\displaystyle \textbf{Question 1. }\int(\sin^{4}x-\cos^{4}x)\,dx\text{ is equal to}
\displaystyle \text{Answer:}
\displaystyle \int(\sin^{4}x-\cos^{4}x)\,dx
\displaystyle =\int\left[(\sin^{2}x)^{2}-(\cos^{2}x)^{2}\right]dx
\displaystyle =\int(\sin^{2}x-\cos^{2}x)(\sin^{2}x+\cos^{2}x)\,dx
\displaystyle =\int(\sin^{2}x-\cos^{2}x)\,dx
\displaystyle =\int(-\cos 2x)\,dx
\displaystyle =-\frac{\sin 2x}{2}+C
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 2. }\text{If }\int x^{-3}5^{\frac{1}{x^{2}}}\,dx=k5^{\frac{1}{x^{2}}}+C,\text{ then }k\text{ is equal to}
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=\frac{1}{x^{2}}
\displaystyle \therefore \frac{dt}{dx}=-\frac{2}{x^{3}}=-2x^{-3}
\displaystyle \therefore x^{-3}dx=-\frac{1}{2}dt
\displaystyle \int x^{-3}5^{\frac{1}{x^{2}}}\,dx=-\frac{1}{2}\int 5^{t}\,dt
\displaystyle =-\frac{1}{2}\cdot\frac{5^{t}}{\log 5}+C
\displaystyle =-\frac{5^{\frac{1}{x^{2}}}}{2\log 5}+C
\displaystyle \therefore k=-\frac{1}{2\log 5}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 3. }\int x\sqrt{1+x^{2}}\,dx\text{ is equal to}
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=1+x^{2}
\displaystyle \therefore dt=2x\,dx
\displaystyle \therefore x\,dx=\frac{1}{2}dt
\displaystyle \int x\sqrt{1+x^{2}}\,dx=\frac{1}{2}\int t^{\frac{1}{2}}\,dt
\displaystyle =\frac{1}{2}\cdot\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+C
\displaystyle =\frac{t^{\frac{3}{2}}}{3}+C
\displaystyle =\frac{(1+x^{2})^{\frac{3}{2}}}{3}+C
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 4. }\int\frac{\sqrt{\tan x}}{\sin x\cos x}\,dx\text{ is equal to}
\displaystyle \text{Answer:}
\displaystyle \int\frac{\sqrt{\tan x}}{\sin x\cos x}\,dx
\displaystyle =\int\frac{\sqrt{\tan x}}{\tan x\cos^{2}x}\,dx
\displaystyle =\int\frac{\sec^{2}x}{\sqrt{\tan x}}\,dx
\displaystyle \text{Let }t=\tan x
\displaystyle \therefore dt=\sec^{2}x\,dx
\displaystyle \therefore \int\frac{\sec^{2}x}{\sqrt{\tan x}}\,dx=\int t^{-\frac{1}{2}}\,dt
\displaystyle =2t^{\frac{1}{2}}+C
\displaystyle =2\sqrt{\tan x}+C
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 5. }\int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin 2x}\,dx\text{ is equal to}
\displaystyle \text{Answer:}
\displaystyle \int_{0}^{\frac{\pi}{2}}\sqrt{1-\sin 2x}\,dx
\displaystyle =\int_{0}^{\frac{\pi}{2}}\sqrt{\sin^{2}x+\cos^{2}x-2\sin x\cos x}\,dx
\displaystyle =\int_{0}^{\frac{\pi}{2}}\sqrt{(\sin x-\cos x)^{2}}\,dx
\displaystyle =\int_{0}^{\frac{\pi}{2}}|\sin x-\cos x|\,dx
\displaystyle =\int_{0}^{\frac{\pi}{4}}(\cos x-\sin x)\,dx+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x)\,dx
\displaystyle =[\sin x+\cos x]_{0}^{\frac{\pi}{4}}+[-\cos x-\sin x]_{\frac{\pi}{4}}^{\frac{\pi}{2}}
\displaystyle =(\sqrt{2}-1)+(\sqrt{2}-1)
\displaystyle =2(\sqrt{2}-1)
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 6. }\int\frac{1}{x^{2}+2x+2}\,dx\text{ is equal to}
\displaystyle \text{Answer:}
\displaystyle \int\frac{1}{x^{2}+2x+2}\,dx
\displaystyle =\int\frac{1}{(x+1)^{2}+1}\,dx
\displaystyle \text{Let }t=x+1
\displaystyle \therefore dt=dx
\displaystyle \therefore \int\frac{1}{(x+1)^{2}+1}\,dx=\int\frac{1}{t^{2}+1}\,dt
\displaystyle =\tan^{-1}t+C
\displaystyle =\tan^{-1}(x+1)+C
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 7. }\int\frac{1}{\sqrt{9x-4x^{2}}}\,dx\text{ is equal to}
\displaystyle \text{Answer:}
\displaystyle 9x-4x^{2}=\frac{81-(8x-9)^{2}}{16}
\displaystyle \therefore \int\frac{1}{\sqrt{9x-4x^{2}}}\,dx=\int\frac{4}{\sqrt{81-(8x-9)^{2}}}\,dx
\displaystyle \text{Let }t=\frac{8x-9}{9}
\displaystyle \therefore dx=\frac{9}{8}\,dt
\displaystyle =\frac{1}{2}\int\frac{1}{\sqrt{1-t^{2}}}\,dt
\displaystyle =\frac{1}{2}\sin^{-1}t+C
\displaystyle =\frac{1}{2}\sin^{-1}\left(\frac{8x-9}{9}\right)+C
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 8. }\int\frac{e^{x}(1+x)}{\cos^{2}(xe^{x})}\,dx\text{ is equal to}
\displaystyle \text{Answer:}
\displaystyle \int\frac{e^{x}(1+x)}{\cos^{2}(xe^{x})}\,dx
\displaystyle =\int e^{x}(1+x)\sec^{2}(xe^{x})\,dx
\displaystyle \text{Let }t=xe^{x}
\displaystyle \therefore dt=e^{x}(1+x)\,dx
\displaystyle \therefore \int e^{x}(1+x)\sec^{2}(xe^{x})\,dx=\int\sec^{2}t\,dt
\displaystyle =\tan t+C
\displaystyle =\tan(xe^{x})+C
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 9. }\text{If }\int_{0}^{1}\frac{e^{t}}{1+t}\,dt=a,\text{ then }\int_{0}^{1}\frac{e^{t}}{(1+t)^{2}}\,dt\text{ is equal to}
\displaystyle \text{Answer:}
\displaystyle \frac{d}{dt}\left(\frac{e^{t}}{1+t}\right)=\frac{e^{t}}{1+t}-\frac{e^{t}}{(1+t)^{2}}
\displaystyle \therefore \frac{e^{t}}{(1+t)^{2}}=\frac{e^{t}}{1+t}-\frac{d}{dt}\left(\frac{e^{t}}{1+t}\right)
\displaystyle \int_{0}^{1}\frac{e^{t}}{(1+t)^{2}}\,dt=\int_{0}^{1}\frac{e^{t}}{1+t}\,dt-\left[\frac{e^{t}}{1+t}\right]_{0}^{1}
\displaystyle =a-\left(\frac{e}{2}-1\right)
\displaystyle =a+1-\frac{e}{2}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 10. }\text{Evaluate }\int\frac{(x+1)(x+\log x)^{2}}{x}\,dx
\displaystyle \text{Answer:}
\displaystyle \int\frac{(x+1)(x+\log x)^{2}}{x}\,dx
\displaystyle =\int\left(1+\frac{1}{x}\right)(x+\log x)^{2}\,dx
\displaystyle \text{Let }t=x+\log x
\displaystyle \therefore \frac{dt}{dx}=1+\frac{1}{x}
\displaystyle \therefore dt=\left(1+\frac{1}{x}\right)dx
\displaystyle \therefore \int\left(1+\frac{1}{x}\right)(x+\log x)^{2}\,dx=\int t^{2}\,dt
\displaystyle =\frac{t^{3}}{3}+C
\displaystyle =\frac{(x+\log x)^{3}}{3}+C
\\

\displaystyle \textbf{Question 11. }\text{Evaluate }\int x\cos x\,dx
\displaystyle \text{Answer:}
\displaystyle \int x\cos x\,dx
\displaystyle \text{Using integration by parts, }
\displaystyle \int x\cos x\,dx=x\int\cos x\,dx-\int\left(\frac{d}{dx}x\right)\left(\int\cos x\,dx\right)dx
\displaystyle =x\sin x-\int\sin x\,dx
\displaystyle =x\sin x+\cos x+C
\\

\displaystyle \textbf{Question 12. }\text{Evaluate }\int\frac{(x^{4}-x)^{\frac{1}{4}}}{x^{5}}\,dx
\displaystyle \text{Answer:}
\displaystyle \int\frac{(x^{4}-x)^{\frac{1}{4}}}{x^{5}}\,dx
\displaystyle =\int\frac{\left[x^{4}\left(1-\frac{1}{x^{3}}\right)\right]^{\frac{1}{4}}}{x^{5}}\,dx
\displaystyle =\int\frac{x\left(1-\frac{1}{x^{3}}\right)^{\frac{1}{4}}}{x^{5}}\,dx
\displaystyle =\int\frac{1}{x^{4}}\left(1-\frac{1}{x^{3}}\right)^{\frac{1}{4}}\,dx
\displaystyle \text{Let }t=1-\frac{1}{x^{3}}
\displaystyle \therefore dt=\frac{3}{x^{4}}\,dx
\displaystyle \therefore \frac{1}{x^{4}}\,dx=\frac{1}{3}dt
\displaystyle \therefore \int\frac{1}{x^{4}}\left(1-\frac{1}{x^{3}}\right)^{\frac{1}{4}}\,dx=\frac{1}{3}\int t^{\frac{1}{4}}\,dt
\displaystyle =\frac{1}{3}\cdot\frac{t^{\frac{5}{4}}}{\frac{5}{4}}+C
\displaystyle =\frac{4}{15}t^{\frac{5}{4}}+C
\displaystyle =\frac{4}{15}\left(1-\frac{1}{x^{3}}\right)^{\frac{5}{4}}+C
\\

\displaystyle \textbf{Question 13. }\text{Evaluate }\int_{0}^{\frac{\pi}{4}}\sqrt{1+\sin 2x}\,dx
\displaystyle \text{Answer:}
\displaystyle \int_{0}^{\frac{\pi}{4}}\sqrt{1+\sin 2x}\,dx
\displaystyle =\int_{0}^{\frac{\pi}{4}}\sqrt{\sin^{2}x+\cos^{2}x+2\sin x\cos x}\,dx
\displaystyle =\int_{0}^{\frac{\pi}{4}}\sqrt{(\sin x+\cos x)^{2}}\,dx
\displaystyle =\int_{0}^{\frac{\pi}{4}}(\sin x+\cos x)\,dx
\displaystyle =\left[-\cos x+\sin x\right]_{0}^{\frac{\pi}{4}}
\displaystyle =\left(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(-1+0)
\displaystyle =1
\\

\displaystyle \textbf{Question 14. }\text{Evaluate }\int_{0}^{\frac{\pi}{2}}\frac{\tan^{7}x}{\cot^{7}x+\tan^{7}x}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\frac{\pi}{2}}\frac{\tan^{7}x}{\cot^{7}x+\tan^{7}x}\,dx
\displaystyle \text{Using }I=\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\tan^{7}\left(\frac{\pi}{2}-x\right)}{\cot^{7}\left(\frac{\pi}{2}-x\right)+\tan^{7}\left(\frac{\pi}{2}-x\right)}\,dx
\displaystyle =\int_{0}^{\frac{\pi}{2}}\frac{\cot^{7}x}{\tan^{7}x+\cot^{7}x}\,dx
\displaystyle \therefore 2I=\int_{0}^{\frac{\pi}{2}}\frac{\tan^{7}x+\cot^{7}x}{\tan^{7}x+\cot^{7}x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}1\,dx=\frac{\pi}{2}
\displaystyle \therefore I=\frac{\pi}{4}
\\

\displaystyle \textbf{Question 15. }\text{Evaluate }\int\sqrt{\tan x}\,(1+\tan^{2}x)\,dx
\displaystyle \text{Answer:}
\displaystyle \int\sqrt{\tan x}\,(1+\tan^{2}x)\,dx
\displaystyle \text{Let }t=\tan x
\displaystyle \therefore dt=(1+\tan^{2}x)\,dx
\displaystyle \therefore \int\sqrt{\tan x}\,(1+\tan^{2}x)\,dx=\int t^{\frac{1}{2}}\,dt
\displaystyle =\frac{t^{\frac{3}{2}}}{\frac{3}{2}}+C
\displaystyle =\frac{2}{3}t^{\frac{3}{2}}+C
\displaystyle =\frac{2}{3}(\tan x)^{\frac{3}{2}}+C
\\

\displaystyle \textbf{Question 16. }\text{Evaluate }\int_{2}^{8}|x-5|\,dx
\displaystyle \text{Answer:}
\displaystyle \int_{2}^{8}|x-5|\,dx=\int_{2}^{5}(5-x)\,dx+\int_{5}^{8}(x-5)\,dx
\displaystyle =\left[5x-\frac{x^{2}}{2}\right]_{2}^{5}+\left[\frac{x^{2}}{2}-5x\right]_{5}^{8}
\displaystyle =\frac{9}{2}+\frac{9}{2}
\displaystyle =9
\\

\displaystyle \textbf{Question 17. }\text{Evaluate }\int_{0}^{\pi}\left(\sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}\right)\,dx
\displaystyle \text{Answer:}
\displaystyle \sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}=-\cos x
\displaystyle \therefore \int_{0}^{\pi}\left(\sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}\right)\,dx=\int_{0}^{\pi}-\cos x\,dx
\displaystyle =[-\sin x]_{0}^{\pi}
\displaystyle =0
\\

\displaystyle \textbf{Question 18. }\text{Evaluate }\int_{0}^{2}x\sqrt{x+2}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=x+2
\displaystyle \therefore x=t-2,\ dx=dt
\displaystyle \text{When }x=0,\ t=2\text{ and when }x=2,\ t=4
\displaystyle \therefore \int_{0}^{2}x\sqrt{x+2}\,dx=\int_{2}^{4}(t-2)t^{\frac{1}{2}}\,dt
\displaystyle =\int_{2}^{4}\left(t^{\frac{3}{2}}-2t^{\frac{1}{2}}\right)dt
\displaystyle =\left[\frac{2}{5}t^{\frac{5}{2}}-\frac{4}{3}t^{\frac{3}{2}}\right]_{2}^{4}
\displaystyle =\left(\frac{64}{5}-\frac{32}{3}\right)-\left(\frac{8\sqrt{2}}{5}-\frac{8\sqrt{2}}{3}\right)
\displaystyle =\frac{32}{15}+\frac{16\sqrt{2}}{15}
\displaystyle =\frac{16}{15}(2+\sqrt{2})
\\

\displaystyle \textbf{Question 19. }\text{Evaluate }\int\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\,dx
\displaystyle \text{Answer:}
\displaystyle \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)=2\tan^{-1}x
\displaystyle \therefore \int\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\,dx=\int 2\tan^{-1}x\,dx
\displaystyle \text{Using integration by parts,}
\displaystyle =2\left[x\tan^{-1}x-\int\frac{x}{1+x^{2}}\,dx\right]
\displaystyle =2x\tan^{-1}x-\log(1+x^{2})+C
\\

\displaystyle \textbf{Question 20. }\text{Evaluate }\int\frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx
\displaystyle \text{Answer:}
\displaystyle \int\frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx
\displaystyle =\int\frac{3(2x-9)+34}{\sqrt{(x-5)(x-4)}}\,dx
\displaystyle =3\int\frac{2x-9}{\sqrt{x^{2}-9x+20}}\,dx+34\int\frac{dx}{\sqrt{x^{2}-9x+20}}
\displaystyle =6\sqrt{x^{2}-9x+20}+34\log\left|2x-9+2\sqrt{x^{2}-9x+20}\right|+C
\displaystyle =6\sqrt{(x-5)(x-4)}+34\log\left|2x-9+2\sqrt{(x-5)(x-4)}\right|+C
\\

\displaystyle \textbf{Question 21. }\text{Evaluate }\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x}{\sin x+\cos x}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x}{\sin x+\cos x}\,dx
\displaystyle \text{Using }I=\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{2}x}{\sin x+\cos x}\,dx
\displaystyle \therefore 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}x+\cos^{2}x}{\sin x+\cos x}\,dx
\displaystyle 2I=\int_{0}^{\frac{\pi}{2}}\frac{dx}{\sin x+\cos x}
\displaystyle =\frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}}\mathrm{cosec}\left(x+\frac{\pi}{4}\right)\,dx
\displaystyle =\frac{1}{\sqrt{2}}\left[\log\tan\left(\frac{x}{2}+\frac{\pi}{8}\right)\right]_{0}^{\frac{\pi}{2}}
\displaystyle =\frac{1}{\sqrt{2}}\log(3+2\sqrt{2})
\displaystyle \therefore I=\frac{1}{2\sqrt{2}}\log(3+2\sqrt{2})
\\

\displaystyle \textbf{Question 22. }\text{Evaluate }\int_{-1}^{2}f(x)\,dx,\text{ where }f(x)=|x+1|+|x|+|x-1|
\displaystyle \text{Answer:}
\displaystyle \int_{-1}^{2}\left(|x+1|+|x|+|x-1|\right)\,dx
\displaystyle =\int_{-1}^{0}\left((x+1)-x+(1-x)\right)\,dx+\int_{0}^{1}\left((x+1)+x+(1-x)\right)\,dx+\int_{1}^{2}\left((x+1)+x+(x-1)\right)\,dx
\displaystyle =\int_{-1}^{0}(2-x)\,dx+\int_{0}^{1}(x+2)\,dx+\int_{1}^{2}3x\,dx
\displaystyle =\left[2x-\frac{x^{2}}{2}\right]_{-1}^{0}+\left[\frac{x^{2}}{2}+2x\right]_{0}^{1}+\left[\frac{3x^{2}}{2}\right]_{1}^{2}
\displaystyle =\frac{5}{2}+\frac{5}{2}+\frac{9}{2}
\displaystyle =\frac{19}{2}
\\

\displaystyle \textbf{Question 23. }\text{Evaluate }\int\tan^{-1}(\sec x+\tan x)\,dx,\ 0\leq x\leq\frac{\pi}{2}
\displaystyle \text{Answer:}
\displaystyle \sec x+\tan x=\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)
\displaystyle \therefore \tan^{-1}(\sec x+\tan x)=\frac{\pi}{4}+\frac{x}{2}
\displaystyle \therefore \int\tan^{-1}(\sec x+\tan x)\,dx=\int\left(\frac{\pi}{4}+\frac{x}{2}\right)dx
\displaystyle =\frac{\pi x}{4}+\frac{x^{2}}{4}+C
\\

\displaystyle \textbf{Question 24. }\text{Evaluate }\int\frac{\sec^{2}(2\tan^{-1}x)}{1+x^{2}}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=2\tan^{-1}x
\displaystyle \therefore dt=\frac{2}{1+x^{2}}\,dx
\displaystyle \therefore \frac{dx}{1+x^{2}}=\frac{dt}{2}
\displaystyle \int\frac{\sec^{2}(2\tan^{-1}x)}{1+x^{2}}\,dx=\frac{1}{2}\int\sec^{2}t\,dt
\displaystyle =\frac{1}{2}\tan t+C
\displaystyle =\frac{1}{2}\tan(2\tan^{-1}x)+C
\displaystyle =\frac{x}{1-x^{2}}+C
\\

\displaystyle \textbf{Question 25. }\text{Evaluate }\int\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\,dx
\displaystyle \text{Answer:}
\displaystyle \sin^{-1}\left(\frac{2x}{1+x^{2}}\right)=2\tan^{-1}x
\displaystyle \therefore \int\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)\,dx=\int2\tan^{-1}x\,dx
\displaystyle \text{Using integration by parts,}
\displaystyle =2\left[x\tan^{-1}x-\int\frac{x}{1+x^{2}}\,dx\right]
\displaystyle =2x\tan^{-1}x-\log(1+x^{2})+C
\\

\displaystyle \textbf{Question 25. }\text{Evaluate }\int\sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right)\,dx
\displaystyle \text{Answer:}
\displaystyle I=\int\sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right)\,dx
\displaystyle \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right)=\tan^{-1}\sqrt{\frac{x}{a}}
\displaystyle \therefore I=\int\tan^{-1}\sqrt{\frac{x}{a}}\,dx
\displaystyle \text{Using integration by parts,}
\displaystyle I=x\tan^{-1}\sqrt{\frac{x}{a}}-\int x\cdot\frac{\sqrt{a}}{2\sqrt{x}(a+x)}\,dx
\displaystyle =x\tan^{-1}\sqrt{\frac{x}{a}}-\frac{\sqrt{a}}{2}\int\frac{\sqrt{x}}{a+x}\,dx
\displaystyle \text{Let }\sqrt{x}=t
\displaystyle \therefore x=t^{2},\ dx=2t\,dt
\displaystyle \int\frac{\sqrt{x}}{a+x}\,dx=\int\frac{2t^{2}}{a+t^{2}}\,dt
\displaystyle =2t-2\sqrt{a}\tan^{-1}\frac{t}{\sqrt{a}}
\displaystyle =2\sqrt{x}-2\sqrt{a}\tan^{-1}\sqrt{\frac{x}{a}}
\displaystyle \therefore I=x\tan^{-1}\sqrt{\frac{x}{a}}-\sqrt{ax}+a\tan^{-1}\sqrt{\frac{x}{a}}+C
\displaystyle =(a+x)\tan^{-1}\sqrt{\frac{x}{a}}-\sqrt{ax}+C
\\

\displaystyle \textbf{Question 27. }\text{Evaluate }\int\frac{3x-2}{(x+1)^{2}(x+3)}\,dx
\displaystyle \text{Answer:}
\displaystyle \frac{3x-2}{(x+1)^{2}(x+3)}=\frac{11}{4(x+1)}-\frac{5}{2(x+1)^{2}}-\frac{11}{4(x+3)}
\displaystyle \therefore \int\frac{3x-2}{(x+1)^{2}(x+3)}\,dx
\displaystyle =\frac{11}{4}\log|x+1|+\frac{5}{2(x+1)}-\frac{11}{4}\log|x+3|+C
\displaystyle =\frac{11}{4}\log\left|\frac{x+1}{x+3}\right|+\frac{5}{2(x+1)}+C
\\

\displaystyle \textbf{Question 28. }\text{Evaluate }\int_{0}^{\frac{\pi}{4}}\sec x\sqrt{\frac{1-\sin x}{1+\sin x}}\,dx
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1-\sin x}{1+\sin x}}=\sqrt{\frac{(1-\sin x)^{2}}{\cos^{2}x}}
\displaystyle =\frac{1-\sin x}{\cos x}
\displaystyle \therefore \int_{0}^{\frac{\pi}{4}}\sec x\sqrt{\frac{1-\sin x}{1+\sin x}}\,dx
\displaystyle =\int_{0}^{\frac{\pi}{4}}\frac{1-\sin x}{\cos^{2}x}\,dx
\displaystyle =\int_{0}^{\frac{\pi}{4}}(\sec^{2}x-\sec x\tan x)\,dx
\displaystyle =[\tan x-\sec x]_{0}^{\frac{\pi}{4}}
\displaystyle =(1-\sqrt{2})-(0-1)
\displaystyle =2-\sqrt{2}
\\

\displaystyle \textbf{Question 29. }\text{Find }\int\left[\log(\log x)+\frac{1}{(\log x)^{2}}\right]\,dx
\displaystyle \text{Answer:}
\displaystyle \int\left[\log(\log x)+\frac{1}{(\log x)^{2}}\right]\,dx
\displaystyle =\int\left[\log(\log x)+\frac{1}{\log x}-\frac{1}{\log x}+\frac{1}{(\log x)^{2}}\right]\,dx
\displaystyle =\int\left[\frac{d}{dx}\left(x\log(\log x)\right)-\frac{d}{dx}\left(\frac{x}{\log x}\right)\right]dx
\displaystyle =x\log(\log x)-\frac{x}{\log x}+C
\\

\displaystyle \textbf{Question 30. }\text{Evaluate }\int\frac{x^{3}+x}{x^{4}-9}\,dx
\displaystyle \text{Answer:}
\displaystyle \int\frac{x^{3}+x}{x^{4}-9}\,dx
\displaystyle =\int\frac{x(x^{2}+1)}{(x^{2})^{2}-9}\,dx
\displaystyle \text{Let }t=x^{2}
\displaystyle \therefore dt=2x\,dx
\displaystyle \therefore x\,dx=\frac{1}{2}dt
\displaystyle \therefore \int\frac{x(x^{2}+1)}{(x^{2})^{2}-9}\,dx=\frac{1}{2}\int\frac{t+1}{t^{2}-9}\,dt
\displaystyle =\frac{1}{2}\int\frac{t+1}{(t-3)(t+3)}\,dt
\displaystyle \frac{t+1}{(t-3)(t+3)}=\frac{2}{3(t-3)}+\frac{1}{3(t+3)}
\displaystyle \therefore \frac{1}{2}\int\frac{t+1}{(t-3)(t+3)}\,dt
\displaystyle =\frac{1}{2}\int\left[\frac{2}{3(t-3)}+\frac{1}{3(t+3)}\right]dt
\displaystyle =\frac{1}{3}\log|t-3|+\frac{1}{6}\log|t+3|+C
\displaystyle =\frac{1}{3}\log|x^{2}-3|+\frac{1}{6}\log|x^{2}+3|+C
\displaystyle =\left(\frac{1}{4}+\frac{1}{12}\right)\log|x^{2}-3|+\left(\frac{1}{4}-\frac{1}{12}\right)\log|x^{2}+3|+C 
\displaystyle =\frac{1}{4}\log|x^{2}-3|+\frac{1}{4}\log|x^{2}+3|+\frac{1}{12}\log|x^{2}-3|-\frac{1}{12}\log|x^{2}+3|+C   
\displaystyle =\frac{1}{4}\log|(x^{2}-3)(x^{2}+3)|+\frac{1}{12}\log\left|\frac{x^{2}-3}{x^{2}+3}\right|+C 
\displaystyle =\frac{1}{4}\log|x^{4}-9|+\frac{1}{12}\log\left|\frac{x^{2}-3}{x^{2}+3}\right|+C 
\displaystyle \therefore \frac{1}{3}\log|x^{2}-3|+\frac{1}{6}\log|x^{2}+3|+C=\frac{1}{4}\log|x^{4}-9|+\frac{1}{12}\log\left|\frac{x^{2}-3}{x^{2}+3}\right|+C 
\\

\displaystyle \textbf{Question 31. }\text{Evaluate }\int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{\pi}\frac{x\sin x}{1+\cos^{2}x}\,dx
\displaystyle \text{Using }I=\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
\displaystyle I=\int_{0}^{\pi}\frac{(\pi-x)\sin x}{1+\cos^{2}x}\,dx
\displaystyle \therefore 2I=\int_{0}^{\pi}\frac{\pi\sin x}{1+\cos^{2}x}\,dx
\displaystyle 2I=\pi\int_{0}^{\pi}\frac{\sin x}{1+\cos^{2}x}\,dx
\displaystyle \text{Let }t=\cos x
\displaystyle \therefore dt=-\sin x\,dx
\displaystyle 2I=\pi\int_{-1}^{1}\frac{dt}{1+t^{2}}
\displaystyle =\pi[\tan^{-1}t]_{-1}^{1}
\displaystyle =\pi\left(\frac{\pi}{4}+\frac{\pi}{4}\right)
\displaystyle =\frac{\pi^{2}}{2}
\displaystyle \therefore I=\frac{\pi^{2}}{4}
\\

\displaystyle \textbf{Question 32. }\text{Evaluate }\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{\tan x}}
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{\tan x}}
\displaystyle \text{Using }I=\int_{a}^{b}f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx
\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{\tan\left(\frac{\pi}{2}-x\right)}}
\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{\cot x}}
\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\,dx
\displaystyle \therefore 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}1\,dx
\displaystyle 2I=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}
\displaystyle \therefore I=\frac{\pi}{12}
\\

\displaystyle \textbf{Question 33. }\text{Evaluate }\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx
\displaystyle \text{Answer:}
\displaystyle \text{Let }I=\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx
\displaystyle \text{Using }I=\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx
\displaystyle I=\int_{0}^{a}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\,dx
\displaystyle \therefore 2I=\int_{0}^{a}\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\,dx
\displaystyle 2I=\int_{0}^{a}1\,dx=a
\displaystyle \therefore I=\frac{a}{2}
\\

\displaystyle \textbf{Question 34. }\text{Evaluate }\int_{-1}^{2}|x^{3}-x|\,dx
\displaystyle \text{Answer:}
\displaystyle x^{3}-x=x(x-1)(x+1)
\displaystyle \int_{-1}^{2}|x^{3}-x|\,dx=\int_{-1}^{0}(x^{3}-x)\,dx+\int_{0}^{1}(x-x^{3})\,dx+\int_{1}^{2}(x^{3}-x)\,dx
\displaystyle =\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0}+\left[\frac{x^{2}}{2}-\frac{x^{4}}{4}\right]_{0}^{1}+\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{1}^{2}
\displaystyle =\frac{1}{4}+\frac{1}{4}+\frac{9}{4}
\displaystyle =\frac{11}{4}
\\

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