^2x
\displaystyle \therefore\ f(-x)=[1-(-x)^2]\sin(-x)\cos^2(-x)
\displaystyle =(1-x^2)(-\sin x)\cos^2x
\displaystyle =-(1-x^2)\sin x\cos^2x=-f(x)
\displaystyle \therefore\ f(x)\text{ is odd function.}
\displaystyle \therefore\ I=0
\displaystyle \left[\because\int_{-a}^a f(x)\,dx=0\text{, if }f(x)\text{ is an odd function}\right]
\\

\displaystyle \textbf{Question 54. }\text{Evaluate }\int_{-1}^2\frac{|x|}{x}\,dx. \hspace{1.2cm} \text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-1}^2\frac{|x|}{x}\,dx=\int_{-1}^0\frac{|x|}{x}\,dx+\int_0^2\frac{|x|}{x}\,dx
\displaystyle =\int_{-1}^0\frac{-x}{x}\,dx+\int_0^2\frac{x}{x}\,dx\qquad\left[\because|x|=\begin{cases}-x,&x<0\\x,&x\geq 0\end{cases}\right]
\displaystyle =\int_{-1}^0-1\,dx+\int_0^2 1\,dx=[-x]_{-1}^0+[x]_0^2
\displaystyle =[0-(-1)]+[2-0]=-1+2=1
\\

\displaystyle \textbf{Question 55. }\text{Evaluate }\int_0^{\pi/4}\frac{\sin x+\cos x}{16+9\sin 2x}\,dx. \hspace{1.2cm}\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi/4}\frac{\sin x+\cos x}{16+9\sin 2x}\,dx
\displaystyle =\int_0^{\pi/4}\frac{\sin x+\cos x}{9(2\sin x\cos x)+16}\,dx
\displaystyle =\int_0^{\pi/4}\frac{\sin x+\cos x}{-9(-2\sin x\cos x)+16}\,dx
\displaystyle =\int_0^{\pi/4}\frac{\sin x+\cos x}{-9(\sin^2 x+\cos^2 x-2\sin x\cos x-1)+16}\,dx\qquad[\because\sin^2\theta+\cos^2\theta=1]
\displaystyle =\int_0^{\pi/4}\frac{\sin x+\cos x}{25-9(\sin x-\cos x)^2}\,dx\qquad[\because a^2+b^2-2ab=(a-b)^2]
\displaystyle \text{Put }\sin x-\cos x=t\Rightarrow(\cos x+\sin x)\,dx=dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=-1
\displaystyle \text{Upper limit: When }x=\frac{\pi}{4}\text{, then }t=0
\displaystyle \therefore\ I=\int_{-1}^{0}\frac{dt}{25-9t^2}=\frac{1}{9}\int_{-1}^{0}\frac{dt}{\left(\frac{5}{3}\right)^2-t^2}
\displaystyle =\frac{1}{9}\times\frac{1}{2\times\frac{5}{3}}\left[\log\frac{\frac{5}{3}+t}{\frac{5}{3}-t}\right]_{-1}^{0}\qquad\left[\because\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|+C\right]
\displaystyle =\frac{1}{30}\left[\log\frac{5+3t}{5-3t}\right]_{-1}^{0}=\frac{1}{30}\left[\log 1-\log\!\left(\frac{2}{8}\right)\right]
\displaystyle =\frac{1}{30}\left[0-\log\frac{1}{4}\right]\qquad[\because\log 1=0]
\displaystyle =\frac{1}{30}\left[-\log 4^{-1}\right]=\frac{1}{30}\log 4\qquad[\because\log m^n=n\log m]
\\

\displaystyle \textbf{Question 56. }\text{Evaluate }\int_2^3 3^x\,dx. \hspace{1.2cm} \text{[CBSE 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_2^3 3^x\,dx=\left(\frac{3^x}{\log 3}\right)_2^3=\frac{1}{\log 3}[3^x]_2^3
\displaystyle =\frac{1}{\log 3}[3^3-3^2]=\frac{1}{\log 3}(27-9)=\frac{18}{\log 3}
\\

\displaystyle \textbf{Question 57. }\text{Evaluate }\int_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx. \hspace{0.2cm} \text{[CBSE 2020, 2017; CBSE 2013, 2012 2011C, 2009C, 2008]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^\pi\frac{x\sin x}{1+\cos^2x}\,dx\qquad\ldots\text{(i)}
\displaystyle \Rightarrow I=\int_0^\pi\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\,dx\qquad\left[\because\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\right]
\displaystyle =\int_0^\pi\frac{(\pi-x)\sin x\,dx}{1+\cos^2x}\qquad\ldots\text{(ii)}
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle 2I=\pi\int_0^\pi\frac{\sin x\,dx}{1+\cos^2x}\Rightarrow I=\frac{\pi}{2}\int_0^\pi\frac{\sin x}{1+\cos^2x}\,dx
\displaystyle \text{Using }\int_0^{2a}f(x)\,dx=2\int_0^a f(x)\,dx\text{, if }f(2a-x)=f(x)
\displaystyle \therefore\ I=\frac{\pi}{2}\times 2\int_0^{\pi/2}\frac{\sin x}{1+\cos^2x}\,dx
\displaystyle \Rightarrow I=\pi\int_0^{\pi/2}\frac{\sin x}{1+\cos^2x}\,dx
\displaystyle \text{Put }\cos x=t\Rightarrow-\sin x\,dx=dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=1
\displaystyle \text{Upper limit: When }x=\frac{\pi}{2}\text{, then }t=0
\displaystyle \therefore\ I=\pi\int_1^0\frac{-dt}{1+t^2}\Rightarrow I=-\pi[\tan^{-1}t]_1^0
\displaystyle \Rightarrow I=-\pi[\tan^{-1}0-\tan^{-1}1]
\displaystyle \Rightarrow I=-\pi\left[0-\frac{\pi}{4}\right]=\frac{\pi^2}{4}
\\

\displaystyle \textbf{Question 58. }\text{Evaluate }\int_1^4(|x-1|+|x-2|+|x-4|)\,dx. \hspace{1.2cm} \text{[CBSE 2017; CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Here, }|x-1|,\ |x-2|\text{ and }|x-4|\text{ occurs.}
\displaystyle \text{Now, define the absolute function as}
\displaystyle |x-1|=\begin{cases}x-1,&x\geq 1\\-(x-1),&x<1\end{cases};\ |x-2|=\begin{cases}x-2,&x\geq 2\\-(x-2),&x<2\end{cases}
\displaystyle \text{and }|x-4|=\begin{cases}x-4,&x\geq 4\\-(x-4),&x<4\end{cases}
\displaystyle \text{Let }I=\int_1^4(|x-1|+|x-2|+|x-4|)\,dx
\displaystyle =\int_1^2(|x-1|+|x-2|+|x-4|)\,dx+\int_2^3(|x-1|+|x-2|+|x-4|)\,dx
\displaystyle +\int_3^4(|x-1|+|x-2|+|x-4|)\,dx
\displaystyle =\int_1^2\{(x-1)-(x-2)-(x-4)\}\,dx+\int_2^3\{(x-1)+(x-2)-(x-4)\}\,dx
\displaystyle +\int_3^4\{(x-1)+(x-2)-(x-4)\}\,dx
\displaystyle =\int_1^2(-x+5)\,dx+\int_2^3(x+1)\,dx+\int_3^4(x+1)\,dx
\displaystyle =\left(\frac{-x^2}{2}+5x\right)_1^2+\left(\frac{x^2}{2}+x\right)_2^3+\left(\frac{x^2}{2}+x\right)_3^4
\displaystyle =\left(\frac{-4}{2}+10\right)-\left(\frac{-1}{2}+5\right)+\left(\frac{9}{2}+3\right)-\left(\frac{4}{2}+2\right)+\left(\frac{16}{4}+4\right)-\left(\frac{9}{2}+3\right)
\displaystyle =8-\frac{9}{2}+\frac{15}{2}-4+12-\frac{15}{2}=16-\frac{9}{2}=\frac{23}{2}
\\

\displaystyle \textbf{Question 59. }\text{Evaluate }\int_0^\pi\frac{x\tan x}{\sec x+\tan x}\,dx. \hspace{0.2cm} \text{[CBSE 2017, 2008; CBSE 2014C, 2010; CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^\pi\frac{x\tan x}{\sec x+\tan x}\,dx\qquad\ldots\text{(i)}
\displaystyle \Rightarrow I=\int_0^\pi\frac{(\pi-x)\tan(\pi-x)}{\sec(\pi-x)+\tan(\pi-x)}\,dx\qquad\left[\because\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\right]
\displaystyle \Rightarrow I=\int_0^\pi\frac{(\pi-x)\tan x}{\sec x+\tan x}\,dx\qquad\ldots\text{(ii)}
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle \Rightarrow 2I=\int_0^\pi\frac{\pi\tan x}{\sec x+\tan x}\,dx
\displaystyle \Rightarrow I=\frac{\pi}{2}\int_0^\pi\frac{\tan x(\sec x-\tan x)}{(\sec x+\tan x)(\sec x-\tan x)}\,dx\qquad\text{[rationalising]}
\displaystyle =\frac{\pi}{2}\int_0^\pi\frac{(\tan x\sec x-\tan^2x)}{(\sec^2x-\tan^2x)}\,dx
\displaystyle [\because(a+b)(a-b)=a^2-b^2]
\displaystyle =\frac{\pi}{2}\int_0^\pi\frac{\tan x\sec x-\sec^2x+1}{1}\,dx
\displaystyle [\because\tan^2\theta=\sec^2\theta-1]
\displaystyle =\frac{\pi}{2}[\sec x-\tan x+x]_0^\pi
\displaystyle =\frac{\pi}{2}[(\sec\pi-\tan\pi+\pi)-(\sec 0-\tan 0+0)]
\displaystyle =\frac{\pi}{2}[\sec\pi-\sec 0-\tan\pi+\tan 0+\pi-0]
\displaystyle =\frac{\pi}{2}[-1-1-0+0+\pi-0]
\displaystyle =\frac{\pi}{2}[\pi-2]
\\

\displaystyle \textbf{Question 60. }\text{Evaluate }\int_{-2}^2\frac{x^2}{1+5^x}\,dx. \hspace{1.2cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-2}^2\frac{x^2}{1+5^x}\,dx\qquad\ldots\text{(i)}
\displaystyle =\int_{-2}^2\frac{(2-2-x)^2}{1+5^{2-2-x}}\,dx\qquad\left[\because\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx\right]
\displaystyle =\int_{-2}^2\frac{x^2}{1+5^{-x}}\,dx
\displaystyle \Rightarrow I=\int_{-2}^2\frac{5^x}{5^x+1}\cdot x^2\,dx\qquad\ldots\text{(ii)}
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle 2I=\int_{-2}^2\left(\frac{1+5^x}{5^x+1}\right)x^2\,dx=\int_{-2}^2 x^2\,dx
\displaystyle \Rightarrow 2I=2\int_0^2 x^2\,dx
\displaystyle \left[\because x^2\text{ is even, so }\int_{-2}^2 x^2\,dx=2\int_0^2 x^2\,dx\right]
\displaystyle \Rightarrow I=\left[\frac{x^3}{3}\right]_0^2
\displaystyle =\frac{1}{3}(2^3-0)=\frac{8}{3}
\\

\displaystyle \textbf{Question 61. }\text{Evaluate }\int_0^{\pi}e^{2x}\cdot\sin\!\left(\frac{\pi}{4}+x\right)dx. \hspace{1.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi}e^{2x}\sin\!\left(\frac{\pi}{4}+x\right)dx
\displaystyle \text{Again, let }I_1=\int e^{2x}\sin\!\left(\frac{\pi}{4}+x\right)dx\qquad\ldots\text{(i)}
\displaystyle =\sin\!\left(\frac{\pi}{4}+x\right)\int e^{2x}\,dx-\int\!\left\{\frac{d}{dx}\sin\!\left(\frac{\pi}{4}+x\right)\int e^{2x}\,dx\right\}dx\quad\text{[using integration by parts]}
\displaystyle =\sin\!\left(\frac{\pi}{4}+x\right)\cdot\frac{e^{2x}}{2}-\int\cos\!\left(\frac{\pi}{4}+x\right)\cdot\frac{e^{2x}}{2}\,dx
\displaystyle =\frac{e^{2x}}{2}\sin\!\left(\frac{\pi}{4}+x\right)-\frac{1}{2}\left[\cos\!\left(\frac{\pi}{4}+x\right)\cdot\frac{e^{2x}}{2}-\int\!\left(-\sin\!\left(\frac{\pi}{4}+x\right)\right)\frac{e^{2x}}{2}\,dx\right]\quad\text{[using integration by parts]}
\displaystyle =\frac{e^{2x}}{2}\sin\!\left(\frac{\pi}{4}+x\right)-\frac{e^{2x}}{4}\cos\!\left(\frac{\pi}{4}+x\right)-\frac{1}{4}\int e^{2x}\sin\!\left(\frac{\pi}{4}+x\right)dx\qquad\ldots\text{(1)}
\displaystyle \Rightarrow I_1=\frac{e^{2x}}{4}\left\{2\sin\!\left(\frac{\pi}{4}+x\right)-\cos\!\left(\frac{\pi}{4}+x\right)\right\}-\frac{1}{4}I_1\quad\text{[from Eq. (i)]}
\displaystyle \Rightarrow I_1+\frac{1}{4}I_1=\frac{e^{2x}}{4}\left\{2\sin\!\left(\frac{\pi}{4}+x\right)-\cos\!\left(\frac{\pi}{4}+x\right)\right\}
\displaystyle \Rightarrow \frac{5}{4}I_1=\frac{e^{2x}}{4}\left\{2\sin\!\left(\frac{\pi}{4}+x\right)-\cos\!\left(\frac{\pi}{4}+x\right)\right\}
\displaystyle \Rightarrow I_1=\frac{e^{2x}}{5}\left\{2\sin\!\left(\frac{\pi}{4}+x\right)-\cos\!\left(\frac{\pi}{4}+x\right)\right\}\qquad\ldots\text{(1)}
\displaystyle \therefore\ I=[I_1]_0^{\pi}=\left[\frac{e^{2x}}{5}\left\{2\sin\!\left(\frac{\pi}{4}+x\right)-\cos\!\left(\frac{\pi}{4}+x\right)\right\}\right]_0^{\pi}
\displaystyle =\frac{1}{5}\left[e^{2\pi}\left\{2\sin\!\left(\frac{\pi}{4}+\pi\right)-\cos\!\left(\frac{\pi}{4}+\pi\right)\right\}-e^{0}\left\{2\sin\!\left(\frac{\pi}{4}+0\right)-\cos\!\left(\frac{\pi}{4}+0\right)\right\}\right]
\displaystyle =\frac{1}{5}\left[e^{2\pi}\left\{-2\sin\frac{\pi}{4}+\cos\frac{\pi}{4}\right\}-e^{0}\left\{2\sin\frac{\pi}{4}-\cos\frac{\pi}{4}\right\}\right]
\displaystyle =\frac{1}{5}\left[e^{2\pi}\left\{-2\times\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right\}-1\cdot\left\{2\times\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right\}\right]
\displaystyle =\frac{1}{5}\left[e^{2\pi}\left\{-\frac{1}{\sqrt{2}}\right\}-\frac{1}{\sqrt{2}}\right]
\displaystyle =-\frac{1}{5\sqrt{2}}\left[e^{2\pi}+1\right]
\\

\displaystyle \textbf{Question 62. }\text{Evaluate }\int_0^{3/2}|x\cos\pi x|\,dx. \hspace{1.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{3/2}|x\cos\pi x|\,dx
\displaystyle \text{Consider }x\cos\pi x=0\Rightarrow x=0\text{ or }\cos\pi x=0
\displaystyle \Rightarrow x=0\text{ or }\pi x=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2}
\displaystyle \Rightarrow x=0\text{ or }x=\frac{1}{2},\frac{3}{2},\frac{5}{2}\qquad\left[\because\cos\frac{n\pi}{2}=0,\ n\text{ being an odd integer}\right]
\displaystyle \Rightarrow x=0\text{ or }x=\frac{1}{2},\frac{3}{2}\qquad\left[\because 0<x<\frac{3}{2}\right]
\displaystyle \Rightarrow x=0,\frac{1}{2},\frac{3}{2}
\displaystyle \text{So, let us divide the integral at }x=\frac{1}{2}
\displaystyle \text{Note that }|x\cos\pi x|=\begin{cases}x\cos\pi x, & 0\leq x\leq\dfrac{1}{2}\\[6pt]-x\cos\pi x, & \dfrac{1}{2}<x\leq\dfrac{3}{2}\end{cases}
\displaystyle \left[\because 0\leq x\leq\frac{1}{2}\Rightarrow 0\leq\pi x\leq\frac{\pi}{2}\Rightarrow\cos\pi x\geq 0\text{ and }\frac{1}{2}<x\leq\frac{3}{2}\Rightarrow\frac{\pi}{2}\leq\pi x\leq\frac{3\pi}{2}\Rightarrow\cos\pi x\leq 0\right]
\displaystyle \text{Now, }I=\int_0^{1/2}|x\cos\pi x|\,dx+\int_{1/2}^{3/2}|x\cos\pi x|\,dx
\displaystyle =\int_0^{1/2}x\cos\pi x\,dx-\int_{1/2}^{3/2}x\cos\pi x\,dx\qquad\ldots\text{(i)}
\displaystyle \text{Let }I_1=\int x\cos\pi x\,dx=x\cdot\frac{\sin\pi x}{\pi}-\int\frac{\sin\pi x}{\pi}\,dx\quad\text{[using integration by parts]}
\displaystyle =\frac{x\sin\pi x}{\pi}+\frac{\cos\pi x}{\pi^2}\qquad\ldots\text{(1)}
\displaystyle \text{Now, from Eq. (i), we have}
\displaystyle I=\left[\frac{x\sin\pi x}{\pi}+\frac{\cos\pi x}{\pi^2}\right]_0^{1/2}-\left[\frac{x\sin\pi x}{\pi}+\frac{\cos\pi x}{\pi^2}\right]_{1/2}^{3/2}
\displaystyle =\left[\left(\frac{1}{2\pi}\sin\frac{\pi}{2}+\frac{1}{\pi^2}\cos\frac{\pi}{2}\right)-\left(0+\frac{\cos 0}{\pi^2}\right)\right]
\displaystyle -\left[\left(\frac{3}{2\pi}\sin\frac{3\pi}{2}+\frac{1}{\pi^2}\cos\frac{3\pi}{2}\right)-\left(\frac{1}{2\pi}\sin\frac{\pi}{2}+\frac{1}{\pi^2}\cos\frac{\pi}{2}\right)\right]
\displaystyle =\frac{1}{2\pi}-\frac{1}{\pi^2}-\left(\frac{-3}{2\pi}-\frac{1}{2\pi}\right)
\displaystyle =\frac{1}{2\pi}-\frac{1}{\pi^2}+\frac{3}{2\pi}+\frac{1}{2\pi}=\frac{5}{2\pi}-\frac{1}{\pi^2}=\frac{5\pi-2}{2\pi^2}
\\

\displaystyle \textbf{Question 63. }\text{Evaluate }\int_0^{\pi}\frac{x}{1+\sin\alpha\sin x}\,dx. \hspace{1.2cm}\text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi}\frac{x}{1+\sin\alpha\sin x}\,dx\qquad\ldots\text{(i)}
\displaystyle =\int_0^{\pi}\frac{(\pi-x)}{1+\sin\alpha\sin(\pi-x)}\,dx\qquad\left[\because\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\right]
\displaystyle =\int_0^{\pi}\frac{(\pi-x)}{1+\sin\alpha\sin x}\,dx\qquad\ldots\text{(ii)}\qquad[\because\sin(\pi-x)=\sin x]
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle 2I=\int_0^{\pi}\frac{\pi}{1+\sin\alpha\sin x}\,dx=\pi\int_0^{\pi}\frac{dx}{1+\sin\alpha\sin x}
\displaystyle =\pi\int_0^{\pi}\frac{dx}{1+\sin\alpha\!\left(\dfrac{2\tan(x/2)}{1+\tan^2(x/2)}\right)}
\displaystyle =\pi\int_0^{\pi}\frac{1+\tan^2(x/2)}{1+\tan^2(x/2)+2\sin\alpha\tan(x/2)}\,dx
\displaystyle =\pi\int_0^{\pi}\frac{\sec^2(x/2)}{\tan^2(x/2)+2\sin\alpha\cdot\tan(x/2)+1}\,dx
\displaystyle \text{Put }\tan\frac{x}{2}=t\Rightarrow\sec^2\frac{x}{2}\cdot\frac{1}{2}\,dx=dt\Rightarrow\sec^2\frac{x}{2}\,dx=2\,dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=0
\displaystyle \text{Upper limit: When }x=\pi\text{, then }t\to\infty
\displaystyle \therefore\ 2I=\pi\int_0^{\infty}\frac{2\,dt}{t^2+2\sin\alpha\cdot t+1}
\displaystyle \Rightarrow I=\pi\int_0^{\infty}\frac{dt}{t^2+2\sin\alpha\cdot t+\sin^2\alpha+\cos^2\alpha}\qquad\left[\because\sin^2\theta+\cos^2\theta=1\right]
\displaystyle =\pi\int_0^{\infty}\frac{dt}{(t+\sin\alpha)^2+(\cos\alpha)^2}
\displaystyle =\frac{\pi}{\cos\alpha}\left[\tan^{-1}\!\left(\frac{t+\sin\alpha}{\cos\alpha}\right)\right]_0^{\infty}\qquad\left[\because\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\!\left(\frac{x}{a}\right)+C\right]
\displaystyle =\frac{\pi}{\cos\alpha}\left[\tan^{-1}(\infty)-\tan^{-1}(\tan\alpha)\right]
\displaystyle =\frac{\pi}{\cos\alpha}\left[\frac{\pi}{2}-\alpha\right]
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\displaystyle \textbf{Question 64. }\text{Evaluate }\int_{-\pi}^{\pi}(\cos ax-\sin bx)^2\,dx. \hspace{1.2cm}\text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{-\pi}^{\pi}(\cos ax-\sin bx)^2\,dx
\displaystyle =\int_{-\pi}^{\pi}(\cos^2 ax+\sin^2 bx-2\cos ax\sin bx)\,dx
\displaystyle =\int_{-\pi}^{\pi}(\cos^2 ax+\sin^2 bx)\,dx-2\int_{-\pi}^{\pi}\cos ax\sin bx\,dx
\displaystyle =I_1-I_2
\displaystyle \text{Now consider, }I_1=\int_{-\pi}^{\pi}(\cos^2 ax+\sin^2 bx)\,dx\quad\text{[be an even function]}
\displaystyle =2\int_0^{\pi}(\cos^2 ax+\sin^2 bx)\,dx\qquad\left[\because\int_{-a}^{a}f(x)\,dx=2\int_0^a f(x)\,dx,\text{ if }f(x)\text{ is even}\right]
\displaystyle =2\int_0^{\pi}\!\left(\frac{1+\cos 2ax}{2}+\frac{1-\cos 2bx}{2}\right)dx
\displaystyle =\int_0^{\pi}(1+\cos 2ax+1-\cos 2bx)\,dx
\displaystyle =\int_0^{\pi}(2+\cos 2ax-\cos 2bx)\,dx
\displaystyle =\left[2x+\frac{\sin 2ax}{2a}-\frac{\sin 2bx}{2b}\right]_0^{\pi}
\displaystyle =\left(2\pi+\frac{\sin 2a\pi}{2a}-\frac{\sin 2b\pi}{2b}\right)-0
\displaystyle =2\pi+\frac{\sin 2a\pi}{2a}-\frac{\sin 2b\pi}{2b}
\displaystyle \text{and }I_2=2\int_{-\pi}^{\pi}(\cos ax\sin bx)\,dx\quad\text{[be an odd function]}
\displaystyle =0\qquad\left[\because\int_{-a}^{a}f(x)\,dx=\begin{cases}2\int_0^a f(x)\,dx,&\text{if }f(x)\text{ is even}\\0,&\text{if }f(x)\text{ is odd}\end{cases}\right]
\displaystyle \therefore\ I=I_1-I_2=2\pi+\frac{\sin 2a\pi}{2a}-\frac{\sin 2b\pi}{2b}
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\displaystyle \textbf{Question 65. }\text{Evaluate }\int_0^1 xe^{x^2}\,dx. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^1 xe^{x^2}\,dx
\displaystyle \text{Put }x^2=t\Rightarrow 2x\,dx=dt\Rightarrow dx=\frac{dt}{2x}
\displaystyle \text{Lower limit: When }x=0\text{, then }t=0
\displaystyle \text{Upper limit: When }x=1\text{, then }t=1
\displaystyle \therefore\ I=\int_0^1 xe^t\frac{dt}{2x}=\frac{1}{2}\int_0^1 e^t\,dt
\displaystyle =\frac{1}{2}[e^t]_0^1=\frac{1}{2}[e^1-e^0]=\frac{1}{2}[e-1]
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\displaystyle \textbf{Question 66. }\text{Evaluate }\int_0^{\pi/4}\sin 2x\,dx. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi/4}\sin 2x\,dx
\displaystyle =\left[\frac{-\cos 2x}{2}\right]_0^{\pi/4}=-\frac{1}{2}[\cos 2x]_0^{\pi/4}
\displaystyle =-\frac{1}{2}\left[\cos 2\left(\frac{\pi}{4}\right)-\cos 0\right]
\displaystyle =-\frac{1}{2}\left[\cos\frac{\pi}{2}-1\right]=-\frac{1}{2}[0-1]=\frac{1}{2}
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\displaystyle \textbf{Question 67. }\text{Evaluate }\int_0^1\frac{1}{\sqrt{1-x^2}}\,dx. \hspace{1.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^1\frac{1}{\sqrt{1-x^2}}\,dx
\displaystyle =[\sin^{-1}x]_0^1\qquad\left[\because\int\frac{1}{\sqrt{1-x^2}}\,dx=\sin^{-1}x+C\right]
\displaystyle =\sin^{-1}1-\sin^{-1}0
\displaystyle =\sin^{-1}\left(\sin\frac{\pi}{2}\right)-\sin^{-1}(\sin 0)=\frac{\pi}{2}-0=\frac{\pi}{2}
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\displaystyle \textbf{Question 68. }\text{If }\int_0^a\frac{1}{4+x^2}\,dx=\frac{\pi}{8}\text{, then find the value of }a. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\int_0^a\frac{1}{4+x^2}\,dx=\frac{\pi}{8}\qquad\ldots\text{(i)}
\displaystyle \text{Now, consider }I=\int_0^a\frac{1}{x^2+(2)^2}\,dx\Rightarrow I=\left[\frac{1}{2}\tan^{-1}\frac{x}{2}\right]_0^a
\displaystyle \left[\because\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C\right]
\displaystyle \Rightarrow I=\frac{1}{2}\tan^{-1}\frac{a}{2}-\frac{1}{2}\tan^{-1}(0)
\displaystyle \Rightarrow I=\frac{1}{2}\tan^{-1}\frac{a}{2}\qquad\ldots\text{(ii)}
\displaystyle \text{From Eqs. (i) and (ii), we get}
\displaystyle \frac{1}{2}\tan^{-1}\frac{a}{2}=\frac{\pi}{8}\Rightarrow\tan^{-1}\frac{a}{2}=\frac{\pi}{4}
\displaystyle \Rightarrow\frac{a}{2}=\tan\frac{\pi}{4}\Rightarrow\frac{a}{2}=1\Rightarrow a=2\qquad\left[\because\tan\frac{\pi}{4}=1\right]
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\displaystyle \textbf{Question 69. }\text{If }f(x)=\int_0^x t\sin t\,dt\text{, then write the value of }f'(x). \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }f(x)=\int_0^x t\sin t\,dt
\displaystyle =\left[t\int\sin t\,dt-\int\left\{\frac{d}{dt}(t)\int\sin t\,dt\right\}dt\right]_0^x\qquad\text{[using integration by parts]}
\displaystyle =[t(-\cos t)]_0^x-\int_0^x(-\cos t)\,dt
\displaystyle =[-t\cos t]_0^x+[\sin t]_0^x
\displaystyle =-x\cos x+0+\sin x-0=\sin x-x\cos x
\displaystyle \text{Thus, }f(x)=\sin x-x\cos x
\displaystyle \text{On differentiating both sides w.r.t. }x\text{, we get}
\displaystyle f'(x)=\cos x-\left[x\frac{d}{dx}(\cos x)+\cos x\frac{d}{dx}(x)\right]\qquad\text{[by product rule of derivative]}
\displaystyle =\cos x-[x(-\sin x)+\cos x]
\displaystyle =\cos x+x\sin x-\cos x=x\sin x
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\displaystyle \textbf{Question 70. }\text{Evaluate }\int_2^4\frac{x}{x^2+1}\,dx. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_2^4\frac{x}{x^2+1}\,dx
\displaystyle \text{Put }x^2+1=t\Rightarrow 2x\,dx=dt\Rightarrow x\,dx=\frac{dt}{2}
\displaystyle \text{Lower limit: When }x=2\text{, then }t=2^2+1=5
\displaystyle \text{Upper limit: When }x=4\text{, then }t=4^2+1=17
\displaystyle \therefore\ I=\int_5^{17}\frac{1}{t}\cdot\frac{dt}{2}=\frac{1}{2}\int_5^{17}\frac{dt}{t}=\frac{1}{2}[\log|t|]_5^{17}
\displaystyle =\frac{1}{2}[\log 17-\log 5]=\frac{1}{2}\log\left(\frac{17}{5}\right)
\displaystyle \left[\because\log m-\log n=\log\left(\frac{m}{n}\right)\right]
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\displaystyle \textbf{Question 71. }\text{Evaluate }\int_0^3\frac{dx}{9+x^2}. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^3\frac{dx}{9+x^2}\Rightarrow I=\int_0^3\frac{dx}{x^2+(3)^2}
\displaystyle \Rightarrow I=\left[\frac{1}{3}\tan^{-1}\frac{x}{3}\right]_0^3\qquad\left[\because\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C\right]
\displaystyle \Rightarrow I=\frac{1}{3}\left[\tan^{-1}\left(\frac{3}{3}\right)-\tan^{-1}(0)\right]
\displaystyle =\frac{1}{3}[\tan^{-1}(1)-0]=\frac{1}{3}\left(\frac{\pi}{4}\right)=\frac{\pi}{12}
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\displaystyle \textbf{Question 72. }\text{Evaluate }\int_0^{\pi/2}e^x(\sin x-\cos x)\,dx. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi/2}e^x(\sin x-\cos x)\,dx
\displaystyle \Rightarrow I=-\int_0^{\pi/2}e^x(\cos x-\sin x)\,dx
\displaystyle \text{Now consider, }f(x)=\cos x
\displaystyle \text{Then, }f'(x)=-\sin x
\displaystyle \text{Now, by using }\int e^x[f(x)+f'(x)]\,dx=e^x f(x)+C\text{,}
\displaystyle \text{we get }I=-[e^x\cos x]_0^{\pi/2}
\displaystyle =-e^{\pi/2}\cos\frac{\pi}{2}+e^0\cos(0)=0+1(1)=1
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\displaystyle \textbf{Question 73. }\text{Evaluate }\int_e^{e^2}\frac{dx}{x\log x}. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_e^{e^2}\frac{dx}{x\log x}
\displaystyle \text{Put }\log x=t\Rightarrow\frac{1}{x}\,dx=dt
\displaystyle \text{Lower limit: When }x=e\text{, then }t=\log e=1
\displaystyle \text{Upper limit: When }x=e^2\text{, then }t=\log e^2=2
\displaystyle \therefore\ I=\int_1^2\frac{dt}{t}=[\log|t|]_1^2=\log 2-\log 1=\log 2
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\displaystyle \textbf{Question 74. }\text{Evaluate }\int_0^1\frac{\tan^{-1}x}{1+x^2}\,dx. \hspace{1.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^1\frac{\tan^{-1}x}{1+x^2}\,dx
\displaystyle \text{Put }\tan^{-1}x=t\Rightarrow\frac{1}{1+x^2}\,dx=dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=0
\displaystyle \text{Upper limit: When }x=1\text{, then }t=\pi/4
\displaystyle \therefore\ I=\int_0^{\pi/4}t\,dt=\left[\frac{t^2}{2}\right]_0^{\pi/4}=\frac{1}{2}\left[\left(\frac{\pi}{4}\right)^2-(0)^2\right]=\frac{\pi^2}{32}
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\displaystyle \textbf{Question 75. }\text{Evaluate }\int_1^2\frac{x^3-1}{x^2}\,dx. \hspace{1.2cm} \text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_1^2\frac{x^3-1}{x^2}\,dx=\int_1^2\left(x-\frac{1}{x^2}\right)dx
\displaystyle =\left[\frac{x^2}{2}+\frac{1}{x}\right]_1^2=\left(\frac{(2)^2}{2}+\frac{1}{2}\right)-\left(\frac{(1)^2}{2}+1\right)
\displaystyle =\left(2+\frac{1}{2}\right)-\left(\frac{1}{2}+1\right)=1
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\displaystyle \textbf{Question 76. }\text{Evaluate }\int_0^{\pi/4}\tan x\,dx. \hspace{1.2cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi/4}\tan x\,dx=[\log|\sec x|]_0^{\pi/4}
\displaystyle =\log\left|\sec\frac{\pi}{4}\right|-\log|\sec 0|
\displaystyle =\log|\sqrt{2}|-\log|1|=\frac{1}{2}\log 2
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\displaystyle \textbf{Question 77. }\text{Evaluate }\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}\,dx. \hspace{1.2cm}\text{[CBSE 2014C; CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}\,dx
\displaystyle \text{Put }\sin x-\cos x=t\Rightarrow(\cos x+\sin x)\,dx=dt
\displaystyle \text{Lower limit: When }x=\frac{\pi}{6}\text{, then }t=\sin\frac{\pi}{6}-\cos\frac{\pi}{6}=\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2}
\displaystyle \text{Upper limit: When }x=\frac{\pi}{3}\text{, then }t=\sin\frac{\pi}{3}-\cos\frac{\pi}{3}=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}
\displaystyle \text{Also, }(\sin x-\cos x)^2=t^2\Rightarrow\sin^2 x+\cos^2 x-2\sin x\cos x=t^2
\displaystyle \Rightarrow 1-\sin 2x=t^2\Rightarrow\sin 2x=1-t^2
\displaystyle \therefore\ I=\int_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}}\frac{dt}{\sqrt{1-t^2}}=\left[\sin^{-1}t\right]_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}}
\displaystyle =\sin^{-1}\!\left(\frac{\sqrt{3}-1}{2}\right)-\sin^{-1}\!\left(\frac{1-\sqrt{3}}{2}\right)
\displaystyle =\sin^{-1}\!\left(\frac{\sqrt{3}-1}{2}\right)+\sin^{-1}\!\left(\frac{\sqrt{3}-1}{2}\right)\qquad[\because\sin^{-1}(-x)=-\sin^{-1}x]
\displaystyle =2\sin^{-1}\!\left(\frac{\sqrt{3}-1}{2}\right)
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\displaystyle \textbf{Question 78. }\text{Evaluate }\int_0^{\pi/2}x^2\sin x\,dx. \hspace{1.2cm}\text{[CBSE 2014C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi/2}x^2\sin x\,dx
\displaystyle \text{Now, }\int x^2\sin x\,dx=-x^2\cos x+2\int x\cos x\,dx\quad\text{[using integration by parts]}
\displaystyle =-x^2\cos x+2\left[x\sin x-\int 1\cdot(\sin x)\,dx\right]
\displaystyle =-x^2\cos x+2(x\sin x+\cos x)
\displaystyle \therefore\ I=\left[-x^2\cos x+2(x\sin x+\cos x)\right]_0^{\pi/2}
\displaystyle =\left[-\!\left(\frac{\pi}{2}\right)^2\cos\frac{\pi}{2}+2\!\left(\frac{\pi}{2}\sin\frac{\pi}{2}+\cos\frac{\pi}{2}\right)\right]-2(0+\cos 0)
\displaystyle =-\frac{\pi^2}{4}\times 0+2\!\left(\frac{\pi}{2}+0\right)-2(0+1)
\displaystyle =\pi-2
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\displaystyle \textbf{Question 79. }\text{Evaluate }\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx. \hspace{1.2cm}\text{[CBSE 2014, 2011; CBSE 2010C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}\,dx\qquad\ldots\text{(i)}
\displaystyle \text{Using }\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\text{, we get}
\displaystyle I=\int_0^{\pi/2}\frac{\left(\frac{\pi}{2}-x\right)\sin\!\left(\frac{\pi}{2}-x\right)\cos\!\left(\frac{\pi}{2}-x\right)}{\sin^4\!\left(\frac{\pi}{2}-x\right)+\cos^4\!\left(\frac{\pi}{2}-x\right)}\,dx
\displaystyle \Rightarrow I=\int_0^{\pi/2}\frac{\left(\frac{\pi}{2}-x\right)\cos x\sin x}{\cos^4 x+\sin^4 x}\,dx\qquad\ldots\text{(ii)}
\displaystyle \left[\because\cos\!\left(\frac{\pi}{2}-\theta\right)=\sin\theta\text{ and }\sin\!\left(\frac{\pi}{2}-\theta\right)=\cos\theta\right]
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle 2I=\frac{\pi}{2}\int_0^{\pi/2}\frac{\cos x\sin x}{\sin^4 x+\cos^4 x}\,dx
\displaystyle \Rightarrow I=\frac{\pi}{4}\int_0^{\pi/2}\frac{\sin x\cos x}{(\sin^2 x)^2+(1-\sin^2 x)^2}\,dx
\displaystyle \text{Put }\sin^2 x=t\Rightarrow 2\sin x\cos x\,dx=dt
\displaystyle \Rightarrow\sin x\cos x\,dx=\frac{dt}{2}
\displaystyle \text{Lower limit: When }x=0\text{, then }t=\sin 0=0
\displaystyle \text{Upper limit: When }x=\frac{\pi}{2}\text{, then }t=\sin^2\frac{\pi}{2}=1
\displaystyle \therefore\ I=\frac{\pi}{4}\int_0^1\frac{1}{t^2+(1-t)^2}\cdot\frac{dt}{2}
\displaystyle \Rightarrow I=\frac{\pi}{8}\int_0^1\frac{1}{t^2+(1+t^2-2t)}\,dt
\displaystyle \Rightarrow I=\frac{\pi}{8}\int_0^1\frac{1}{2t^2-2t+1}\,dt
\displaystyle \Rightarrow I=\frac{\pi}{16}\int_0^1\frac{1}{t^2-t+\frac{1}{2}}\,dt
\displaystyle \Rightarrow I=\frac{\pi}{16}\int_0^1\frac{1}{t^2-t+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2+\frac{1}{2}}\,dt
\displaystyle \Rightarrow I=\frac{\pi}{16}\int_0^1\frac{1}{\left(t-\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}\,dt
\displaystyle \Rightarrow I=\frac{\pi}{16}\cdot\frac{1}{\frac{1}{2}}\left[\tan^{-1}\frac{t-\frac{1}{2}}{\frac{1}{2}}\right]_0^1\qquad\left[\because\int\frac{1}{x^2+a^2}\,dx=\frac{1}{a}\tan^{-1}\frac{x}{a}+C\right]
\displaystyle \Rightarrow I=\frac{\pi}{8}\left[\tan^{-1}2\!\left(1-\frac{1}{2}\right)-\tan^{-1}2\!\left(0-\frac{1}{2}\right)\right]
\displaystyle \Rightarrow I=\frac{\pi}{8}\left[\tan^{-1}(1)-\tan^{-1}(-1)\right]
\displaystyle \left[\because\tan^{-1}(-1)=-\tan^{-1}(1)=-\frac{\pi}{4}\right]
\displaystyle \Rightarrow I=\frac{\pi}{8}\left[\frac{\pi}{4}+\frac{\pi}{4}\right]=\frac{\pi^2}{16}
\\

\displaystyle \textbf{Question 80. }\text{Evaluate }\int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx. \qquad \text{[CBSE 2014; CBSE 2014C, 2011]}
\displaystyle \text{Answer:}
\displaystyle \int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx
\displaystyle \text{Since }\sin 2x=1-(\sin x-\cos x)^{2}:
\displaystyle =\int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\left[1-(\sin x-\cos x)^{2}\right]}\,dx
\displaystyle =\int_{0}^{\pi/4}\frac{\sin x+\cos x}{25-16(\sin x-\cos x)^{2}}\,dx
\displaystyle \text{Let }t=\sin x-\cos x,\text{ so }dt=(\cos x+\sin x)\,dx
\displaystyle \text{When }x=0,\ t=-1;\quad\text{when }x=\frac{\pi}{4},\ t=0
\displaystyle =\int_{-1}^{0}\frac{dt}{25-16t^{2}}
\displaystyle =\frac{1}{16}\int_{-1}^{0}\frac{dt}{\frac{25}{16}-t^{2}}
\displaystyle =\frac{1}{16}\int_{-1}^{0}\frac{dt}{\left(\frac{5}{4}\right)^{2}-t^{2}}
\displaystyle =\frac{1}{16}\cdot\frac{1}{2\cdot\frac{5}{4}}\left[\log\left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|\right]_{-1}^{0}
\displaystyle =\frac{1}{40}\left[\log\left|\frac{5+4t}{5-4t}\right|\right]_{-1}^{0}
\displaystyle =\frac{1}{40}\left[\log\left|\frac{5}{5}\right|-\log\left|\frac{5-4}{5+4}\right|\right]
\displaystyle =\frac{1}{40}\left[\log 1-\log\frac{1}{9}\right]
\displaystyle =\frac{1}{40}\left[0+\log 9\right]
\displaystyle \therefore \int_{0}^{\pi/4}\frac{\sin x+\cos x}{9+16\sin 2x}\,dx=\frac{\log 9}{40}
\\

\displaystyle \textbf{Question 81. }\text{Evaluate }\int_0^{\pi}\frac{x}{a^2\cos^2 x+b^2\sin^2 x}\,dx. \hspace{1.2cm}\text{[CBSE 2014; CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi}\frac{x}{a^2\cos^2 x+b^2\sin^2 x}\,dx\qquad\ldots\text{(i)}
\displaystyle \Rightarrow I=\int_0^{\pi}\frac{(\pi-x)}{a^2\cos^2(\pi-x)+b^2\sin^2(\pi-x)}\,dx\qquad\left[\because\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx\right]
\displaystyle \Rightarrow I=\int_0^{\pi}\frac{(\pi-x)}{a^2\cos^2 x+b^2\sin^2 x}\,dx\qquad\ldots\text{(ii)}
\displaystyle \text{On adding Eqs. (i) and (ii), we get}
\displaystyle 2I=\int_0^{\pi}\frac{(x+\pi-x)}{a^2\cos^2 x+b^2\sin^2 x}\,dx
\displaystyle \Rightarrow 2I=\pi\int_0^{\pi}\frac{dx}{a^2\cos^2 x+b^2\sin^2 x}
\displaystyle \text{Now, we know that }\int_0^{2a}f(x)\,dx=2\int_0^a f(x)\,dx\text{, if }f(2a-x)=f(x)
\displaystyle \text{Here, }a^2\cos^2(\pi-x)+b^2\sin^2(\pi-x)=a^2\cos^2 x+b^2\sin^2 x
\displaystyle \therefore\ 2I=2\pi\int_0^{\pi/2}\frac{dx}{a^2\cos^2 x+b^2\sin^2 x}
\displaystyle \text{On dividing numerator and denominator by }\cos^2 x\text{, we get}
\displaystyle 2I=2\pi\int_0^{\pi/2}\frac{\sec^2 x}{a^2+b^2\tan^2 x}\,dx
\displaystyle \text{Put }\tan x=t\Rightarrow\sec^2 x\,dx=dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=\tan 0=0
\displaystyle \text{Upper limit: When }x=\frac{\pi}{2}\text{, then }t=\tan\frac{\pi}{2}=\infty
\displaystyle \therefore\ I=\pi\int_0^{\infty}\frac{dt}{a^2+b^2t^2}=\frac{\pi}{b^2}\int_0^{\infty}\frac{dt}{\left(\frac{a}{b}\right)^2+t^2}
\displaystyle \Rightarrow I=\frac{\pi}{ab}\left[\tan^{-1}\frac{bt}{a}\right]_0^{\infty}\qquad\left[\because\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C\right]
\displaystyle \Rightarrow I=\frac{\pi}{ab}\left[\tan^{-1}\infty-\tan^{-1}0\right]
\displaystyle \Rightarrow I=\frac{\pi}{ab}\left[\frac{\pi}{2}-0\right]\qquad\left[\because\tan^{-1}\infty=\tan^{-1}\!\left(\tan\frac{\pi}{2}\right)=\frac{\pi}{2}\text{ and }\tan^{-1}0=\tan^{-1}(\tan 0^\circ)=0\right]
\displaystyle \therefore\ I=\frac{\pi^2}{2ab}
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\displaystyle \textbf{Question 82. }\text{Evaluate }\int_2^5\left[|x-2|+|x-3|+|x-5|\right]dx. \hspace{1.2cm}\text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{For }2\leq x<5,\ |x-2|=x-2
\displaystyle 2\leq x<3,\ |x-3|=-x-3
\displaystyle 3\leq x<5,\ |x-3|=x-3
\displaystyle \text{and }2\leq x<5,\ |x-5|=5-x
\displaystyle \therefore\ I=\int_2^5\left[|x-2|+|x-3|+|x-5|\right]dx
\displaystyle =\int_2^3(x-2)\,dx+\int_2^3(3-x)\,dx+\int_3^5(x-3)\,dx+\int_2^5(5-x)\,dx
\displaystyle =\left[\frac{x^2}{2}-2x\right]_2^3+\left[3x-\frac{x^2}{2}\right]_2^3+\left[\frac{x^2}{2}-3x\right]_3^5+\left[5x-\frac{x^2}{2}\right]_2^5
\displaystyle =\left[\left(\frac{25}{2}-10\right)-(2-4)\right]+\left[\left(9-\frac{9}{2}\right)-(6-2)\right]
\displaystyle +\left[\left(\frac{25}{2}-15\right)-\left(\frac{9}{2}-9\right)\right]+\left[\left(25-\frac{25}{2}\right)-(10-2)\right]
\displaystyle =\left[\frac{25}{2}-8\right]+\left[\frac{9}{2}-4\right]+[8-6]+\left[\frac{25}{2}-8\right]
\displaystyle =\frac{9}{2}+\frac{1}{2}+2+\frac{9}{2}=\frac{19}{2}+2=\frac{23}{2}
\\

\displaystyle \textbf{Question 83. }\text{Evaluate }\int_{0}^{4}\left[|x|+|x-2|+|x-4|\right]dx. \qquad \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \int_{0}^{4}\left[|x|+|x-2|+|x-4|\right]dx
\displaystyle \text{For }0\leq x\leq 4:|x|=x
\displaystyle \text{For }|x-2|=\begin{cases}2-x, & 0\leq x<2\\x-2, & 2\leq x\leq 4\end{cases}
\displaystyle \text{For }|x-4|=\begin{cases}4-x, & 0\leq x<4\\x-4, & x=4\end{cases}
\displaystyle =\int_{0}^{2}\left[x+(2-x)+(4-x)\right]dx+\int_{2}^{4}\left[x+(x-2)+(4-x)\right]dx
\displaystyle =\int_{0}^{2}(6-x)\,dx+\int_{2}^{4}(x+2)\,dx
\displaystyle =\left[6x-\frac{x^{2}}{2}\right]_{0}^{2}+\left[\frac{x^{2}}{2}+2x\right]_{2}^{4}
\displaystyle =\left(12-2\right)-0+\left(8+8\right)-\left(2+4\right)
\displaystyle =10+16-6
\displaystyle =20
\displaystyle \therefore \int_{0}^{4}\left[|x|+|x-2|+|x-4|\right]dx=20
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\displaystyle \textbf{Question 84. }\text{Evaluate }\int_{1}^{3}\left[|x-1|+|x-2|+|x-3|\right]dx. \qquad \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \int_{1}^{3}\left[|x-1|+|x-2|+|x-3|\right]dx
\displaystyle \text{For }|x-1|=\begin{cases}x-1, & 1\leq x\leq 3\end{cases}
\displaystyle \text{For }|x-2|=\begin{cases}2-x, & 1\leq x<2\\x-2, & 2\leq x\leq 3\end{cases}
\displaystyle \text{For }|x-3|=\begin{cases}3-x, & 1\leq x<3\\x-3, & x=3\end{cases}
\displaystyle =\int_{1}^{2}\left[(x-1)+(2-x)+(3-x)\right]dx+\int_{2}^{3}\left[(x-1)+(x-2)+(3-x)\right]dx
\displaystyle =\int_{1}^{2}(4-x)\,dx+\int_{2}^{3}(x)\,dx
\displaystyle =\left[4x-\frac{x^{2}}{2}\right]_{1}^{2}+\left[\frac{x^{2}}{2}\right]_{2}^{3}
\displaystyle =\left(8-2\right)-\left(4-\frac{1}{2}\right)+\frac{9}{2}-2
\displaystyle =6-\frac{7}{2}+\frac{9}{2}-2
\displaystyle =4+\frac{2}{2}
\displaystyle =4+1=5
\displaystyle \therefore \int_{1}^{3}\left[|x-1|+|x-2|+|x-3|\right]dx=5
\\

\displaystyle \textbf{Question 85. }\text{Evaluate }\int_0^2\sqrt{4-x^2}\,dx. \hspace{1.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \int_0^2\sqrt{4-x^2}\,dx=\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\sin^{-1}\frac{x}{2}\right]_0^2
\displaystyle \left[\because\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C\right]
\displaystyle =0+2\sin^{-1}1=2\sin^{-1}\left(\sin\frac{\pi}{2}\right)=2\times\frac{\pi}{2}=\pi
\\

\displaystyle \textbf{Question 86. }\text{Write the value of }\int_0^1\frac{e^x}{1+e^{2x}}\,dx. \hspace{1.2cm} \text{[CBSE 2012C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^1\frac{e^x}{1+e^{2x}}\,dx=\int_0^1\frac{e^x}{1+(e^x)^2}\,dx
\displaystyle \text{Put }e^x=t\Rightarrow e^x\,dx=dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=1
\displaystyle \text{Upper limit: When }x=1\text{, then }t=e
\displaystyle \text{Now, }I=\int_1^e\frac{dt}{1+t^2}=(\tan^{-1}t)_1^e
\displaystyle =\tan^{-1}e-\tan^{-1}1
\displaystyle =\tan^{-1}\left(\frac{e-1}{1+e}\right)
\displaystyle \left[\because\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)\right]
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\displaystyle \textbf{Question 87. }\text{Evaluate }\int_2^3\frac{1}{x}\,dx. \hspace{1.2cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \int_2^3\frac{1}{x}\,dx=[\log|x|]_2^3=\log 3-\log 2=\log\frac{3}{2}
\displaystyle \left[\because\log m-\log n=\log\frac{m}{n}\right]
\\

\displaystyle \textbf{Question 88. }\text{Prove that }\int_0^{\pi/4}\left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx=\sqrt{2}\cdot\frac{\pi}{2}. \hspace{1.2cm}\text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi/4}\left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx
\displaystyle =\int_0^{\pi/4}\left(\frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}}\right)dx
\displaystyle =\int_0^{\pi/4}\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}\,dx
\displaystyle =\sqrt{2}\int_0^{\pi/4}\frac{\sin x+\cos x}{\sqrt{2\sin x\cos x}}\,dx
\displaystyle =\sqrt{2}\int_0^{\pi/4}\frac{\sin x+\cos x}{\sqrt{1+2\sin x\cos x-1}}\,dx
\displaystyle =\sqrt{2}\int_0^{\pi/4}\frac{\sin x+\cos x}{\sqrt{1-(1-2\sin x\cos x)}}\,dx
\displaystyle =\sqrt{2}\int_0^{\pi/4}\frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}}\,dx
\displaystyle \text{Now, put }\sin x-\cos x=t\Rightarrow(\cos x+\sin x)\,dx=dt
\displaystyle \text{Lower limit: When }x=0\text{, then }t=-1
\displaystyle \text{Upper limit: When }x=\frac{\pi}{4}\text{, then }t=0
\displaystyle \therefore\ I=\sqrt{2}\int_{-1}^{0}\frac{dt}{\sqrt{1-t^2}}=\sqrt{2}\left[\sin^{-1}t\right]_{-1}^{0}
\displaystyle =\sqrt{2}\left[\sin^{-1}(0)-\sin^{-1}(-1)\right]
\displaystyle =\sqrt{2}\left[\sin^{-1}(0)+\sin^{-1}(1)\right]
\displaystyle =\sqrt{2}\left[\frac{\pi}{2}\right]\qquad\textbf{Hence proved.}
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\displaystyle \textbf{Question 89. }\text{Evaluate }\int_{0}^{1} \frac{2x}{1+x^{2}}\,dx. \qquad \text{[CBSE 2011C, 2008]}
\displaystyle \text{Answer:}
\displaystyle \int_{0}^{1} \frac{2x}{1+x^{2}}\,dx
\displaystyle \text{Let }1+x^{2}=t,\text{ so }2x\,dx=dt
\displaystyle \text{When }x=0,\ t=1;\quad\text{when }x=1,\ t=2
\displaystyle =\int_{1}^{2}\frac{dt}{t}
\displaystyle =\Big[\log|t|\Big]_{1}^{2}
\displaystyle =\log 2-\log 1
\displaystyle =\log 2-0
\displaystyle \therefore \int_{0}^{1} \frac{2x}{1+x^{2}}\,dx=\log 2
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\displaystyle \textbf{Question 90. }\text{Evaluate }\int_0^1\frac{x^4+1}{x^2+1}\,dx. \hspace{1.2cm}\text{[CBSE 2011C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^1\frac{x^4+1}{x^2+1}\,dx\Rightarrow I=\int_0^1\frac{(x^4-1)+2}{x^2+1}\,dx
\displaystyle =\int_0^1\frac{(x^2-1)(x^2+1)+2}{x^2+1}\,dx\qquad[\because(a^2-b^2)=(a-b)(a+b)]
\displaystyle =\int_0^1\left[\frac{(x^2-1)(x^2+1)}{x^2+1}+\frac{2}{x^2+1}\right]dx
\displaystyle \Rightarrow I=\int_0^1\left[x^2-1+\frac{2}{x^2+1}\right]dx
\displaystyle \Rightarrow I=\left[\frac{x^3}{3}-x+2\tan^{-1}x\right]_0^1
\displaystyle \therefore\ I=\frac{1}{3}-1+2\tan^{-1}1-0
\displaystyle =-\frac{2}{3}+2\times\frac{\pi}{4}=\frac{3\pi-4}{6}
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\displaystyle \textbf{Question 91. }\text{Evaluate }\int_0^{\pi/2}\frac{x+\sin x}{1+\cos x}\,dx. \hspace{1.2cm}\text{[CBSE 2011]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_0^{\pi/2}\frac{x+\sin x}{1+\cos x}\,dx
\displaystyle \Rightarrow I=\int_0^{\pi/2}\frac{x+2\sin\frac{x}{2}\cdot\cos\frac{x}{2}}{2\cos^2\frac{x}{2}}\,dx\qquad\left[\because\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\text{ and }1+\cos\theta=2\cos^2\frac{\theta}{2}\right]
\displaystyle \Rightarrow I=\frac{1}{2}\int_0^{\pi/2}x\sec^2\frac{x}{2}\,dx+\int_0^{\pi/2}\tan\frac{x}{2}\,dx
\displaystyle \Rightarrow I=\frac{1}{2}\left[\left[x\int\sec^2\frac{x}{2}\,dx\right]_0^{\pi/2}-\int_0^{\pi/2}\left[\frac{d}{dx}(x)\int\!\left(\sec^2\frac{x}{2}\,dx\right)\right]dx\right]+\int_0^{\pi/2}\tan\frac{x}{2}\,dx
\displaystyle \text{[using integration by parts]}
\displaystyle \Rightarrow I=\frac{1}{2}\left[\left[x\cdot\frac{\tan\frac{x}{2}}{\frac{1}{2}}\right]_0^{\pi/2}-\int_0^{\pi/2}\tan\frac{x}{2}\,dx\right]+\int_0^{\pi/2}\tan\frac{x}{2}\,dx
\displaystyle =\left[x\cdot\tan\frac{x}{2}\right]_0^{\pi/2}-\int_0^{\pi/2}\tan\frac{x}{2}\,dx+\int_0^{\pi/2}\tan\frac{x}{2}\,dx
\displaystyle =\frac{\pi}{2}\cdot\tan\frac{\pi}{4}-0
\displaystyle \therefore\ I=\frac{\pi}{2}\qquad\left[\because\tan\frac{\pi}{4}=1\right]
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