\displaystyle \textbf{Question 1: }\quad \text{Evaluate:}
\displaystyle \int \frac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha}\,dx \hspace{8.0cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \frac{\cos 2x-\cos 2\alpha}{\cos x-\cos \alpha}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{(2\cos^{2}x-1)-(2\cos^{2}\alpha-1)}{\cos x-\cos \alpha}\,dx
\displaystyle \Rightarrow\ I=\int \frac{2(\cos^{2}x-\cos^{2}\alpha)}{\cos x-\cos \alpha}\,dx
\displaystyle \Rightarrow\ I=2\int (\cos x+\cos \alpha)\,dx=2\int \cos x\,dx+2\int \cos \alpha\,dx
\displaystyle \Rightarrow\ I=2\int \cos x\,dx+2\cos \alpha\int 1\,dx=2\sin x+2x\cos \alpha+C

\displaystyle \textbf{Question 2: }\quad \text{Evaluate:}
\displaystyle \int \frac{\sin^{6}x+\cos^{6}x}{\sin^{2}x\cos^{2}x}\,dx \hspace{8.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \frac{\sin^{6}x+\cos^{6}x}{\sin^{2}x\cos^{2}x}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{(\sin^{2}x+\cos^{2}x)^{3}-3\sin^{2}x\cos^{2}x(\sin^{2}x+\cos^{2}x)}{\sin^{2}x\cos^{2}x}\,dx
\displaystyle \text{Using: }a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)
\displaystyle \Rightarrow\ I=\int \frac{1-3\sin^{2}x\cos^{2}x}{\sin^{2}x\cos^{2}x}\,dx=\int \left\{\frac{1}{\sin^{2}x\cos^{2}x}-3\right\}\,dx
\displaystyle \Rightarrow\ I=\int \left\{\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}-3\right\}\,dx=\int (\sec^{2}x+\mathrm{cosec}^{2}x-3)\,dx
\displaystyle =\tan x-\cot x-3x+C

\displaystyle \textbf{Question 3: }\quad \text{Evaluate:}
\displaystyle \int \tan^{-1}(\sec x+\tan x)\,dx,\ -\frac{\pi}{2}<x<\frac{\pi}{2} \hspace{5.0cm} \text{[CBSE 2003]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \tan^{-1}(\sec x+\tan x)\,dx.\ \text{Then,}
\displaystyle I=\int \tan^{-1}\left(\frac{1+\sin x}{\cos x}\right)dx
\displaystyle \Rightarrow\ I=\int \tan^{-1}\left\{\frac{1-\cos\left(\frac{\pi}{2}+x\right)}{\sin\left(\frac{\pi}{2}+x\right)}\right\}dx
\displaystyle =\int \tan^{-1}\left\{\frac{2\sin^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)}{2\sin\left(\frac{\pi}{4}+\frac{x}{2}\right)\cos\left(\frac{\pi}{4}+\frac{x}{2}\right)}\right\}dx
\displaystyle \Rightarrow\ I=\int \tan^{-1}\left\{\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right\}dx
\displaystyle =\int \left(\frac{\pi}{4}+\frac{x}{2}\right)dx=\frac{\pi}{4}\int 1\,dx+\frac{1}{2}\int x\,dx
\displaystyle =\frac{\pi}{4}x+\frac{x^{2}}{4}+C

\displaystyle \textbf{Question 4: }\quad \text{Evaluate:}\ \int \tan^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}\,dx,\ -\frac{\pi}{2}<x<\frac{\pi}{2} \hspace{1.0cm} \text{[CBSE 2003, 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \tan^{-1}\sqrt{\frac{1-\sin x}{1+\sin x}}\,dx.\ \text{Then,}
\displaystyle I=\int \tan^{-1}\sqrt{\frac{1-\cos\left(\frac{\pi}{2}-x\right)}{1+\cos\left(\frac{\pi}{2}-x\right)}}\,dx
\displaystyle =\int \tan^{-1}\sqrt{\frac{2\sin^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2\cos^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right)}}\,dx
\displaystyle \Rightarrow\ I=\int \tan^{-1}\left\{\tan\left(\frac{\pi}{4}-\frac{x}{2}\right)\right\}dx
\displaystyle =\int \left(\frac{\pi}{4}-\frac{x}{2}\right)dx=\frac{\pi}{4}\int 1\,dx-\frac{1}{2}\int x\,dx
\displaystyle =\frac{\pi}{4}x-\frac{x^{2}}{4}+C

\displaystyle \textbf{Question 5: }\quad \text{Evaluate:}
\displaystyle \int \sec^{2}(7-4x)\,dx \hspace{8.0cm} \text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle  \quad \int \sec^{2}(7-4x)\,dx=-\frac{1}{4}\tan(7-4x)+C

\displaystyle \textbf{Question 6: }\quad \text{Evaluate:}
\displaystyle \int \frac{x^{2}+1}{(x+1)^{2}}\,dx \hspace{8.0cm} \text{[CBSE 2006]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \frac{x^{2}+1}{(x+1)^{2}}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{x^{2}+1+2x-2x}{(x+1)^{2}}\,dx
\displaystyle \Rightarrow\ I=\int \frac{(x+1)^{2}-2x}{(x+1)^{2}}\,dx
\displaystyle \Rightarrow\ I=\int \left\{1-\frac{2x}{(x+1)^{2}}\right\}dx
\displaystyle \Rightarrow\ I=\int 1\cdot dx-2\int \frac{x}{(x+1)^{2}}\,dx
\displaystyle \Rightarrow\ I=\int 1\cdot dx-2\int \frac{(x+1)-1}{(x+1)^{2}}\,dx
\displaystyle \Rightarrow\ I=\int 1\cdot dx-2\int \left\{\frac{1}{x+1}-\frac{1}{(x+1)^{2}}\right\}dx
\displaystyle \Rightarrow\ I=\int 1\cdot dx-2\int \frac{1}{x+1}\,dx+2\int \frac{1}{(x+1)^{2}}\,dx
\displaystyle =x-2\log|x+1|-\frac{2}{x+1}+C

\displaystyle \textbf{Question 7: }\quad \text{Evaluate:}
\displaystyle (i)\ \int \sin^{4}x\,dx \hspace{8.0cm} \text{[CBSE 2000, 2004]}
\displaystyle (ii)\ \int \cos^{4}x\,dx \hspace{8.0cm} \text{[CBSE 2000, 2003]}
\displaystyle \text{Answer:}
\displaystyle  \quad (i)\ \text{Let }I=\int \sin^{4}x\,dx.\ \text{Then,}
\displaystyle I=\int \left(\frac{1-\cos 2x}{2}\right)^{2}dx\qquad \left[\because\ \sin^{2}\theta=\frac{1-\cos 2\theta}{2}\right]
\displaystyle \Rightarrow\ I=\frac{1}{4}\int (1-2\cos 2x+\cos^{2}2x)\,dx
\displaystyle \Rightarrow\ I=\frac{1}{4}\int \left(1-2\cos 2x+\frac{1+\cos 4x}{2}\right)dx
\displaystyle \Rightarrow\ I=\frac{1}{8}\int (2-4\cos 2x+1+\cos 4x)\,dx
\displaystyle \Rightarrow\ I=\frac{1}{8}\int (3-4\cos 2x+\cos 4x)\,dx=\frac{1}{8}\left(3x-2\sin 2x+\frac{\sin 4x}{4}\right)+C
\displaystyle (ii)\ \text{Let }I=\int \cos^{4}x\,dx.\ \text{Then,}
\displaystyle I=\int \left(\frac{1+\cos 2x}{2}\right)^{2}dx\qquad \left[\because\ \cos^{2}\theta=\frac{1+\cos 2\theta}{2}\right]
\displaystyle \Rightarrow\ I=\frac{1}{4}\int (1+2\cos 2x+\cos^{2}2x)\,dx
\displaystyle \Rightarrow\ I=\frac{1}{4}\int \left(1+2\cos 2x+\frac{1+\cos 4x}{2}\right)dx
\displaystyle \Rightarrow\ I=\frac{1}{8}\int (3+4\cos 2x+\cos 4x)\,dx=\frac{1}{8}\left(3x+2\sin 2x+\frac{\sin 4x}{4}\right)+C

\displaystyle \textbf{Question 8: }\quad \text{Evaluate:}
\displaystyle \int \sin x\sin 2x\sin 3x\,dx \hspace{8.0cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \sin x\sin 2x\sin 3x\,dx.\ \text{Then,}
\displaystyle I=\frac{1}{2}\int (2\sin 2x\sin x)\sin 3x\,dx
\displaystyle \Rightarrow\ I=\frac{1}{2}\int (\cos x-\cos 3x)\sin 3x\,dx
\displaystyle \Rightarrow\ I=\frac{1}{4}\int (2\sin 3x\cos x-2\sin 3x\cos 3x)\,dx
\displaystyle \Rightarrow\ I=\frac{1}{4}\int (\sin 4x+\sin 2x-\sin 6x)\,dx
\displaystyle =\frac{1}{4}\left(-\frac{\cos 4x}{4}-\frac{\cos 2x}{2}+\frac{\cos 6x}{6}\right)+C

\displaystyle \textbf{Question 9: }\quad \text{Evaluate:}
\displaystyle \int \frac{\sin x}{\sin(x-a)}\,dx \hspace{8.0cm} \text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \frac{\sin x}{\sin(x-a)}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{\sin\{(x-a)+a\}}{\sin(x-a)}\,dx
\displaystyle \Rightarrow\ I=\int \frac{\sin(x-a)\cos a+\cos(x-a)\sin a}{\sin(x-a)}\,dx
\displaystyle \Rightarrow\ I=\int \{\cos a+\cot(x-a)\sin a\}\,dx
\displaystyle \Rightarrow\ I=\cos a\int 1\cdot dx+\sin a\int \cot(x-a)\,dx=x\cos a+\sin a\log|\sin(x-a)|+C

\displaystyle \textbf{Question 10: }\quad \text{Evaluate:}
\displaystyle (i)\ \int \frac{\sin 2x}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}\,dx \hspace{6.0cm} \text{[CBSE 2005]}
\displaystyle (ii)\ \int \frac{\sin x-x\cos x}{x(x+\sin x)}\,dx \hspace{8.0cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle (i)\ \text{Let }I=\int \frac{\sin 2x}{a^{2}\sin^{2}x+b^{2}\cos^{2}x}\,dx
\displaystyle \text{Let }a^{2}\sin^{2}x+b^{2}\cos^{2}x=t.\ \text{Then,}
\displaystyle d(a^{2}\sin^{2}x+b^{2}\cos^{2}x)=dt
\displaystyle \Rightarrow\ (a^{2}-b^{2})\sin 2x\,dx=dt
\displaystyle \Rightarrow\ dx=\frac{dt}{(a^{2}-b^{2})\sin 2x}
\displaystyle \text{Putting }a^{2}\sin^{2}x+b^{2}\cos^{2}x=t\ \text{and }dx=\frac{dt}{(a^{2}-b^{2})\sin 2x},\ \text{we get}
\displaystyle I=\int \frac{\sin 2x}{t}\times \frac{dt}{2\sin x\cos x\,(a^{2}-b^{2})}
\displaystyle \Rightarrow\ I=\frac{1}{(a^{2}-b^{2})}\int \frac{1}{t}\,dt
\displaystyle =\frac{1}{(a^{2}-b^{2})}\log|t|+C
\displaystyle =\frac{1}{(a^{2}-b^{2})}\log|a^{2}\sin^{2}x+b^{2}\cos^{2}x|+C
\displaystyle (ii)\ \text{Let }I=\int \frac{\sin x-x\cos x}{x(x+\sin x)}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{(x+\sin x)-x-x\cos x}{x(x+\sin x)}\,dx
\displaystyle \Rightarrow\ I=\int \frac{(x+\sin x)-x(1+\cos x)}{x(x+\sin x)}\,dx
\displaystyle \Rightarrow\ I=\int \left\{\frac{1}{x}-\frac{1+\cos x}{x+\sin x}\right\}dx
\displaystyle \Rightarrow\ I=\log|x|-\log(x+\sin x)+C
\displaystyle \Rightarrow\ I=\log\left|\frac{x}{x+\sin x}\right|+C

\displaystyle \textbf{Question 11: }\quad \text{Evaluate:}
\displaystyle \int \frac{(1+\log x)^{2}}{x}\,dx \hspace{8.0cm} \text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle  \quad \text{Let }I=\int \frac{(1+\log x)^{2}}{x}\,dx
\displaystyle \text{Let }1+\log x=t.\ \text{Then, }d(1+\log x)=dt\Rightarrow \frac{1}{x}dx=dt\Rightarrow dx=x\,dt
\displaystyle \text{Putting }1+\log x=t\ \text{and }dx=x\,dt,\ \text{we get}
\displaystyle I=\int \frac{t^{2}}{x}\times x\,dt=\int t^{2}\,dt=\frac{t^{3}}{3}+C=\frac{(1+\log x)^{3}}{3}+C

\displaystyle \textbf{Question 12: }\quad \text{Evaluate:}\ \int \frac{x^{2}-3x+1}{\sqrt{1-x^{2}}}\,dx \hspace{3.0cm} \text{[CBSE 2015]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int \frac{x^{2}-3x+1}{\sqrt{1-x^{2}}}\,dx\ \text{and }x=\sin\theta.\ \text{Then, }dx=d(\sin\theta)=\cos\theta\,d\theta
\displaystyle \therefore\ I=\int \frac{\sin^{2}\theta-3\sin\theta+1}{\sqrt{1-\sin^{2}\theta}}\cos\theta\,d\theta=\int (\sin^{2}\theta-3\sin\theta+1)\,d\theta
\displaystyle \Rightarrow\ I=\int \left(\frac{1-\cos 2\theta}{2}-3\sin\theta+1\right)d\theta=\frac{1}{2}\int (3-6\sin\theta-\cos 2\theta)\,d\theta
\displaystyle \Rightarrow\ I=\frac{1}{2}\left(3\theta+6\cos\theta-\frac{1}{2}\sin 2\theta\right)+C
\displaystyle \Rightarrow\ I=\frac{1}{2}\left(3\theta+6\sqrt{1-\sin^{2}\theta}-\sin\theta\sqrt{1-\sin^{2}\theta}\right)+C
\displaystyle \Rightarrow\ I=\frac{1}{2}\left(3\sin^{-1}x+6\sqrt{1-x^{2}}-x\sqrt{1-x^{2}}\right)+C

\displaystyle \textbf{Question 13: }\quad \text{Evaluate:}
\displaystyle \int \frac{1}{x^{2}+4x+8}\,dx \hspace{8.0cm} \text{[CBSE 2002, 2017]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \frac{1}{x^{2}+4x+8}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{1}{x^{2}+4x+4+4}\,dx=\int \frac{1}{(x+2)^{2}+2^{2}}\,dx =\frac{1}{2}\tan^{-1}\left(\frac{x+2}{2}\right)+C

\displaystyle \textbf{Question 14: }\quad \text{Evaluate:}
\displaystyle \int \frac{1}{x(x^{n}+1)}\,dx \hspace{8.0cm} \text{[CBSE 2000C]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle I=\int \frac{1}{x(x^{n}+1)}\,dx=\int \frac{x^{n-1}}{x^{n}(x^{n}+1)}\,dx
\displaystyle \text{Let }x^{n}+1=t.\ \text{Then, }d(x^{n}+1)=dt\Rightarrow n x^{n-1}dx=dt
\displaystyle \Rightarrow\ dx=\frac{dt}{n x^{n-1}}
\displaystyle \therefore\ I=\int \frac{1}{n x^{n}t}\,dt=\frac{1}{n}\int \frac{1}{(t-1)t}\,dt
\displaystyle \left[\because\ x^{n}+1=t\ \therefore\ x^{n}=t-1\right]
\displaystyle \Rightarrow\ I=\frac{1}{n}\int \frac{1}{t^{2}-t}\,dt=\frac{1}{n}\int \frac{dt}{t^{2}-t+\frac{1}{4}-\frac{1}{4}}
\displaystyle =\frac{1}{n}\int \frac{dt}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}
\displaystyle \Rightarrow\ I=\frac{1}{n}\times\frac{1}{2\left(\frac{1}{2}\right)} \log\left|\frac{t-\frac{1}{2}-\frac{1}{2}}{t-\frac{1}{2}+\frac{1}{2}}\right|+C
\displaystyle =\frac{1}{n}\log\left|\frac{t-1}{t}\right|+C=\frac{1}{n}\log\left|\frac{x^{n}}{x^{n}+1}\right|+C

\displaystyle \textbf{Question 15: }\quad \text{Evaluate:}
\displaystyle \int \frac{1}{\sqrt{(x-a)(x-b)}}\,dx \hspace{8.0cm} \text{[CBSE 2001]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int \frac{1}{\sqrt{(x-a)(x-b)}}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{1}{\sqrt{x^{2}-x(a+b)+ab}}\,dx
\displaystyle \Rightarrow\ I=\int \frac{1}{\sqrt{x^{2}-x(a+b)+\left(\frac{a+b}{2}\right)^{2}-\left(\frac{a+b}{2}\right)^{2}+ab}}\,dx
\displaystyle \Rightarrow\ I=\int \frac{1}{\sqrt{\left\{x-\left(\frac{a+b}{2}\right)\right\}^{2}-\left(\frac{a-b}{2}\right)^{2}}}\,dx
\displaystyle \Rightarrow\ I=\log\left|x-\left(\frac{a+b}{2}\right)+\sqrt{\left\{x-\left(\frac{a+b}{2}\right)\right\}^{2}-\left(\frac{a-b}{2}\right)^{2}}\right|+C
\displaystyle \Rightarrow\ I=\log\left|\frac{2x-a-b}{2}+\sqrt{(x-a)(x-b)}\right|+C
\displaystyle \Rightarrow\ I=\log\left|\frac{(x-a)+(x-b)+2\sqrt{(x-a)(x-b)}}{2}\right|+C
\displaystyle \Rightarrow\ I=\log\left|\left(\sqrt{x-a}+\sqrt{x-b}\right)^{2}\right|-\log 2+C
\displaystyle \Rightarrow\ I=2\log\left|\sqrt{x-a}+\sqrt{x-b}\right|+C_{1},\ \text{where }C_{1}=C-\log 2

\displaystyle \textbf{Question 16:}\quad \text{Evaluate:}
\displaystyle (i)\ \int \frac{2x}{\sqrt{1-x^{2}-x^{4}}}\,dx \hspace{8.0cm} \text{[CBSE 2005]}
\displaystyle (ii)\ \int \frac{e^{x}}{\sqrt{5-4e^{x}-e^{2x}}}\,dx \hspace{8.0cm} \text{[CBSE 2009]}
\displaystyle (iii)\ \int \sqrt{\frac{x}{a^{3}-x^{3}}}\,dx \hspace{8.0cm} \text{[CBSE 2016]}
\displaystyle \text{Answer:}
\displaystyle (i)\ \text{Let }I=\int \frac{2x}{\sqrt{1-x^{2}-x^{4}}}\,dx
\displaystyle =\int \frac{2x}{\sqrt{1-x^{2}-(x^{2})^{2}}}\,dx
\displaystyle \text{Let }x^{2}=t.\ \text{Then, }d(x^{2})=dt\Rightarrow 2x\,dx=dt
\displaystyle \Rightarrow\ dx=\frac{dt}{2x}
\displaystyle \therefore\ I=\int \frac{1}{\sqrt{1-t-t^{2}}}\,dt
\displaystyle =\int \frac{1}{\sqrt{-(t^{2}+t-1)}}\,dt
\displaystyle =\int \frac{1}{\sqrt{-\left(t^{2}+t+\frac{1}{4}-\frac{1}{4}-1\right)}}\,dt
\displaystyle =\int \frac{1}{\sqrt{\frac{5}{4}-\left(t+\frac{1}{2}\right)^{2}}}\,dt
\displaystyle =\int \frac{1}{\sqrt{\left(\frac{\sqrt{5}}{2}\right)^{2}-\left(t+\frac{1}{2}\right)^{2}}}\,dt
\displaystyle =\sin^{-1}\left(\frac{t+\frac{1}{2}}{\frac{\sqrt{5}}{2}}\right)+C
\displaystyle =\sin^{-1}\left(\frac{2t+1}{\sqrt{5}}\right)+C =\sin^{-1}\left(\frac{2x^{2}+1}{\sqrt{5}}\right)+C
\displaystyle (ii)\ \text{Let }I=\int \frac{e^{x}}{\sqrt{5-4e^{x}-e^{2x}}}\,dx
\displaystyle =\int \frac{e^{x}}{\sqrt{5-4e^{x}-(e^{x})^{2}}}\,dx
\displaystyle \text{Let }e^{x}=t.\ \text{Then, }d(e^{x})=dt\Rightarrow e^{x}dx=dt
\displaystyle \Rightarrow\ dx=\frac{dt}{e^{x}}
\displaystyle \therefore\ I=\int \frac{1}{\sqrt{5-4t-t^{2}}}\,dt
\displaystyle =\int \frac{1}{\sqrt{-(t^{2}+4t-5)}}\,dt
\displaystyle =\int \frac{1}{\sqrt{-\left(t^{2}+4t+4-9\right)}}\,dt
\displaystyle =\int \frac{1}{\sqrt{9-(t+2)^{2}}}\,dt =\sin^{-1}\left(\frac{t+2}{3}\right)+C
\displaystyle =\sin^{-1}\left(\frac{e^{x}+2}{3}\right)+C
\displaystyle (iii)\ \text{Let }I=\int \sqrt{\frac{x}{a^{3}-x^{3}}}\,dx
\displaystyle =\int \frac{\sqrt{x}}{\sqrt{(a^{\frac{3}{2}})^{2}-(x^{\frac{3}{2}})^{2}}}\,dx
\displaystyle \text{Let }x^{\frac{3}{2}}=t.\ \text{Then, }d(x^{\frac{3}{2}})=dt \Rightarrow \frac{3}{2}x^{\frac{1}{2}}dx=dt\Rightarrow dx=\frac{2}{3\sqrt{x}}\,dt
\displaystyle \therefore\ I=\frac{2}{3}\int \frac{1}{\sqrt{(a^{\frac{3}{2}})^{2}-t^{2}}}\,dt
\displaystyle =\frac{2}{3}\sin^{-1}\left(\frac{t}{a^{\frac{3}{2}}}\right)+C =\frac{2}{3}\sin^{-1}\left(\frac{x^{\frac{3}{2}}}{a^{\frac{3}{2}}}\right)+C

\displaystyle \textbf{Question 17: }\quad \text{Evaluate:}
\displaystyle \int \frac{x+2}{\sqrt{x^{2}+5x+6}}\,dx \hspace{8.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x+2=\lambda \frac{d}{dx}(x^{2}+5x+6)+\mu.\ \text{Then, }x+2=\lambda(2x+5)+\mu
\displaystyle \text{Comparing the coefficients of like powers of }x,\ \text{we get}
\displaystyle 1=2\lambda\ \text{and }5\lambda+\mu=2\ \Rightarrow\ \lambda=\frac{1}{2}\ \text{and }\mu=-\frac{1}{2}
\displaystyle \therefore\ I=\int \frac{x+2}{\sqrt{x^{2}+5x+6}}\,dx
\displaystyle \Rightarrow\ I=\int \frac{\frac{1}{2}(2x+5)-\frac{1}{2}}{\sqrt{x^{2}+5x+6}}\,dx
\displaystyle \Rightarrow\ I=\frac{1}{2}\int \frac{2x+5}{\sqrt{x^{2}+5x+6}}\,dx-\frac{1}{2}\int \frac{1}{\sqrt{x^{2}+5x+6}}\,dx
\displaystyle \Rightarrow\ I=\frac{1}{2}\int \frac{dt}{\sqrt{t}}-\frac{1}{2}\int \frac{1}{\sqrt{\left(x+\frac{5}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}}\,dx,\ \text{where }t=x^{2}+5x+6
\displaystyle \Rightarrow\ I=\sqrt{t}-\frac{1}{2}\log\left|\left(x+\frac{5}{2}\right)+\sqrt{x^{2}+5x+6}\right|+C
\displaystyle \Rightarrow\ I=\sqrt{x^{2}+5x+6}-\frac{1}{2}\log\left|\left(x+\frac{5}{2}\right)+\sqrt{x^{2}+5x+6}\right|+C

\displaystyle \textbf{Question 18: }\quad \text{Evaluate:}
\displaystyle \int (\sin^{-1}x)^{2}\,dx \hspace{8.0cm} \text{[CBSE 2004]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\sin^{-1}x=t.\ \text{Then, }x=\sin t\Rightarrow dx=\cos t\,dt
\displaystyle \therefore\ I=\int (\sin^{-1}x)^{2}\,dx
\displaystyle \Rightarrow\ I=\int t^{2}\cos t\,dt=t^{2}(\sin t)-\int 2t\sin t\,dt=t^{2}\sin t-2\int t\sin t\,dt
\displaystyle \Rightarrow\ I=t^{2}\sin t-2\left\{t(-\cos t)-\int 1\times(-\cos t)\,dt\right\}
\displaystyle \Rightarrow\ I=t^{2}\sin t-2\left\{-t\cos t+\int \cos t\,dt\right\}
\displaystyle \Rightarrow\ I=t^{2}\sin t-2(-t\cos t+\sin t)+C
\displaystyle \Rightarrow\ I=t^{2}\sin t-2\left\{-t\sqrt{1-\sin^{2}t}+\sin t\right\}+C
\displaystyle \Rightarrow\ I=x(\sin^{-1}x)^{2}-2\left\{-\sqrt{1-x^{2}}\sin^{-1}x+x\right\}+C

\displaystyle \textbf{Question 19: }\quad \text{Evaluate:}
\displaystyle \int x\sin^{-1}x\,dx \hspace{8.0cm} \text{[CBSE 2000, 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int x\sin^{-1}x\,dx.\ \text{Then,}
\displaystyle I=(\sin^{-1}x)\frac{x^{2}}{2}-\int \frac{1}{\sqrt{1-x^{2}}}\times\frac{x^{2}}{2}\,dx
\displaystyle \Rightarrow\ I=\frac{x^{2}}{2}\sin^{-1}x+\frac{1}{2}\int \frac{-x^{2}}{\sqrt{1-x^{2}}}\,dx
\displaystyle =\frac{x^{2}}{2}\sin^{-1}x+\frac{1}{2}\int \frac{1-x^{2}-1}{\sqrt{1-x^{2}}}\,dx
\displaystyle \Rightarrow\ I=\frac{x^{2}}{2}\sin^{-1}x+\frac{1}{2}\left\{\int \frac{1-x^{2}}{\sqrt{1-x^{2}}}\,dx-\int \frac{1}{\sqrt{1-x^{2}}}\,dx\right\}
\displaystyle \Rightarrow\ I=\frac{x^{2}}{2}\sin^{-1}x+\frac{1}{2}\left\{\int \sqrt{1-x^{2}}\,dx-\int \frac{1}{\sqrt{1-x^{2}}}\,dx\right\}
\displaystyle \Rightarrow\ I=\frac{x^{2}}{2}\sin^{-1}x+\frac{1}{2}\left\{\frac{1}{2}x\sqrt{1-x^{2}}+\frac{1}{2}\sin^{-1}x\right\}-\sin^{-1}x+C
\displaystyle \Rightarrow\ I=\frac{1}{2}x^{2}\sin^{-1}x+\frac{1}{4}x\sqrt{1-x^{2}}-\frac{1}{4}\sin^{-1}x+C

\displaystyle \textbf{Question 20: }\quad \text{Evaluate:}\ \int \frac{\sqrt{x^{2}+1}\{\log(x^{2}+1)-2\log x\}}{x^{4}}\,dx\qquad \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let}
\displaystyle I=\int \frac{\sqrt{x^{2}+1}\{\log(x^{2}+1)-2\log x\}}{x^{4}}\,dx
\displaystyle \Rightarrow\ I=\int \frac{\sqrt{x^{2}+1}}{x^{4}}\{\log(x^{2}+1)-\log x^{2}\}\,dx
\displaystyle \Rightarrow\ I=\int \sqrt{1+\frac{1}{x^{2}}}\ \log\left(1+\frac{1}{x^{2}}\right)\frac{1}{x^{3}}\,dx
\displaystyle \text{Let }1+\frac{1}{x^{2}}=t.\ \text{Then, }d\left(1+\frac{1}{x^{2}}\right)=dt\Rightarrow -\frac{2}{x^{3}}dx=dt
\displaystyle \Rightarrow\ \frac{1}{x^{3}}dx=-\frac{1}{2}dt
\displaystyle \therefore\ I=-\frac{1}{2}\int \sqrt{t}\ \log t\,dt
\displaystyle \Rightarrow\ I=-\frac{1}{2}\left\{\frac{2}{3}(\log t)t^{\frac{3}{2}}-\frac{2}{3}\int \frac{1}{t}\times t^{\frac{3}{2}}\,dt\right\} =-\frac{1}{2}\left\{\frac{2}{3}(\log t)t^{\frac{3}{2}}-\frac{4}{9}t^{\frac{3}{2}}\right\}+C
\displaystyle \Rightarrow\ I=-\frac{1}{3}t^{\frac{3}{2}}\left\{\log t-\frac{2}{3}\right\}+C =-\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{\frac{3}{2}}\left\{\log\left(1+\frac{1}{x^{2}}\right)-\frac{2}{3}\right\}+C

\displaystyle \textbf{Question 21: }\quad \text{Evaluate:}
\displaystyle \int \left\{\log(\log x)+\frac{1}{(\log x)^{2}}\right\}dx \hspace{5.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int \left\{\log(\log x)+\frac{1}{(\log x)^{2}}\right\}dx.\ \text{Let }\log x=t.\ \text{Then, }x=e^{t}\Rightarrow dx=d(e^{t})=e^{t}\,dt
\displaystyle \therefore\ I=\int \left\{\log t+\frac{1}{t^{2}}\right\}e^{t}\,dt
\displaystyle \Rightarrow\ I=\int \left\{\log t+\frac{1}{t}-\frac{1}{t}+\frac{1}{t^{2}}\right\}e^{t}\,dt
\displaystyle \Rightarrow\ I=\int \left(\log t+\frac{1}{t}\right)e^{t}\,dt+\int \left(-\frac{1}{t}+\frac{1}{t^{2}}\right)e^{t}\,dt
\displaystyle \Rightarrow\ I=\int e^{t}\log t\,dt+\int e^{t}\frac{1}{t}\,dt+\int e^{t}\left(-\frac{1}{t}\right)dt+\int e^{t}\frac{1}{t^{2}}\,dt
\displaystyle \Rightarrow\ I=(\log t)e^{t}-\int \frac{1}{t}\cdot e^{t}\,dt+\int e^{t}\frac{1}{t}\,dt+\left(-\frac{1}{t}\right)e^{t}-\int \frac{1}{t^{2}}\cdot e^{t}\,dt+\int e^{t}\frac{1}{t^{2}}\,dt+C
\displaystyle \Rightarrow\ I=e^{t}\cdot\log t-\frac{1}{t}e^{t}+C=x\log(\log x)-\frac{x}{\log x}+C

\displaystyle \textbf{Question 22: }\quad \text{Evaluate:}\ \int e^{2x}\left(\frac{1+\sin 2x}{1+\cos 2x}\right)dx \hspace{3.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int e^{2x}\left(\frac{1+\sin 2x}{1+\cos 2x}\right)dx
\displaystyle \Rightarrow\ I=\int e^{2x}\left\{\frac{1+2\sin x\cos x}{2\cos^{2}x}\right\}dx
\displaystyle \Rightarrow\ I=\int e^{2x}\left\{\frac{1}{2}\sec^{2}x+\tan x\right\}dx=\int e^{2x}\left\{2\left(\frac{1}{2}\tan x\right)'+\frac{1}{2}\sec^{2}x\right\}dx
\displaystyle \Rightarrow\ I=\int e^{2x}\cdot\tan x\,dx+\frac{1}{2}\int e^{2x}\cdot\sec^{2}x\,dx
\displaystyle \Rightarrow\ I=(\tan x)\frac{e^{2x}}{2}-\int \sec^{2}x\cdot\frac{e^{2x}}{2}\,dx+\frac{1}{2}\int e^{2x}\sec^{2}x\,dx+C
\displaystyle \Rightarrow\ I=\frac{1}{2}e^{2x}\tan x-\frac{1}{2}\int e^{2x}\sec^{2}x\,dx+\frac{1}{2}\int e^{2x}\sec^{2}x\,dx+C=\frac{1}{2}e^{2x}\tan x+C

\displaystyle \textbf{Question 23: }\quad \text{Prove that:}
\displaystyle \int e^{ax}\sin bx\,dx=\frac{e^{ax}}{a^{2}+b^{2}}(a\sin bx-b\cos bx)+C \hspace{3.0cm} \text{[CBSE 2002]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int e^{ax}\sin bx\,dx.\ \text{Then,}
\displaystyle I=\int e^{ax}\sin bx\,dx
\displaystyle \Rightarrow\ I=-e^{ax}\frac{\cos bx}{b}-\int ae^{ax}\left(-\frac{\cos bx}{b}\right)dx
\displaystyle \Rightarrow\ I=-\frac{1}{b}e^{ax}\cos bx+\frac{a}{b}\int e^{ax}\cos bx\,dx
\displaystyle \Rightarrow\ I=-\frac{1}{b}e^{ax}\cos bx+\frac{a}{b}\left\{\frac{e^{ax}\sin bx}{b}-\int ae^{ax}\frac{\sin bx}{b}\,dx\right\}
\displaystyle \Rightarrow\ I=-\frac{1}{b}e^{ax}\cos bx+\frac{a}{b^{2}}e^{ax}\sin bx-\frac{a^{2}}{b^{2}}\int e^{ax}\sin bx\,dx
\displaystyle \Rightarrow\ I=-\frac{1}{b}e^{ax}\cos bx+\frac{a}{b^{2}}e^{ax}\sin bx-\frac{a^{2}}{b^{2}}I
\displaystyle \Rightarrow\ I+I\cdot\frac{a^{2}}{b^{2}}=\frac{e^{ax}}{b^{2}}(a\sin bx-b\cos bx)
\displaystyle \Rightarrow\ I\left(\frac{a^{2}+b^{2}}{b^{2}}\right)=\frac{e^{ax}}{b^{2}}(a\sin bx-b\cos bx)
\displaystyle \Rightarrow\ I=\frac{e^{ax}}{a^{2}+b^{2}}(a\sin bx-b\cos bx)+C

\displaystyle \textbf{Question 24: }\quad \text{Evaluate:}
\displaystyle  \int (3x-2)\sqrt{x^{2}+x+1}\,dx \hspace{5.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }3x-2=\lambda\frac{d}{dx}(x^{2}+x+1)+\mu\ \text{i.e. }3x-2=\lambda(2x+1)+\mu
\displaystyle \text{Comparing the coefficients of like powers of }x,\ \text{we get}
\displaystyle 2\lambda=3\ \text{and }\lambda+\mu=-2\Rightarrow \lambda=\frac{3}{2}\ \text{and }\mu=-\frac{7}{2}
\displaystyle \therefore\ I=\int (3x-2)\sqrt{x^{2}+x+1}\,dx
\displaystyle \Rightarrow\ I=\int \left\{\frac{3}{2}(2x+1)-\frac{7}{2}\right\}\sqrt{x^{2}+x+1}\,dx
\displaystyle \Rightarrow\ I=\frac{3}{2}\int (2x+1)\sqrt{x^{2}+x+1}\,dx-\frac{7}{2}\int \sqrt{x^{2}+x+1}\,dx
\displaystyle \Rightarrow\ I=\frac{3}{2}\int \sqrt{t}\,dt-\frac{7}{2}\int \sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\,dx,\ \text{where }t=x^{2}+x+1
\displaystyle \Rightarrow\ I=t^{\frac{3}{2}}-\frac{7}{4}\left\{\left(x+\frac{1}{2}\right)\sqrt{x^{2}+x+1}+\left(\frac{\sqrt{3}}{2}\right)^{2}\log\left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|\right\}+C
\displaystyle \Rightarrow\ I=(x^{2}+x+1)^{\frac{3}{2}}-\frac{7}{2}\left\{\left(x+\frac{1}{2}\right)\sqrt{x^{2}+x+1}+\frac{3}{8}\log\left|x+\frac{1}{2}+\sqrt{x^{2}+x+1}\right|\right\}+C

\displaystyle \textbf{Question 25: }\quad \text{Evaluate:}
\displaystyle \int \frac{2x-1}{(x-1)(x+2)(x-3)}\,dx \hspace{5.0cm} \text{[CBSE 2005]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{2x-1}{(x-1)(x+2)(x-3)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{x-3}\qquad \cdots(i)
\displaystyle \Rightarrow\ 2x-1=A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2)\qquad \cdots(ii)
\displaystyle \text{Putting }x+2=0\ \text{or, }x=-2\ \text{in (ii), we get}
\displaystyle -5=B(-3)(-5)\Rightarrow B=-\frac{1}{3}
\displaystyle \text{Putting }x-3=0\ \text{or, }x=3\ \text{in (ii), we get}
\displaystyle 5=C(2)(5)\Rightarrow C=\frac{1}{2}
\displaystyle \text{Putting }x-1=0\ \text{or, }x=1\ \text{in (ii), we get}
\displaystyle 1=A(3)(-2)\Rightarrow A=-\frac{1}{6}
\displaystyle \text{Substituting the values of }A,B\ \text{and }C\ \text{in (i), we obtain}
\displaystyle \frac{2x-1}{(x-1)(x+2)(x-3)}=-\frac{1}{6}\cdot\frac{1}{x-1}-\frac{1}{3}\cdot\frac{1}{x+2}+\frac{1}{2}\cdot\frac{1}{x-3}
\displaystyle \therefore\ I=\int \frac{2x-1}{(x-1)(x+2)(x-3)}\,dx
\displaystyle \Rightarrow\ I=-\frac{1}{6}\int \frac{1}{x-1}\,dx-\frac{1}{3}\int \frac{1}{x+2}\,dx+\frac{1}{2}\int \frac{1}{x-3}\,dx
\displaystyle \Rightarrow\ I=-\frac{1}{6}\log|x-1|-\frac{1}{3}\log|x+2|+\frac{1}{2}\log|x-3|+C

\displaystyle \textbf{Question 26: }\quad \text{Evaluate:}
\displaystyle  \int \frac{1}{\sin x-\sin 2x}\,dx \hspace{8.0cm} \text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle  \text{We have,}
\displaystyle I=\int \frac{1}{\sin x-\sin 2x}\,dx
\displaystyle \Rightarrow\ I=\int \frac{1}{(\sin x-2\sin x\cos x)}\,dx=\int \frac{1}{\sin x(1-2\cos x)}\,dx=\int \frac{\sin x}{\sin^{2}x(1-2\cos x)}\,dx
\displaystyle \Rightarrow\ I=\int \frac{\sin x}{(1-\cos^{2}x)(1-2\cos x)}\,dx
\displaystyle \text{Putting }\cos x=t,\ \text{and }-\sin x\,dx=dt\ \text{or, }\sin x\,dx=-dt,\ \text{we get}
\displaystyle I=\int \frac{-dt}{(1-t^{2})(1-2t)}=\int \frac{-1}{(1-t)(1+t)(1-2t)}\,dt
\displaystyle \text{Let }\frac{-1}{(1-t)(1+t)(1-2t)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{1-2t}.\ \text{Then,}
\displaystyle -1=A(1+t)(1-2t)+B(1-t)(1-2t)+C(1-t)(1+t)\qquad \cdots(i)
\displaystyle \text{Putting }t+1=0\ \text{or, }t=-1\ \text{in (i), we get}
\displaystyle -1=6B\Rightarrow B=-\frac{1}{6}
\displaystyle \text{Putting }1-t=0\ \text{or, }t=1\ \text{in (i), we get}
\displaystyle -1=-2A\Rightarrow A=\frac{1}{2}
\displaystyle \text{Putting }1-2t=0\ \text{or, }t=\frac{1}{2}\ \text{in (i), we get}
\displaystyle -1=C\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\Rightarrow C=-\frac{4}{3}
\displaystyle \therefore\ \frac{-1}{(1-t)(1+t)(1-2t)}=\frac{1}{2}\cdot\frac{1}{1-t}-\frac{1}{6}\cdot\frac{1}{1+t}-\frac{4}{3}\cdot\frac{1}{1-2t}
\displaystyle \Rightarrow\ I=\int \frac{-dt}{(1-t)(1+t)(1-2t)}
\displaystyle \Rightarrow\ I=\frac{1}{2}\int \frac{1}{1-t}\,dt-\frac{1}{6}\int \frac{1}{1+t}\,dt-\frac{4}{3}\int \frac{1}{1-2t}\,dt
\displaystyle \Rightarrow\ I=-\frac{1}{2}\log|1-t|-\frac{1}{6}\log|1+t|-\frac{4}{3}\times\left(-\frac{1}{2}\right)\log|1-2t|+C
\displaystyle \Rightarrow\ I=-\frac{1}{2}\log|1-\cos x|-\frac{1}{6}\log|1+\cos x|+\frac{2}{3}\log|1-2\cos x|+C

\displaystyle \textbf{Question 27: }\quad \text{Evaluate:}
\displaystyle (i)\ \int \frac{3x+1}{(x-2)^{2}(x+2)}\,dx \hspace{8.0cm} \text{[CBSE 2007]}
\displaystyle (ii)\ \int \frac{x^{2}+1}{(x-1)^{2}(x+3)}\,dx \hspace{8.0cm} \text{[CBSE 2012]}
\displaystyle \text{Answer:}
\displaystyle  (i)\ \text{Let }\frac{3x+1}{(x-2)^{2}(x+2)}=\frac{A}{x-2}+\frac{B}{(x-2)^{2}}+\frac{C}{x+2}\qquad \cdots(i)
\displaystyle \Rightarrow\ 3x+1=A(x-2)(x+2)+B(x+2)+C(x-2)^{2}\qquad \cdots(ii)
\displaystyle \text{Putting }x-2=0\ \text{i.e. }x=2\ \text{in (ii), we get}
\displaystyle 7=4B\Rightarrow B=\frac{7}{4}
\displaystyle \text{Putting }x+2=0\ \text{i.e. }x=-2\ \text{in (ii), we get}
\displaystyle -5=16C\Rightarrow C=-\frac{5}{16}
\displaystyle \text{Comparing coefficients of }x^{2}\ \text{on both sides of the identity (ii), we get}
\displaystyle A+C=0\Rightarrow A=-C=\frac{5}{16}
\displaystyle \text{Substituting the values of }A,B\ \text{and }C\ \text{in (i), we get}
\displaystyle \frac{3x+1}{(x-2)^{2}(x+2)}=\frac{5}{16}\cdot\frac{1}{x-2}+\frac{7}{4}\cdot\frac{1}{(x-2)^{2}}-\frac{5}{16}\cdot\frac{1}{x+2}
\displaystyle \therefore\ I=\int \frac{3x+1}{(x-2)^{2}(x+2)}\,dx
\displaystyle \Rightarrow\ I=\frac{5}{16}\int \frac{1}{x-2}\,dx+\frac{7}{4}\int \frac{1}{(x-2)^{2}}\,dx-\frac{5}{16}\int \frac{1}{x+2}\,dx
\displaystyle \Rightarrow\ I=\frac{5}{16}\log|x-2|-\frac{7}{4(x-2)}-\frac{5}{16}\log|x+2|+C
\displaystyle (ii)\ \text{We have,}
\displaystyle I=\int \frac{x^{2}+1}{(x-1)^{2}(x+3)}\,dx
\displaystyle \text{Let }\frac{x^{2}+1}{(x-1)^{2}(x+3)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+3}\qquad \cdots(i)
\displaystyle \Rightarrow\ x^{2}+1=A(x-1)(x+3)+B(x+3)+C(x-1)^{2}\qquad \cdots(ii)
\displaystyle \text{Putting }x-1=0\ \text{i.e. }x=1\ \text{in (ii), we get}
\displaystyle 2=4B\Rightarrow B=\frac{1}{2}
\displaystyle \text{Putting }x+3=0\ \text{i.e. }x=-3\ \text{in (ii), we get}
\displaystyle 10=16C\Rightarrow C=\frac{5}{8}
\displaystyle \text{Equating the coefficients of }x^{2}\ \text{on both sides of the identity (ii), we get}
\displaystyle 1=A+C\Rightarrow A=1-C=1-\frac{5}{8}=\frac{3}{8}
\displaystyle \text{Substituting the values of }A,B\ \text{and }C\ \text{in (i), we get}
\displaystyle \frac{x^{2}+1}{(x-1)^{2}(x+3)}=\frac{3}{8}\cdot\frac{1}{x-1}+\frac{1}{2}\cdot\frac{1}{(x-1)^{2}}+\frac{5}{8}\cdot\frac{1}{x+3}
\displaystyle \Rightarrow\ I=\int \frac{x^{2}+1}{(x-1)^{2}(x+3)}\,dx=\frac{3}{8}\int \frac{1}{x-1}\,dx+\frac{1}{2}\int \frac{1}{(x-1)^{2}}\,dx+\frac{5}{8}\int \frac{1}{x+3}\,dx
\displaystyle \Rightarrow\ I=\frac{3}{8}\log|x-1|-\frac{1}{2(x-1)}+\frac{5}{8}\log|x+3|+C

\displaystyle \textbf{Question 28: }\quad \text{Evaluate:}
\displaystyle \int \frac{8}{(x+2)(x^{2}+4)}\,dx \hspace{8.0cm} \text{[CBSE 2013]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{8}{(x+2)(x^{2}+4)}=\frac{A}{x+2}+\frac{Bx+C}{x^{2}+4}\qquad \cdots(i)
\displaystyle \text{Then, }8=A(x^{2}+4)+(Bx+C)(x+2)\qquad \cdots(ii)
\displaystyle \text{Putting }x+2=0\ \text{i.e. }x=-2\ \text{in (ii), we get}
\displaystyle 8=8A\Rightarrow A=1
\displaystyle \text{Putting }x=0\ \text{and }1\ \text{respectively in (ii), we get}
\displaystyle 8=4A+2C\ \text{and }8=5A+3B+3C
\displaystyle \text{Solving these equations, we obtain}
\displaystyle A=1,\ C=2\ \text{and }B=-1
\displaystyle \text{Substituting the values of }A,B\ \text{and }C\ \text{in (i), we obtain}
\displaystyle \frac{8}{(x+2)(x^{2}+4)}=\frac{1}{x+2}+\frac{-x+2}{x^{2}+4}
\displaystyle \therefore\ I=\int \frac{8}{(x+2)(x^{2}+4)}\,dx
\displaystyle \Rightarrow\ I=\int \frac{1}{x+2}\,dx+\int \frac{-x+2}{x^{2}+4}\,dx
\displaystyle \Rightarrow\ I=\int \frac{1}{x+2}\,dx-\int \frac{x}{x^{2}+4}\,dx+2\int \frac{1}{x^{2}+4}\,dx
\displaystyle \Rightarrow\ I=\log|x+2|-\frac{1}{2}\int \frac{1}{t}\,dt+2\times\frac{1}{2}\tan^{-1}\frac{x}{2}+C,\ \text{where }t=x^{2}+4
\displaystyle \Rightarrow\ I=\log|x+2|-\frac{1}{2}\log t+\tan^{-1}\frac{x}{2}+C =\log|x+2|-\frac{1}{2}\log(x^{2}+4)+\tan^{-1}\frac{x}{2}+C

\displaystyle \textbf{Question 29: }\quad \text{Evaluate:}\ \int \left\{\sqrt{\tan \theta}+\sqrt{\cot \theta}\right\}d\theta\qquad \text{[CBSE 2010, 2013, 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int \left\{\sqrt{\tan \theta}+\sqrt{\cot \theta}\right\}d\theta.\ \text{Then,}
\displaystyle I=\int \left\{\sqrt{\tan \theta}+\frac{1}{\sqrt{\tan \theta}}\right\}d\theta=\int \frac{\tan \theta+1}{\sqrt{\tan \theta}}\,d\theta
\displaystyle \text{Let }\tan \theta=x^{2}.\ \text{Then, }d(\tan \theta)=d(x^{2})\Rightarrow \sec^{2}\theta\,d\theta=2x\,dx
\displaystyle \Rightarrow\ d\theta=\frac{2x\,dx}{\sec^{2}\theta}=\frac{2x\,dx}{1+\tan^{2}\theta}=\frac{2x\,dx}{1+x^{4}}
\displaystyle \therefore\ I=\int \frac{x^{2}+1}{x}\cdot\frac{2x\,dx}{1+x^{4}}=2\int \frac{x^{2}+1}{1+x^{4}}\,dx=2\int \frac{1+\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}}\,dx
\displaystyle \Rightarrow\ I=2\int \frac{1+\frac{1}{x^{2}}}{\left(x-\frac{1}{x}\right)^{2}+2}\,dx
\displaystyle \text{Let }u=x-\frac{1}{x}.\ \text{Then, }du=\left(1+\frac{1}{x^{2}}\right)dx
\displaystyle \Rightarrow\ I=2\int \frac{du}{u^{2}+(\sqrt{2})^{2}}=\frac{2}{\sqrt{2}}\tan^{-1}\left(\frac{u}{\sqrt{2}}\right)+C
\displaystyle \Rightarrow\ I=\sqrt{2}\tan^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+C =\sqrt{2}\tan^{-1}\left(\frac{x^{2}-1}{\sqrt{2}x}\right)+C
\displaystyle \Rightarrow\ I=\sqrt{2}\tan^{-1}\left(\frac{\tan \theta-1}{\sqrt{2\tan \theta}}\right)+C

\displaystyle \textbf{Question 30: }\quad \text{Evaluate:}\ \int \frac{1}{\sin^{4}x+\cos^{4}x}\,dx \hspace{5.0cm} \text{[CBSE 2014]}
\displaystyle \text{Answer:}
\displaystyle  \text{Let }I=\int \frac{1}{\sin^{4}x+\cos^{4}x}\,dx.\ \text{Then,}
\displaystyle I=\int \frac{\frac{1}{\cos^{4}x}}{\frac{\sin^{4}x+\cos^{4}x}{\cos^{4}x}}\,dx=\int \frac{\sec^{4}x}{\tan^{4}x+1}\,dx
\displaystyle =\int \frac{\sec^{2}x\cdot\sec^{2}x}{\tan^{4}x+1}\,dx=\int \frac{1+\tan^{2}x}{1+\tan^{4}x}\sec^{2}x\,dx
\displaystyle \text{Putting }\tan x=t\ \text{and }\sec^{2}x\,dx=dt,\ \text{we get}
\displaystyle I=\int \frac{1+t^{2}}{1+t^{4}}\,dt=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{t^{2}-1}{\sqrt{2}t}\right)+C
\displaystyle \Rightarrow\ I=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{\tan^{2}x-1}{\sqrt{2}\tan x}\right)+C

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