Class 12: Definite Integrals

$\displaystyle \text{Evaluate the following integrals:}$

$\displaystyle \textbf{Question 1: }~\int_{2}^{4}\frac{x}{x^2+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x^2=t$
$\displaystyle \text{Then }2x\,dx=dt$
$\displaystyle \text{When }x=2,\;t=4\text{ and when }x=4,\;t=16$
$\displaystyle I=\int_2^4\frac{x}{x^2+1}\,dx$
$\displaystyle I=\int_4^{16}\frac{1}{2(t+1)}\,dt$
$\displaystyle I=\frac12\left[\log(t+1)\right]_4^{16}$
$\displaystyle I=\frac12(\log17-\log5)$
$\displaystyle I=\frac12\log\frac{17}{5}$

$\displaystyle \textbf{Question 2: }~\int_{1}^{2}\frac{1}{x(1+\log x)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }1+\log x=t$
$\displaystyle \text{Then }\frac{1}{x}\,dx=dt$
$\displaystyle \text{When }x=1,\;t=1\text{ and when }x=2,\;t=1+\log2$
$\displaystyle I=\int_1^2\frac{1}{x(1+\log x)^2}\,dx$
$\displaystyle I=\int_1^{1+\log2}\frac{1}{t^2}\,dt$
$\displaystyle I=\left[-\frac{1}{t}\right]_1^{1+\log2}$
$\displaystyle I=-\frac{1}{1+\log2}+1$
$\displaystyle I=\frac{\log2}{1+\log2}$
$\displaystyle I=\frac{\log2}{\log e+\log2}$
$\displaystyle I=\frac{\log2}{\log(2e)}$

$\displaystyle \textbf{Question 3: }~\int_{1}^{2}\frac{3x}{9x^2-1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } x^2=t.$
$\displaystyle \text{Then } 2x\,dx=dt.$
$\displaystyle \text{When } x=1,\ t=1 \text{ and when } x=2,\ t=4.$
$\displaystyle \therefore I=\int_{1}^{2}\frac{3x}{9x^2-1}\,dx.$
$\displaystyle I=\frac{3}{2}\int_{1}^{4}\frac{dt}{9t-1}.$
$\displaystyle I=\frac{3}{2}\cdot\frac{1}{9}\left[\log(9t-1)\right]_{1}^{4}.$
$\displaystyle I=\frac{1}{6}\left[\log(35)-\log(8)\right].$

$\displaystyle \textbf{Question 4: }~\int_{0}^{\pi/2}\frac{1}{5\cos x+3\sin x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{1}{5\cos x+3\sin x}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{5\left(\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)+3\left(\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)}\,dx.$
$\displaystyle \text{Using } \sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}},\ \cos x=\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1+\tan^2\frac{x}{2}}{5-5\tan^2\frac{x}{2}+6\tan\frac{x}{2}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sec^2\frac{x}{2}}{5-5\tan^2\frac{x}{2}+6\tan\frac{x}{2}}\,dx.$
$\displaystyle \text{Let } \tan\frac{x}{2}=t.$
$\displaystyle \text{Then } \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{2},\ t=1.$
$\displaystyle I=\int_{0}^{1}\frac{2\,dt}{5-5t^2+6t}.$
$\displaystyle I=\frac{1}{5}\int_{0}^{1}\frac{2\,dt}{1-t^2+\frac{6}{5}t}.$
$\displaystyle I=\frac{2}{5}\int_{0}^{1}\frac{dt}{-\left(t-\frac{3}{5}\right)^2+\frac{34}{25}}.$
$\displaystyle I=\frac{2}{5}\cdot\frac{5}{\sqrt{34}}\left[-\log\left(\frac{t-\frac{3}{5}-\frac{\sqrt{34}}{5}}{t-\frac{3}{5}+\frac{\sqrt{34}}{5}}\right)\right]_{0}^{1}.$
$\displaystyle I=\frac{1}{\sqrt{34}}\left[\log\left(\frac{8+\sqrt{34}}{8-\sqrt{34}}\right)\right].$

$\displaystyle \textbf{Question 5: }~\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } x=a\tan t.$
$\displaystyle \text{Then } dx=a\sec^2 t\,dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=a,\ t=\frac{\pi}{4}.$
$\displaystyle \therefore I=\int_{0}^{a}\frac{x}{\sqrt{a^2+x^2}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{a\tan t}{\sqrt{a^2+a^2\tan^2 t}}\cdot a\sec^2 t\,dt.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{a\tan t\sec^2 t}{\sec t}\,dt.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}a\tan t\sec t\,dt.$
$\displaystyle I=a\left[\sec t\right]_{0}^{\frac{\pi}{4}}.$
$\displaystyle I=a(\sqrt{2}-1).$

$\displaystyle \textbf{Question 6: }~\int_{0}^{1}\frac{e^x}{1+e^{2x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } e^x=t.$
$\displaystyle \text{Then } e^x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=1 \text{ and when } x=1,\ t=e.$
$\displaystyle \therefore I=\int_{0}^{1}\frac{e^x}{1+e^{2x}}\,dx.$
$\displaystyle I=\int_{1}^{e}\frac{dt}{1+t^2}.$
$\displaystyle I=\left[\tan^{-1}t\right]_{1}^{e}.$
$\displaystyle I=\tan^{-1}e-\tan^{-1}1.$
$\displaystyle I=\tan^{-1}e-\frac{\pi}{4}.$

$\displaystyle \textbf{Question 7: }~\int_{0}^{1}xe^{x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}x e^{x^2}\,dx.$
$\displaystyle \text{Let } x^2=t.$
$\displaystyle \text{Then } 2x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=1,\ t=1.$
$\displaystyle \therefore I=\frac{1}{2}\int_{0}^{1}e^{t}\,dt.$
$\displaystyle I=\frac{1}{2}\left[e^{t}\right]_{0}^{1}.$
$\displaystyle I=\frac{1}{2}(e-1).$

$\displaystyle \textbf{Question 8: }~\int_{1}^{3}\frac{\cos(\log x)}{x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{1}^{3}\frac{\cos(\log x)}{x}\,dx.$
$\displaystyle \text{Let } \log x=t.$
$\displaystyle \text{Then } \frac{1}{x}\,dx=dt.$
$\displaystyle \text{When } x=1,\ t=0 \text{ and when } x=3,\ t=\log 3.$
$\displaystyle \therefore I=\int_{0}^{\log 3}\cos t\,dt.$
$\displaystyle I=\left[\sin t\right]_{0}^{\log 3}.$
$\displaystyle I=\sin(\log 3).$

$\displaystyle \textbf{Question 9: }~\int_{0}^{1}\frac{2x}{1+x^4}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}\frac{2x}{1+x^4}\,dx.$
$\displaystyle \text{Let } x^2=t.$
$\displaystyle \text{Then } 2x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=1,\ t=1.$
$\displaystyle \therefore I=\int_{0}^{1}\frac{dt}{1+t^2}.$
$\displaystyle I=\left[\tan^{-1}t\right]_{0}^{1}.$
$\displaystyle I=\tan^{-1}1-\tan^{-1}0.$
$\displaystyle I=\frac{\pi}{4}.$

$\displaystyle \textbf{Question 10: }~\int_{0}^{a}\sqrt{a^2-x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{a}\sqrt{a^2-x^2}\,dx.$
$\displaystyle \text{Let } x=a\sin t.$
$\displaystyle \text{Then } dx=a\cos t\,dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=a,\ t=\frac{\pi}{2}.$
$\displaystyle \therefore I=\int_{0}^{\frac{\pi}{2}}\sqrt{a^2-a^2\sin^2 t}\,a\cos t\,dt.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}a^2\cos^2 t\,dt.$
$\displaystyle I=a^2\int_{0}^{\frac{\pi}{2}}\frac{1+\cos 2t}{2}\,dt.$
$\displaystyle I=\frac{a^2}{2}\left[t+\frac{\sin 2t}{2}\right]_{0}^{\frac{\pi}{2}}.$
$\displaystyle I=\frac{a^2}{2}\left(\frac{\pi}{2}-0\right).$
$\displaystyle I=\frac{a^2\pi}{4}.$

$\displaystyle \textbf{Question 11: }~\int_{0}^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi.$
$\displaystyle \text{Let } \sin\phi=t.$
$\displaystyle \text{Then } \cos\phi\,d\phi=dt.$
$\displaystyle \text{When } \phi=0,\ t=0 \text{ and when } \phi=\frac{\pi}{2},\ t=1.$
$\displaystyle \text{Also, } \cos^5\phi=\cos^4\phi\cos\phi=(1-\sin^2\phi)^2\cos\phi.$
$\displaystyle \therefore I=\int_{0}^{1}\sqrt{t}(1-t^2)^2\,dt.$
$\displaystyle I=\int_{0}^{1}\sqrt{t}(1-2t^2+t^4)\,dt.$
$\displaystyle I=\int_{0}^{1}\left(t^{\frac12}-2t^{\frac52}+t^{\frac92}\right)\,dt.$
$\displaystyle I=\left[\frac{2}{3}t^{\frac32}-\frac{4}{7}t^{\frac72}+\frac{2}{11}t^{\frac{11}{2}}\right]_{0}^{1}.$
$\displaystyle I=\frac{2}{3}-\frac{4}{7}+\frac{2}{11}.$
$\displaystyle I=\frac{64}{231}.$

$\displaystyle \textbf{Question 12: }~\int_{0}^{\pi/2}\frac{\cos x}{1+\sin^2 x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{1+\sin^2 x}\,dx.$
$\displaystyle \text{Let } \sin x=t.$
$\displaystyle \text{Then } \cos x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{2},\ t=1.$
$\displaystyle \therefore I=\int_{0}^{1}\frac{1}{1+t^2}\,dt.$
$\displaystyle I=\left[\tan^{-1}t\right]_{0}^{1}.$
$\displaystyle I=\tan^{-1}1-\tan^{-1}0.$
$\displaystyle I=\frac{\pi}{4}.$

$\displaystyle \textbf{Question 13: }~\int_{0}^{\pi/2}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\,d\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{\sin\theta}{\sqrt{1+\cos\theta}}\,d\theta.$
$\displaystyle \text{Let } \cos\theta=t.$
$\displaystyle \text{Then } -\sin\theta\,d\theta=dt.$
$\displaystyle \text{When } \theta=0,\ t=1 \text{ and when } \theta=\frac{\pi}{2},\ t=0.$
$\displaystyle \therefore I=\int_{1}^{0}\frac{-dt}{\sqrt{1+t}}.$
$\displaystyle I=\int_{0}^{1}\frac{dt}{\sqrt{1+t}}.$
$\displaystyle I=\left[2\sqrt{1+t}\right]_{0}^{1}.$
$\displaystyle I=2(\sqrt{2}-1).$

$\displaystyle \textbf{Question 14: }~\int_{0}^{\pi/3}\frac{\cos x}{3+4\sin x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{3}}\frac{\cos x}{3+4\sin x}\,dx.$
$\displaystyle \text{Let } \sin x=t.$
$\displaystyle \text{Then } \cos x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{3},\ t=\frac{\sqrt{3}}{2}.$
$\displaystyle \therefore I=\int_{0}^{\frac{\sqrt{3}}{2}}\frac{1}{3+4t}\,dt.$
$\displaystyle I=\frac{1}{4}\left[\log(3+4t)\right]_{0}^{\frac{\sqrt{3}}{2}}.$
$\displaystyle I=\frac{1}{4}\left(\log(3+2\sqrt{3})-\log 3\right).$
$\displaystyle I=\frac{1}{4}\log\left(\frac{3+2\sqrt{3}}{3}\right).$

$\displaystyle \textbf{Question 15: }~\int_{0}^{1}\frac{\sqrt{\tan^{-1}x}}{1+x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}\frac{\sqrt{\tan^{-1}x}}{1+x^2}\,dx.$
$\displaystyle \text{Let } \tan^{-1}x=t.$
$\displaystyle \text{Then } \frac{1}{1+x^2}\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=1,\ t=\frac{\pi}{4}.$
$\displaystyle \therefore I=\int_{0}^{\frac{\pi}{4}}\sqrt{t}\,dt.$
$\displaystyle I=\left[\frac{2}{3}t^{\frac{3}{2}}\right]_{0}^{\frac{\pi}{4}}.$
$\displaystyle I=\frac{2}{3}\left(\frac{\pi}{4}\right)^{\frac{3}{2}}.$
$\displaystyle I=\frac{\pi^{\frac{3}{2}}}{12}.$

$\displaystyle \textbf{Question 16: }~\int_{0}^{2}x\sqrt{x+2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{2}x\sqrt{x+2}\,dx.$
$\displaystyle \text{Let } x+2=t^2.$
$\displaystyle \text{Then } dx=2t\,dt.$
$\displaystyle \text{When } x=0,\ t=\sqrt{2} \text{ and when } x=2,\ t=2.$
$\displaystyle \therefore I=\int_{\sqrt{2}}^{2}(t^2-2)t\cdot2t\,dt.$
$\displaystyle I=2\int_{\sqrt{2}}^{2}(t^4-2t^2)\,dt.$
$\displaystyle I=2\left[\frac{t^5}{5}-\frac{2t^3}{3}\right]_{\sqrt{2}}^{2}.$
$\displaystyle I=2\left[\left(\frac{32}{5}-\frac{16}{3}\right)-\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\right].$
$\displaystyle I=2\left(\frac{16}{15}+\frac{8\sqrt{2}}{15}\right).$
$\displaystyle I=\frac{16}{15}(2+\sqrt{2}).$

$\displaystyle \textbf{Question 17: }~\int_{0}^{1}\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int_{0}^{1}\tan^{-1}\left(\frac{2x}{1-x^2}\right)\,dx.$
$\displaystyle I=\int_{0}^{1}2\tan^{-1}x\,dx.$
$\displaystyle I=2\left[x\tan^{-1}x\right]_{0}^{1}-2\int_{0}^{1}\frac{x}{1+x^2}\,dx.$
$\displaystyle I=2\left[x\tan^{-1}x\right]_{0}^{1}-\left[\log(1+x^2)\right]_{0}^{1}.$
$\displaystyle I=2\left(\frac{\pi}{4}-0\right)-(\log 2-0).$
$\displaystyle I=\frac{\pi}{2}-\log 2.$

$\displaystyle \textbf{Question 18: }~\int_{0}^{\pi/2}\frac{\sin x\cos x}{1+\sin^4 x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos x}{1+\sin^4 x}\,dx.$
$\displaystyle \text{Let } \sin x=t.$
$\displaystyle \text{Then } \cos x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{2},\ t=1.$
$\displaystyle \therefore I=\int_{0}^{1}\frac{t}{1+t^4}\,dt.$
$\displaystyle \text{Let } t^2=u.$
$\displaystyle \text{Then } 2t\,dt=du.$
$\displaystyle I=\frac{1}{2}\int_{0}^{1}\frac{1}{1+u^2}\,du.$
$\displaystyle I=\frac{1}{2}\left[\tan^{-1}u\right]_{0}^{1}.$
$\displaystyle I=\frac{\pi}{8}.$

$\displaystyle \textbf{Question 19: }~\int_{0}^{\pi/2}\frac{dx}{a\cos x+b\sin x},\ a,b>0.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{dx}{a\cos x+b\sin x}.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{dx}{a\left(\frac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\right)+b\left(\frac{2\tan\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}\right)}.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1+\tan^{2}\frac{x}{2}}{a-a\tan^{2}\frac{x}{2}+2b\tan\frac{x}{2}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}\frac{x}{2}}{a-a\tan^{2}\frac{x}{2}+2b\tan\frac{x}{2}}\,dx.$
$\displaystyle \text{Let } \tan\frac{x}{2}=t.$
$\displaystyle \text{Then } \frac{1}{2}\sec^{2}\frac{x}{2}\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{2},\ t=1.$
$\displaystyle \therefore I=2\int_{0}^{1}\frac{dt}{a-at^{2}+2bt}.$
$\displaystyle I=2\int_{0}^{1}\frac{dt}{-a\left(t^{2}-\frac{2b}{a}t-1\right)}.$
$\displaystyle I=\frac{2}{a}\int_{0}^{1}\frac{dt}{\left(\frac{b^{2}}{a^{2}}+1\right)-\left(t-\frac{b}{a}\right)^{2}}.$
$\displaystyle I=\frac{2}{a}\int_{0}^{1}\frac{dt}{\left(\frac{a^{2}+b^{2}}{a^{2}}\right)-\left(t-\frac{b}{a}\right)^{2}}.$
$\displaystyle I=\frac{2}{\sqrt{a^{2}+b^{2}}}\left[\log\left(\frac{\sqrt{a^{2}+b^{2}}+a t-b}{\sqrt{a^{2}+b^{2}}-a t+b}\right)\right]_{0}^{1}.$
$\displaystyle I=\frac{1}{\sqrt{a^{2}+b^{2}}}\log\left(\frac{a+b+\sqrt{a^{2}+b^{2}}}{a+b-\sqrt{a^{2}+b^{2}}}\right).$

$\displaystyle \textbf{Question 20: }~\int_{0}^{\pi/2}\frac{1}{5+4\sin x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\sin x}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1}{5+4\left(\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{1+\tan^2\frac{x}{2}}{5(1+\tan^2\frac{x}{2})+8\tan\frac{x}{2}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sec^2\frac{x}{2}}{5\tan^2\frac{x}{2}+8\tan\frac{x}{2}+5}\,dx.$
$\displaystyle \text{Let } \tan\frac{x}{2}=t.$
$\displaystyle \text{Then } \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{2},\ t=1.$
$\displaystyle \therefore I=2\int_{0}^{1}\frac{1}{5t^2+8t+5}\,dt.$
$\displaystyle I=2\int_{0}^{1}\frac{1}{5\left(t+\frac{4}{5}\right)^2+\frac{9}{5}}\,dt.$
$\displaystyle I=\frac{2}{5}\int_{0}^{1}\frac{1}{\left(t+\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2}\,dt.$
$\displaystyle I=\frac{2}{5}\cdot\frac{5}{3}\left[\tan^{-1}\left(\frac{t+\frac{4}{5}}{\frac{3}{5}}\right)\right]_{0}^{1}.$
$\displaystyle I=\frac{2}{3}\left[\tan^{-1}3-\tan^{-1}\frac{4}{3}\right].$
$\displaystyle I=\frac{2}{3}\tan^{-1}\frac{1}{3}.$

$\displaystyle \textbf{Question 21: }~\int_{0}^{\pi}\frac{\sin x}{\sin x+\cos x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle I=\int_{0}^{\pi}\frac{\sin x}{\sin x+\cos x}\,dx.$
$\displaystyle I=\frac{1}{2}\int_{0}^{\pi}\frac{2\sin x}{\sin x+\cos x}\,dx.$
$\displaystyle I=\frac{1}{2}\int_{0}^{\pi}\frac{(\sin x+\cos x)-(\cos x-\sin x)}{\sin x+\cos x}\,dx.$
$\displaystyle I=\frac{1}{2}\int_{0}^{\pi}dx-\frac{1}{2}\int_{0}^{\pi}\frac{\cos x-\sin x}{\sin x+\cos x}\,dx.$
$\displaystyle I=\frac{1}{2}\left[x\right]_{0}^{\pi}-\frac{1}{2}\left[\log|\sin x+\cos x|\right]_{0}^{\pi}.$
$\displaystyle I=\frac{1}{2}(\pi-0)-\frac{1}{2}(\log1-\log1).$
$\displaystyle I=\frac{\pi}{2}.$

$\displaystyle \textbf{Question 22: }~\int_{0}^{\pi}\frac{1}{3+2\sin x+\cos x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\pi}\frac{1}{3+2\sin x+\cos x}\,dx.$
$\displaystyle I=\int_{0}^{\pi}\frac{1}{3+2\left(\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)+\left(\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)}\,dx.$
$\displaystyle I=\int_{0}^{\pi}\frac{1+\tan^2\frac{x}{2}}{2\tan^2\frac{x}{2}+4\tan\frac{x}{2}+4}\,dx.$
$\displaystyle \text{Let } \tan\frac{x}{2}=t.$
$\displaystyle \text{Then } \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\pi,\ t=\infty.$
$\displaystyle \therefore I=\int_{0}^{\infty}\frac{2\,dt}{2t^2+4t+4}.$
$\displaystyle I=\int_{0}^{\infty}\frac{dt}{(t+1)^2+1}.$
$\displaystyle I=\left[\tan^{-1}(t+1)\right]_{0}^{\infty}.$
$\displaystyle I=\frac{\pi}{2}-\frac{\pi}{4}.$
$\displaystyle I=\frac{\pi}{4}.$

$\displaystyle \textbf{Question 23: }~\int_{0}^{1}\tan^{-1}x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}\tan^{-1}x\,dx.$
$\displaystyle I=\int_{0}^{1}1\cdot\tan^{-1}x\,dx.$
$\displaystyle \text{Integrating by parts.}$
$\displaystyle I=\left[x\tan^{-1}x\right]_{0}^{1}-\int_{0}^{1}\frac{x}{1+x^2}\,dx.$
$\displaystyle I=\left[x\tan^{-1}x\right]_{0}^{1}-\frac{1}{2}\left[\log(1+x^2)\right]_{0}^{1}.$
$\displaystyle I=\frac{\pi}{4}-0-\frac{1}{2}(\log2-0).$
$\displaystyle I=\frac{\pi}{4}-\frac{1}{2}\log2.$

$\displaystyle \textbf{Question 24: }~\int_{0}^{1/2}\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac12}\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Put } x=\sin\theta.$
$\displaystyle \text{Then } dx=\cos\theta\,d\theta.$
$\displaystyle \text{When } x\to0,\ \theta\to0.$
$\displaystyle \text{When } x\to\frac12,\ \theta\to\frac{\pi}{6}.$
$\displaystyle \therefore I=\int_{0}^{\frac{\pi}{6}}\frac{\sin\theta\sin^{-1}(\sin\theta)}{\cos\theta}\cos\theta\,d\theta.$
$\displaystyle I=\int_{0}^{\frac{\pi}{6}}\theta\sin\theta\,d\theta.$
$\displaystyle \text{Applying integration by parts.}$
$\displaystyle I=\left[-\theta\cos\theta\right]_{0}^{\frac{\pi}{6}}+\int_{0}^{\frac{\pi}{6}}\cos\theta\,d\theta.$
$\displaystyle I=-\frac{\pi}{6}\cos\frac{\pi}{6}+\left[\sin\theta\right]_{0}^{\frac{\pi}{6}}.$
$\displaystyle I=-\frac{\pi}{6}\cdot\frac{\sqrt{3}}{2}+\left(\sin\frac{\pi}{6}-\sin0\right).$
$\displaystyle I=\frac{1}{2}-\frac{\pi}{4\sqrt{3}}.$

$\displaystyle \textbf{Question 25: }~\int_{0}^{\pi/4}\left(\sqrt{\tan x}+\sqrt{\cot x}\right)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{4}}\left(\sqrt{\tan x}+\sqrt{\cot x}\right)\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\left(\sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}}\right)\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{\sin x+\cos x}{\sqrt{\sin x\cos x}}\,dx.$
$\displaystyle I=\sqrt{2}\int_{0}^{\frac{\pi}{4}}\frac{\sin x+\cos x}{\sqrt{2\sin x\cos x}}\,dx.$
$\displaystyle I=\sqrt{2}\int_{0}^{\frac{\pi}{4}}\frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x)^2}}\,dx.$
$\displaystyle \text{Let } \sin x-\cos x=t.$
$\displaystyle \text{Then } \cos x+\sin x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=-1 \text{ and when } x=\frac{\pi}{4},\ t=0.$
$\displaystyle \therefore I=\sqrt{2}\int_{-1}^{0}\frac{dt}{\sqrt{1-t^2}}.$
$\displaystyle I=\sqrt{2}\left[\sin^{-1}t\right]_{-1}^{0}.$
$\displaystyle I=\sqrt{2}\left(\sin^{-1}0-\sin^{-1}(-1)\right).$
$\displaystyle I=\frac{\pi}{\sqrt{2}}.$

$\displaystyle \textbf{Question 26: }~\int_{0}^{\pi/4}\frac{\tan^3 x}{1+\cos 2x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{4}}\frac{\tan^3 x}{1+\cos2x}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{\tan^3 x}{2\cos^2 x}\,dx.$
$\displaystyle I=\frac{1}{2}\int_{0}^{\frac{\pi}{4}}\tan^3 x\sec^2 x\,dx.$
$\displaystyle \text{Let } \tan x=t.$
$\displaystyle \text{Then } \sec^2 x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{4},\ t=1.$
$\displaystyle \therefore I=\frac{1}{2}\int_{0}^{1}t^3\,dt.$
$\displaystyle I=\frac{1}{2}\left[\frac{t^4}{4}\right]_{0}^{1}.$
$\displaystyle I=\frac{1}{2}\left(\frac{1}{4}-0\right).$
$\displaystyle I=\frac{1}{8}.$

$\displaystyle \textbf{Question 27: }~\int_{0}^{\pi}\frac{1}{5+3\cos x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\pi}\frac{1}{5+3\cos x}\,dx.$
$\displaystyle I=\int_{0}^{\pi}\frac{1}{5+3\left(\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\right)}\,dx.$
$\displaystyle I=\int_{0}^{\pi}\frac{1+\tan^2\frac{x}{2}}{5+5\tan^2\frac{x}{2}-3+3\tan^2\frac{x}{2}}\,dx.$
$\displaystyle I=\int_{0}^{\pi}\frac{\sec^2\frac{x}{2}}{2+8\tan^2\frac{x}{2}}\,dx.$
$\displaystyle \text{Let } \tan\frac{x}{2}=t.$
$\displaystyle \text{Then } \frac{1}{2}\sec^2\frac{x}{2}\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\pi,\ t=\infty.$
$\displaystyle \therefore I=\int_{0}^{\infty}\frac{dt}{1+4t^2}.$
$\displaystyle I=\frac{1}{2}\int_{0}^{\infty}\frac{dt}{\frac{1}{4}+t^2}.$
$\displaystyle I=\frac{1}{2}\left[\tan^{-1}(2t)\right]_{0}^{\infty}.$
$\displaystyle I=\frac{1}{2}\left(\frac{\pi}{2}-0\right).$
$\displaystyle I=\frac{\pi}{4}.$

$\displaystyle \textbf{Question 28: }~\int_{0}^{\pi/2}\frac{1}{a^2\sin^2 x+b^2\cos^2 x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{1}{a^2\sin^2 x+b^2\cos^2 x}\,dx.$
$\displaystyle \text{Dividing the numerator and denominator by } \cos^2 x.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sec^2 x}{a^2\tan^2 x+b^2}\,dx.$
$\displaystyle \text{Let } \tan x=t.$
$\displaystyle \text{Then } \sec^2 x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{2},\ t=\infty.$
$\displaystyle \therefore I=\int_{0}^{\infty}\frac{1}{a^2t^2+b^2}\,dt.$
$\displaystyle I=\frac{1}{a^2}\int_{0}^{\infty}\frac{1}{t^2+\frac{b^2}{a^2}}\,dt.$
$\displaystyle I=\frac{1}{a^2}\cdot\frac{a}{b}\left[\tan^{-1}\left(\frac{at}{b}\right)\right]_{0}^{\infty}.$
$\displaystyle I=\frac{1}{ab}\cdot\frac{\pi}{2}.$
$\displaystyle I=\frac{\pi}{2ab}.$

$\displaystyle \textbf{Question 29: }~\int_{0}^{\pi/2}\frac{x+\sin x}{1+\cos x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{x+\sin x}{1+\cos x}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{x+\sin x}{2\cos^2\frac{x}{2}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\left(\frac{x}{2\cos^2\frac{x}{2}}+\frac{\sin x}{2\cos^2\frac{x}{2}}\right)\,dx.$
$\displaystyle I=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}x\sec^2\frac{x}{2}\,dx+\int_{0}^{\frac{\pi}{2}}\tan\frac{x}{2}\,dx.$
$\displaystyle I=\left[x\tan\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\tan\frac{x}{2}\,dx+\int_{0}^{\frac{\pi}{2}}\tan\frac{x}{2}\,dx.$
$\displaystyle I=\left[x\tan\frac{x}{2}\right]_{0}^{\frac{\pi}{2}}.$
$\displaystyle I=\frac{\pi}{2}\tan\frac{\pi}{4}.$
$\displaystyle I=\frac{\pi}{2}.$

$\displaystyle \textbf{Question 30: }~\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}\frac{\tan^{-1}x}{1+x^2}\,dx.$
$\displaystyle \text{Let } \tan^{-1}x=t.$
$\displaystyle \text{Then } \frac{1}{1+x^2}\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=1,\ t=\frac{\pi}{4}.$
$\displaystyle \therefore I=\int_{0}^{\frac{\pi}{4}} t\,dt.$
$\displaystyle I=\left[\frac{t^2}{2}\right]_{0}^{\frac{\pi}{4}}.$
$\displaystyle I=\frac{\pi^2}{32}.$

$\displaystyle \textbf{Question 31: }~\int_{0}^{\pi/4}\frac{\sin x+\cos x}{3+\sin 2x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{4}}\frac{\sin x+\cos x}{3+\sin2x}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{\sin x+\cos x}{4-(1-\sin2x)}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{\sin x+\cos x}{4-(\sin^2 x+\cos^2 x-2\sin x\cos x)}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{\sin x+\cos x}{4-(\sin x-\cos x)^2}\,dx.$
$\displaystyle \text{Put } \sin x-\cos x=z.$
$\displaystyle \text{Then } (\cos x+\sin x)\,dx=dz.$
$\displaystyle \text{When } x=0,\ z=\sin0-\cos0=-1.$
$\displaystyle \text{When } x=\frac{\pi}{4},\ z=\sin\frac{\pi}{4}-\cos\frac{\pi}{4}=0.$
$\displaystyle \therefore I=\int_{-1}^{0}\frac{dz}{4-z^2}.$
$\displaystyle I=\frac{1}{4}\left[\log\left(\frac{2+z}{2-z}\right)\right]_{-1}^{0}.$
$\displaystyle I=\frac{1}{4}\left(\log1-\log\frac{1}{3}\right).$
$\displaystyle I=\frac{1}{4}\log3.$

$\displaystyle \textbf{Question 32: }~\int_{0}^{1}x\tan^{-1}x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}x\tan^{-1}x\,dx.$
$\displaystyle \text{Integrating by parts.}$
$\displaystyle I=\left[\frac{x^2\tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_{0}^{1}\frac{x^2}{1+x^2}\,dx.$
$\displaystyle I=\left[\frac{x^2\tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\int_{0}^{1}\left(1-\frac{1}{1+x^2}\right)\,dx.$
$\displaystyle I=\left[\frac{x^2\tan^{-1}x}{2}\right]_{0}^{1}-\frac{1}{2}\left[x-\tan^{-1}x\right]_{0}^{1}.$
$\displaystyle I=\frac{\pi}{8}-\frac{1}{2}\left(1-\frac{\pi}{4}\right).$
$\displaystyle I=\frac{\pi}{4}-\frac{1}{2}.$

$\displaystyle \textbf{Question 33: }~\int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int\frac{1-x^2}{x^4+x^2+1}\,dx.$
$\displaystyle I=-\int\frac{x^2-1}{x^4+x^2+1}\,dx.$
$\displaystyle I=-\int\frac{1-\frac{1}{x^2}}{x^2+1+\frac{1}{x^2}}\,dx.$
$\displaystyle I=-\int\frac{1-\frac{1}{x^2}}{x^2+2+\frac{1}{x^2}-1}\,dx.$
$\displaystyle I=-\int\frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-1}\,dx.$
$\displaystyle \text{Let } x+\frac{1}{x}=t.$
$\displaystyle \text{Then } \left(1-\frac{1}{x^2}\right)\,dx=dt.$
$\displaystyle \therefore I=-\int\frac{dt}{t^2-1}.$
$\displaystyle I=-\frac{1}{2}\log\left|\frac{t-1}{t+1}\right|.$
$\displaystyle I=\frac{1}{2}\log\left|\frac{t+1}{t-1}\right|.$
$\displaystyle I=\frac{1}{2}\log\left|\frac{x+\frac{1}{x}+1}{x+\frac{1}{x}-1}\right|.$
$\displaystyle I=\frac{1}{2}\log\left|\frac{x^2+x+1}{x^2-x+1}\right|.$
$\displaystyle \text{i.e., } \int\frac{1-x^2}{x^4+x^2+1}\,dx=\frac{1}{2}\log\left|\frac{x^2+x+1}{x^2-x+1}\right|.$
$\displaystyle \int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}\,dx=\left[\frac{1}{2}\log\left(\frac{x^2+x+1}{x^2-x+1}\right)\right]_{0}^{1}.$
$\displaystyle \int_{0}^{1}\frac{1-x^2}{x^4+x^2+1}\,dx=\frac{1}{2}\log3.$

$\displaystyle \textbf{Question 34: }~\int_{0}^{1}\frac{24x^3}{(1+x^2)^4}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}\frac{24x^3}{(1+x^2)^4}\,dx.$
$\displaystyle \text{Let } x^2=t.$
$\displaystyle \text{Then } 2x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=1,\ t=1.$
$\displaystyle \therefore I=\int_{0}^{1}\frac{12t}{(1+t)^4}\,dt.$
$\displaystyle \text{Integrating by parts.}$
$\displaystyle I=12\left[\frac{t}{-3(1+t)^3}\right]_{0}^{1}+12\int_{0}^{1}\frac{1}{3(1+t)^3}\,dt.$
$\displaystyle I=12\left\{\left[\frac{t}{-3(1+t)^3}\right]_{0}^{1}-\left[\frac{1}{6(1+t)^2}\right]_{0}^{1}\right\}.$
$\displaystyle I=12\left(-\frac{1}{24}-0-\frac{1}{24}+\frac{1}{6}\right).$
$\displaystyle I=12\cdot\frac{1}{12}.$
$\displaystyle I=1.$

$\displaystyle \textbf{Question 35: }~\int_{4}^{12}x(x-4)^{1/3}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{4}^{12}x(x-4)^{\frac{1}{3}}\,dx.$
$\displaystyle \text{Let } x-4=t.$
$\displaystyle \text{Then } dx=dt.$
$\displaystyle \text{When } x=4,\ t=0 \text{ and when } x=12,\ t=8.$
$\displaystyle \therefore I=\int_{0}^{8}(t+4)t^{\frac{1}{3}}\,dt.$
$\displaystyle I=\int_{0}^{8}\left(t^{\frac{4}{3}}+4t^{\frac{1}{3}}\right)\,dt.$
$\displaystyle I=\left[\frac{3}{7}t^{\frac{7}{3}}+3t^{\frac{4}{3}}\right]_{0}^{8}.$
$\displaystyle I=\frac{3}{7}\cdot8^{\frac{7}{3}}+3\cdot8^{\frac{4}{3}}.$
$\displaystyle I=\frac{384}{7}+48.$
$\displaystyle I=\frac{720}{7}.$

$\displaystyle \textbf{Question 36: }~\int_{0}^{\pi/2}x^2\sin x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}x^2\sin x\,dx.$
$\displaystyle \text{Integrating by parts.}$
$\displaystyle I=\left[-x^2\cos x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}(-2x\cos x)\,dx.$
$\displaystyle I=\left[-x^2\cos x\right]_{0}^{\frac{\pi}{2}}+2\int_{0}^{\frac{\pi}{2}}x\cos x\,dx.$
$\displaystyle \text{Again, integrating by parts.}$
$\displaystyle I=\left[-x^2\cos x\right]_{0}^{\frac{\pi}{2}}+2\left(\left[x\sin x\right]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\sin x\,dx\right).$
$\displaystyle I=\left[-x^2\cos x\right]_{0}^{\frac{\pi}{2}}+2\left[x\sin x\right]_{0}^{\frac{\pi}{2}}-2\left[-\cos x\right]_{0}^{\frac{\pi}{2}}.$
$\displaystyle I=0+2\cdot\frac{\pi}{2}-2(0-1).$
$\displaystyle I=\pi-2.$

$\displaystyle \textbf{Question 37: }~\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}\,dx.$
$\displaystyle I=\int_{0}^{1}\sqrt{\frac{1-x}{1+x}}\cdot\frac{\sqrt{1-x}}{\sqrt{1-x}}\,dx.$
$\displaystyle I=\int_{0}^{1}\frac{1-x}{\sqrt{1-x^2}}\,dx.$
$\displaystyle I=\int_{0}^{1}\frac{1}{\sqrt{1-x^2}}\,dx-\int_{0}^{1}\frac{x}{\sqrt{1-x^2}}\,dx.$
$\displaystyle I=\left[\sin^{-1}x\right]_{0}^{1}+\frac{1}{2}\int_{0}^{1}\frac{-2x}{\sqrt{1-x^2}}\,dx.$
$\displaystyle I=\left[\sin^{-1}x\right]_{0}^{1}+\frac{1}{2}\left[2\sqrt{1-x^2}\right]_{0}^{1}.$
$\displaystyle I=\frac{\pi}{2}-0+0-1.$
$\displaystyle I=\frac{\pi}{2}-1.$

$\displaystyle \textbf{Question 38: }~\int_{0}^{1}\frac{1-x^2}{(1+x^2)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}\frac{1-x^2}{(1+x^2)^2}\,dx.$
$\displaystyle I=\int_{0}^{1}\frac{\frac{1}{x^2}-1}{\left(x+\frac{1}{x}\right)^2}\,dx.$
$\displaystyle \text{Let } x+\frac{1}{x}=t.$
$\displaystyle \text{Then } \left(1-\frac{1}{x^2}\right)\,dx=dt.$
$\displaystyle \text{When } x\to0,\ t\to\infty \text{ and when } x=1,\ t=2.$
$\displaystyle \therefore I=\int_{\infty}^{2}\frac{-dt}{t^2}.$
$\displaystyle I=\int_{2}^{\infty}\frac{dt}{t^2}.$
$\displaystyle I=\left[-\frac{1}{t}\right]_{2}^{\infty}.$
$\displaystyle I=\frac{1}{2}-0.$
$\displaystyle I=\frac{1}{2}.$

$\displaystyle \textbf{Question 39: }~\int_{-1}^{1}5x^4\sqrt{x^5+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{-1}^{1}5x^4\sqrt{x^5+1}\,dx.$
$\displaystyle \text{Let } x^5+1=t.$
$\displaystyle \text{Then } 5x^4\,dx=dt.$
$\displaystyle \text{When } x=-1,\ t=0 \text{ and when } x=1,\ t=2.$
$\displaystyle \therefore I=\int_{0}^{2}\sqrt{t}\,dt.$
$\displaystyle I=\left[\frac{2}{3}t^{\frac{3}{2}}\right]_{0}^{2}.$
$\displaystyle I=\frac{2}{3}\sqrt{8}.$
$\displaystyle I=\frac{4\sqrt{2}}{3}.$

$\displaystyle \textbf{Question 40: }~\int_{0}^{\pi/2}\frac{\cos^2 x}{1+3\sin^2 x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{\cos^2 x}{1+3\sin^2 x}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos^2 x}{1+3(1-\cos^2 x)}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos^2 x}{4-3\cos^2 x}\,dx.$
$\displaystyle I=-\frac{1}{3}\int_{0}^{\frac{\pi}{2}}\frac{4-3\cos^2 x-4}{4-3\cos^2 x}\,dx.$
$\displaystyle I=-\frac{1}{3}\int_{0}^{\frac{\pi}{2}}dx+\frac{4}{3}\int_{0}^{\frac{\pi}{2}}\frac{1}{4-3\cos^2 x}\,dx.$
$\displaystyle I=-\frac{1}{3}\left[x\right]_{0}^{\frac{\pi}{2}}+\frac{4}{3}\int_{0}^{\frac{\pi}{2}}\frac{\sec^2 x}{4\sec^2 x-3}\,dx.$
$\displaystyle I=-\frac{\pi}{6}+\frac{4}{3}\int_{0}^{\frac{\pi}{2}}\frac{\sec^2 x}{4\tan^2 x+1}\,dx.$
$\displaystyle \text{Put } \tan x=z.$
$\displaystyle \text{Then } \sec^2 x\,dx=dz.$
$\displaystyle \text{When } x\to0,\ z\to0.$
$\displaystyle \text{When } x\to\frac{\pi}{2},\ z\to\infty.$
$\displaystyle I=-\frac{\pi}{6}+\frac{4}{3}\int_{0}^{\infty}\frac{dz}{4z^2+1}.$
$\displaystyle I=-\frac{\pi}{6}+\frac{4}{3}\int_{0}^{\infty}\frac{dz}{(2z)^2+1}.$
$\displaystyle I=-\frac{\pi}{6}+\frac{4}{3}\left[\frac{1}{2}\tan^{-1}(2z)\right]_{0}^{\infty}.$
$\displaystyle I=-\frac{\pi}{6}+\frac{2}{3}\left(\frac{\pi}{2}-0\right).$
$\displaystyle I=\frac{\pi}{6}.$

$\displaystyle \textbf{Question 41: }~\int_{0}^{\pi/4}\sin^3 2t\,\cos 2t\,dt.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{4}}\sin^3 2t\cos 2t\,dt.$
$\displaystyle \text{Let } \sin2t=u.$
$\displaystyle \text{Then } 2\cos2t\,dt=du.$
$\displaystyle \text{When } t=0,\ u=0 \text{ and when } t=\frac{\pi}{4},\ u=1.$
$\displaystyle \therefore I=\frac{1}{2}\int_{0}^{1}u^3\,du.$
$\displaystyle I=\frac{1}{2}\left[\frac{u^4}{4}\right]_{0}^{1}.$
$\displaystyle I=\frac{1}{2}\left(\frac{1}{4}-0\right).$
$\displaystyle I=\frac{1}{8}.$

$\displaystyle \textbf{Question 42: }~\int_{0}^{\pi}5(5-4\cos\theta)^{1/4}\sin\theta\,d\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\pi}5(5-4\cos\theta)^{\frac{1}{4}}\sin\theta\,d\theta.$
$\displaystyle \text{Let } 5-4\cos\theta=t.$
$\displaystyle \text{Then } 4\sin\theta\,d\theta=dt.$
$\displaystyle \text{When } \theta=0,\ t=1 \text{ and when } \theta=\pi,\ t=9.$
$\displaystyle \therefore I=\frac{5}{4}\int_{1}^{9}t^{\frac{1}{4}}\,dt.$
$\displaystyle I=\frac{5}{4}\left[\frac{4}{5}t^{\frac{5}{4}}\right]_{1}^{9}.$
$\displaystyle I=\left[t^{\frac{5}{4}}\right]_{1}^{9}.$
$\displaystyle I=9^{\frac{5}{4}}-1.$
$\displaystyle I=9\sqrt{3}-1.$

$\displaystyle \textbf{Question 43: }~\int_{0}^{\pi/6}\cos^{-3}2\theta\,\sin 2\theta\,d\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{6}}\cos^{-3}2\theta\sin2\theta\,d\theta.$
$\displaystyle I=\int_{0}^{\frac{\pi}{6}}\frac{\sin2\theta}{\cos^32\theta}\,d\theta.$
$\displaystyle \text{Let } \cos2\theta=t.$
$\displaystyle \text{Then } -2\sin2\theta\,d\theta=dt.$
$\displaystyle \text{When } \theta=0,\ t=1 \text{ and when } \theta=\frac{\pi}{6},\ t=\frac{1}{2}.$
$\displaystyle \therefore I=-\frac{1}{2}\int_{1}^{\frac{1}{2}}\frac{dt}{t^3}.$
$\displaystyle I=\frac{1}{2}\int_{\frac{1}{2}}^{1}t^{-3}\,dt.$
$\displaystyle I=\frac{1}{2}\left[\frac{t^{-2}}{-2}\right]_{\frac{1}{2}}^{1}.$
$\displaystyle I=\frac{1}{2}\left(\frac{1}{2}\right)\left(4-1\right).$
$\displaystyle I=\frac{3}{4}.$

$\displaystyle \textbf{Question 44: }~\int_{0}^{(\pi)^{2/3}}\sqrt{x}\,\cos^2\!\left(x^{3/2}\right)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\pi^{\frac{2}{3}}}\sqrt{x}\cos^2 x^{\frac{3}{2}}\,dx.$
$\displaystyle \text{Let } x^{\frac{3}{2}}=t.$
$\displaystyle \text{Then } \frac{3}{2}\sqrt{x}\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\pi^{\frac{2}{3}},\ t=\pi.$
$\displaystyle \therefore I=\frac{2}{3}\int_{0}^{\pi}\cos^2 t\,dt.$
$\displaystyle I=\frac{2}{3}\int_{0}^{\pi}\frac{1+\cos2t}{2}\,dt.$
$\displaystyle I=\frac{1}{3}\int_{0}^{\pi}(1+\cos2t)\,dt.$
$\displaystyle I=\frac{1}{3}\left[t+\frac{\sin2t}{2}\right]_{0}^{\pi}.$
$\displaystyle I=\frac{1}{3}(\pi+0).$
$\displaystyle I=\frac{\pi}{3}.$

$\displaystyle \textbf{Question 45: }~\int_{1}^{2}\frac{1}{x(1+\log x)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{1}^{2}\frac{1}{x(1+\log x)^2}\,dx.$
$\displaystyle \text{Let } 1+\log x=t.$
$\displaystyle \text{Then } \frac{1}{x}\,dx=dt.$
$\displaystyle \text{When } x=1,\ t=1 \text{ and when } x=2,\ t=1+\log2.$
$\displaystyle \therefore I=\int_{1}^{1+\log2}\frac{1}{t^2}\,dt.$
$\displaystyle I=\left[-\frac{1}{t}\right]_{1}^{1+\log2}.$
$\displaystyle I=-\frac{1}{1+\log2}+1.$
$\displaystyle I=\frac{\log2}{1+\log2}.$

$\displaystyle \textbf{Question 46: }~\int_{0}^{\pi/2}\cos^5 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\cos^5 x\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\cos^4 x\cos x\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}(1-\sin^2 x)^2\cos x\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}(1-2\sin^2 x+\sin^4 x)\cos x\,dx.$
$\displaystyle \text{Let } \sin x=t.$
$\displaystyle \text{Then } \cos x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{2},\ t=1.$
$\displaystyle \therefore I=\int_{0}^{1}(1-2t^2+t^4)\,dt.$
$\displaystyle I=\left[t-\frac{2t^3}{3}+\frac{t^5}{5}\right]_{0}^{1}.$
$\displaystyle I=1-\frac{2}{3}+\frac{1}{5}.$
$\displaystyle I=\frac{8}{15}.$

$\displaystyle \textbf{Question 47: }~\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{3/2})^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{4}^{9}\frac{\sqrt{x}}{(30-x^{\frac{3}{2}})^2}\,dx.$
$\displaystyle \text{Let } 30-x^{\frac{3}{2}}=t.$
$\displaystyle \text{Then } -\frac{3}{2}\sqrt{x}\,dx=dt.$
$\displaystyle \text{When } x=4,\ t=22 \text{ and when } x=9,\ t=3.$
$\displaystyle \therefore I=-\frac{2}{3}\int_{22}^{3}\frac{1}{t^2}\,dt.$
$\displaystyle I=\frac{2}{3}\int_{3}^{22}t^{-2}\,dt.$
$\displaystyle I=\frac{2}{3}\left[-\frac{1}{t}\right]_{3}^{22}.$
$\displaystyle I=\frac{2}{3}\left(\frac{1}{3}-\frac{1}{22}\right).$
$\displaystyle I=\frac{19}{99}.$

$\displaystyle \textbf{Question 48: }~\int_{0}^{\pi}\sin^3 x\,(1+2\cos x)(1+\cos x)^2\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\pi}\sin^{3}x(1+2\cos x)(1+\cos x)^2\,dx.$
$\displaystyle I=\int_{0}^{\pi}\sin x\sin^{2}x(1+2\cos x)(1+\cos x)^2\,dx.$
$\displaystyle I=\int_{0}^{\pi}\sin x(1-\cos^{2}x)(1+2\cos x)(1+\cos x)^2\,dx.$
$\displaystyle I=\int_{0}^{\pi}\sin x(1-\cos x)(1+\cos x)(1+2\cos x)(1+\cos x)^2\,dx.$
$\displaystyle I=\int_{0}^{\pi}\sin x(1-\cos x)(1+2\cos x)(1+\cos x)^3\,dx.$
$\displaystyle \text{Let } \cos x=t.$
$\displaystyle \text{Then } -\sin x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=1 \text{ and when } x=\pi,\ t=-1.$
$\displaystyle \therefore I=\int_{-1}^{1}(1-t)(1+2t)(1+t)^3\,dt.$
$\displaystyle I=\int_{-1}^{1}(1+t-2t^2)(1+3t+3t^2+t^3)\,dt.$
$\displaystyle I=\int_{-1}^{1}(1+4t+4t^2-2t^3-5t^4-2t^5)\,dt.$
$\displaystyle I=\left[t+2t^2+\frac{4t^3}{3}-\frac{t^4}{2}-t^5-\frac{t^6}{3}\right]_{-1}^{1}.$
$\displaystyle I=\frac{8}{3}.$

$\displaystyle \textbf{Question 49: }~\int_{0}^{\pi/2}2\sin x\cos x\,\tan^{-1}(\sin x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}2\sin x\cos x\tan^{-1}(\sin x)\,dx.$
$\displaystyle \text{Let } \sin x=t.$
$\displaystyle \text{Then } \cos x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{2},\ t=1.$
$\displaystyle \therefore I=\int_{0}^{1}2t\tan^{-1}t\,dt.$
$\displaystyle I=2\left[\frac{t^{2}\tan^{-1}t}{2}\right]_{0}^{1}-2\int_{0}^{1}\frac{t^{2}}{1+t^{2}}\,dt.$
$\displaystyle I=2\left[\frac{t^{2}\tan^{-1}t}{2}\right]_{0}^{1}-2\int_{0}^{1}\left(1-\frac{1}{1+t^{2}}\right)\,dt.$
$\displaystyle I=2\left[\frac{t^{2}\tan^{-1}t}{2}\right]_{0}^{1}-2\left[t-\tan^{-1}t\right]_{0}^{1}.$
$\displaystyle I=\tan^{-1}1-2\left(1-\tan^{-1}1\right).$
$\displaystyle I=\frac{\pi}{4}-2+\frac{\pi}{2}.$
$\displaystyle I=\frac{\pi}{2}-1.$

$\displaystyle \textbf{Question 50: }~\int_{0}^{\pi/2}\sin 2x\,\tan^{-1}(\sin x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\sin2x\tan^{-1}(\sin x)\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}2\sin x\cos x\tan^{-1}(\sin x)\,dx.$
$\displaystyle \text{Let } \sin x=t.$
$\displaystyle \text{Then } \cos x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=0 \text{ and when } x=\frac{\pi}{2},\ t=1.$
$\displaystyle \therefore I=2\int_{0}^{1}t\tan^{-1}t\,dt.$
$\displaystyle I=2\left[\frac{t^{2}\tan^{-1}t}{2}\right]_{0}^{1}-2\int_{0}^{1}\frac{t^{2}}{1+t^{2}}\,dt.$
$\displaystyle I=2\left[\frac{t^{2}\tan^{-1}t}{2}\right]_{0}^{1}-2\int_{0}^{1}\left(1-\frac{1}{1+t^{2}}\right)\,dt.$
$\displaystyle I=2\left[\frac{t^{2}\tan^{-1}t}{2}\right]_{0}^{1}-2\left[t-\tan^{-1}t\right]_{0}^{1}.$
$\displaystyle I=\tan^{-1}1-2\left(1-\tan^{-1}1\right).$
$\displaystyle I=\frac{\pi}{4}-2+\frac{\pi}{2}.$
$\displaystyle I=\frac{\pi}{2}-1.$

$\displaystyle \textbf{Question 51: }~\int_{0}^{1}(\cos^{-1}x)^2\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{1}(\cos^{-1}x)^2\,dx.$
$\displaystyle I=\int_{0}^{1}1\cdot(\cos^{-1}x)^2\,dx.$
$\displaystyle \text{Integrating by parts.}$
$\displaystyle I=\left[x(\cos^{-1}x)^2\right]_{0}^{1}-\int_{0}^{1}x\cdot2\cos^{-1}x\left(-\frac{1}{\sqrt{1-x^2}}\right)\,dx.$
$\displaystyle I=\left[x(\cos^{-1}x)^2\right]_{0}^{1}+2\int_{0}^{1}\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Again, integrating the second term by parts.}$
$\displaystyle I=\left[x(\cos^{-1}x)^2\right]_{0}^{1}+2\left(\left[\sqrt{1-x^2}\cos^{-1}x\right]_{0}^{1}-\int_{0}^{1}\frac{1}{\sqrt{1-x^2}}\sqrt{1-x^2}\,dx\right).$
$\displaystyle I=\left[x(\cos^{-1}x)^2\right]_{0}^{1}+2\left[\sqrt{1-x^2}\cos^{-1}x\right]_{0}^{1}-2\left[x\right]_{0}^{1}.$
$\displaystyle I=0+\frac{2\pi}{2}-2.$
$\displaystyle I=\pi-2.$

$\displaystyle \textbf{Question 52: }~\int_{0}^{a}\sin^{-1}\!\sqrt{\frac{x}{a+x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{a}\sin^{-1}\sqrt{\frac{x}{a+x}}\,dx.$
$\displaystyle \text{Let } x=a\tan^2\theta.$
$\displaystyle \text{Then } \theta=\tan^{-1}\sqrt{\frac{x}{a}}.$
$\displaystyle \text{When } x=0,\ \theta=0 \text{ and when } x=a,\ \theta=\frac{\pi}{4}.$
$\displaystyle \text{Also } dx=2a\tan\theta\sec^2\theta\,d\theta.$
$\displaystyle \therefore I=\int_{0}^{\frac{\pi}{4}}\sin^{-1}\sqrt{\frac{a\tan^2\theta}{a+a\tan^2\theta}}\cdot2a\tan\theta\sec^2\theta\,d\theta.$
$\displaystyle I=2a\int_{0}^{\frac{\pi}{4}}\sin^{-1}(\sin\theta)\tan\theta\sec^2\theta\,d\theta.$
$\displaystyle I=2a\int_{0}^{\frac{\pi}{4}}\theta\tan\theta\sec^2\theta\,d\theta.$
$\displaystyle \text{Let } \tan\theta=t.$
$\displaystyle \text{Then } \sec^2\theta\,d\theta=dt.$
$\displaystyle \text{When } \theta=0,\ t=0 \text{ and when } \theta=\frac{\pi}{4},\ t=1.$
$\displaystyle \therefore I=2a\int_{0}^{1}t\tan^{-1}t\,dt.$
$\displaystyle I=2a\left[\frac{t^2\tan^{-1}t}{2}\right]_{0}^{1}-2a\int_{0}^{1}\frac{t^2}{1+t^2}\,dt.$
$\displaystyle I=2a\left[\frac{t^2\tan^{-1}t}{2}\right]_{0}^{1}-2a\int_{0}^{1}\left(1-\frac{1}{1+t^2}\right)\,dt.$
$\displaystyle I=2a\left[\frac{t^2\tan^{-1}t}{2}\right]_{0}^{1}-2a\left[t-\tan^{-1}t\right]_{0}^{1}.$
$\displaystyle I=2a\left(\frac{\pi}{8}\right)-2a\left(1-\frac{\pi}{4}\right).$
$\displaystyle I=\frac{\pi a}{4}-2a+\frac{\pi a}{2}.$
$\displaystyle I=\frac{\pi a}{2}-a.$
$\displaystyle I=a\left(\frac{\pi}{2}-1\right).$

$\displaystyle \textbf{Question 53: }~\int_{\pi/3}^{\pi/2}\frac{\sqrt{1+\cos x}}{(1-\cos x)^{3/2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}}\,dx.$
$\displaystyle =\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1+\cos x}}{(1-\cos x)^{\frac{3}{2}}}\cdot\frac{\sqrt{1-\cos x}}{\sqrt{1-\cos x}}\,dx.$
$\displaystyle =\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sqrt{1-\cos^2 x}}{(1-\cos x)^2}\,dx.$
$\displaystyle =\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sin x}{(1-\cos x)^2}\,dx.$
$\displaystyle \text{Let } 1-\cos x=t.$
$\displaystyle \text{Then } \sin x\,dx=dt.$
$\displaystyle \text{When } x=\frac{\pi}{3},\ t=\frac{1}{2} \text{ and when } x=\frac{\pi}{2},\ t=1.$
$\displaystyle \therefore \int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\frac{\sin x}{(1-\cos x)^2}\,dx=\int_{\frac{1}{2}}^{1}\frac{dt}{t^2}.$
$\displaystyle =\left[-\frac{1}{t}\right]_{\frac{1}{2}}^{1}.$
$\displaystyle =-1+2.$
$\displaystyle =1.$

$\displaystyle \textbf{Question 54: }~\int_{0}^{a}x\sqrt{\frac{a^2-x^2}{a^2+x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{a}x\sqrt{\frac{a^{2}-x^{2}}{a^{2}+x^{2}}}\,dx.$
$\displaystyle \text{Consider } x^{2}=a^{2}\cos2\theta.$
$\displaystyle \Rightarrow 2x\,dx=-2a^{2}\sin2\theta\,d\theta.$
$\displaystyle \Rightarrow x\,dx=-a^{2}\sin2\theta\,d\theta.$
$\displaystyle \text{When } x=0,\ \theta=\frac{\pi}{4} \text{ and when } x=a,\ \theta=0.$
$\displaystyle \text{Now, the integral becomes}$
$\displaystyle I=\int_{\frac{\pi}{4}}^{0}-a^{2}\sin2\theta\sqrt{\frac{a^{2}-a^{2}\cos2\theta}{a^{2}+a^{2}\cos2\theta}}\,d\theta.$
$\displaystyle I=\int_{\frac{\pi}{4}}^{0}-a^{2}\sin2\theta\tan\theta\,d\theta.$
$\displaystyle I=a^{2}\int_{0}^{\frac{\pi}{4}}\sin2\theta\tan\theta\,d\theta.$
$\displaystyle I=a^{2}\int_{0}^{\frac{\pi}{4}}2\sin\theta\cos\theta\frac{\sin\theta}{\cos\theta}\,d\theta.$
$\displaystyle I=a^{2}\int_{0}^{\frac{\pi}{4}}2\sin^{2}\theta\,d\theta.$
$\displaystyle I=a^{2}\int_{0}^{\frac{\pi}{4}}(1-\cos2\theta)\,d\theta.$
$\displaystyle I=a^{2}\left[\theta-\frac{\sin2\theta}{2}\right]_{0}^{\frac{\pi}{4}}.$
$\displaystyle I=a^{2}\left(\frac{\pi}{4}-\frac{1}{2}\right).$

$\displaystyle \text{Evaluate the following integrals:}$

$\displaystyle \textbf{Question 55: }~\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{-a}^{a}\sqrt{\frac{a-x}{a+x}}\,dx.$
$\displaystyle \text{Consider } x=a\cos2y.$
$\displaystyle \text{Then } y=\frac{1}{2}\cos^{-1}\left(\frac{x}{a}\right).$
$\displaystyle \Rightarrow dx=-2a\sin2y\,dy.$
$\displaystyle \text{When } x=-a,\ y=\frac{\pi}{2} \text{ and when } x=a,\ y=0.$
$\displaystyle \text{Now, the integral becomes}$
$\displaystyle I=\int_{\frac{\pi}{2}}^{0}-2a\sin2y\sqrt{\frac{a-a\cos2y}{a+a\cos2y}}\,dy.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}2a\sin2y\tan y\,dy.$
$\displaystyle I=2a\int_{0}^{\frac{\pi}{2}}2\sin y\cos y\frac{\sin y}{\cos y}\,dy.$
$\displaystyle I=2a\int_{0}^{\frac{\pi}{2}}2\sin^{2}y\,dy.$
$\displaystyle I=2a\int_{0}^{\frac{\pi}{2}}(1-\cos2y)\,dy.$
$\displaystyle I=2a\left[y-\frac{\sin2y}{2}\right]_{0}^{\frac{\pi}{2}}.$
$\displaystyle I=2a\left(\frac{\pi}{2}-0\right).$
$\displaystyle I=\pi a.$

$\displaystyle \textbf{Question 56: }~\int_{0}^{\pi/2}\frac{\sin x\cos x}{\cos^2 x+3\cos x+2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos x}{\cos^{2}x+3\cos x+2}\,dx.$
$\displaystyle \text{Let } \cos x=t.$
$\displaystyle \text{Then } -\sin x\,dx=dt.$
$\displaystyle \text{When } x=0,\ t=1 \text{ and when } x=\frac{\pi}{2},\ t=0.$
$\displaystyle \therefore I=-\int_{1}^{0}\frac{t\,dt}{t^{2}+3t+2}.$
$\displaystyle I=\int_{1}^{0}\frac{-t\,dt}{(t+1)(t+2)}.$
$\displaystyle I=\int_{1}^{0}\left(\frac{1}{t+1}-\frac{2}{t+2}\right)dt.$
$\displaystyle I=\left[\log(t+1)-2\log(t+2)\right]_{1}^{0}.$
$\displaystyle I=\left[\log\frac{t+1}{(t+2)^{2}}\right]_{1}^{0}.$
$\displaystyle I=\log\frac{1/4}{2/9}.$
$\displaystyle I=\log\frac{9}{8}.$

$\displaystyle \textbf{Question 57: }~\int_{0}^{\pi/2}\frac{\tan x}{1+m^2\tan^2 x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{\tan x}{1+m^{2}\tan^{2}x}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\frac{\sin x}{\cos x}}{1+m^{2}\frac{\sin^{2}x}{\cos^{2}x}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin x\cos x}{\cos^{2}x+m^{2}\sin^{2}x}\,dx.$
$\displaystyle \text{Put } z=\cos^{2}x+m^{2}\sin^{2}x.$
$\displaystyle \text{Then } dz=2\cos x(-\sin x)\,dx+2m^{2}\sin x\cos x\,dx.$
$\displaystyle dz=2(m^{2}-1)\sin x\cos x\,dx.$
$\displaystyle \sin x\cos x\,dx=\frac{dz}{2(m^{2}-1)}.$
$\displaystyle \text{When } x=0,\ z=1.$
$\displaystyle \text{When } x=\frac{\pi}{2},\ z=m^{2}.$
$\displaystyle \therefore I=\frac{1}{2(m^{2}-1)}\int_{1}^{m^{2}}\frac{dz}{z}.$
$\displaystyle I=\frac{1}{2(m^{2}-1)}\left[\log z\right]_{1}^{m^{2}}.$
$\displaystyle I=\frac{1}{2(m^{2}-1)}(\log m^{2}-\log 1).$
$\displaystyle I=\frac{1}{2(m^{2}-1)}(2\log|m|).$
$\displaystyle I=\frac{\log|m|}{m^{2}-1}.$

$\displaystyle \textbf{Question 58: }~\int_{0}^{1/2}\frac{1}{(1+x^2)\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{1}{2}}\frac{1}{(1+x^{2})\sqrt{1-x^{2}}}\,dx.$
$\displaystyle \text{Put } x=\sin\theta.$
$\displaystyle \text{Then } dx=\cos\theta\,d\theta.$
$\displaystyle \text{When } x=0,\ \theta=0.$
$\displaystyle \text{When } x=\frac{1}{2},\ \theta=\frac{\pi}{6}.$
$\displaystyle \therefore I=\int_{0}^{\frac{\pi}{6}}\frac{1}{(1+\sin^{2}\theta)\cos\theta}\cdot\cos\theta\,d\theta.$
$\displaystyle I=\int_{0}^{\frac{\pi}{6}}\frac{1}{1+\sin^{2}\theta}\,d\theta.$
$\displaystyle \text{Dividing numerator and denominator by }\cos^{2}\theta.$
$\displaystyle I=\int_{0}^{\frac{\pi}{6}}\frac{\sec^{2}\theta}{\sec^{2}\theta+\tan^{2}\theta}\,d\theta.$
$\displaystyle I=\int_{0}^{\frac{\pi}{6}}\frac{\sec^{2}\theta}{1+2\tan^{2}\theta}\,d\theta.$
$\displaystyle \text{Now, put } \tan\theta=u.$
$\displaystyle \text{Then } \sec^{2}\theta\,d\theta=du.$
$\displaystyle \text{When } \theta=0,\ u=0.$
$\displaystyle \text{When } \theta=\frac{\pi}{6},\ u=\frac{1}{\sqrt{3}}.$
$\displaystyle \therefore I=\int_{0}^{\frac{1}{\sqrt{3}}}\frac{du}{1+2u^{2}}.$
$\displaystyle I=\int_{0}^{\frac{1}{\sqrt{3}}}\frac{du}{1+(\sqrt{2}u)^{2}}.$
$\displaystyle I=\left[\frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}u)\right]_{0}^{\frac{1}{\sqrt{3}}}.$
$\displaystyle I=\frac{1}{\sqrt{2}}\left(\tan^{-1}\sqrt{\frac{2}{3}}-0\right).$
$\displaystyle I=\frac{1}{\sqrt{2}}\tan^{-1}\sqrt{\frac{2}{3}}.$

$\displaystyle \textbf{Question 59: }~\int_{1/3}^{1}\frac{(x-x^3)^{1/3}}{x^4}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{\frac{1}{3}}^{1}\frac{(x-x^{3})^{\frac{1}{3}}}{x^{4}}\,dx.$
$\displaystyle I=\int_{\frac{1}{3}}^{1}\frac{\left[x^{3}\left(\frac{1}{x^{2}}-1\right)\right]^{\frac{1}{3}}}{x^{4}}\,dx.$
$\displaystyle I=\int_{\frac{1}{3}}^{1}\frac{x\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{4}}\,dx.$
$\displaystyle I=\int_{\frac{1}{3}}^{1}\frac{\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{3}}\,dx.$
$\displaystyle \text{Put } \left(\frac{1}{x^{2}}-1\right)=z.$
$\displaystyle \text{Then } dz=-\frac{2}{x^{3}}\,dx.$
$\displaystyle \Rightarrow \frac{dx}{x^{3}}=-\frac{dz}{2}.$
$\displaystyle \text{When } x=\frac{1}{3},\ z=8.$
$\displaystyle \text{When } x=1,\ z=0.$
$\displaystyle \therefore I=-\frac{1}{2}\int_{8}^{0}z^{\frac{1}{3}}\,dz.$
$\displaystyle I=-\frac{1}{2}\left[\frac{z^{\frac{4}{3}}}{\frac{4}{3}}\right]_{8}^{0}.$
$\displaystyle I=-\frac{3}{8}\left[0-8^{\frac{4}{3}}\right].$
$\displaystyle I=-\frac{3}{8}(-16).$
$\displaystyle I=6.$

$\displaystyle \textbf{Question 60: }~\int_{0}^{\pi/4}\frac{\sin^2 x\cos^2 x}{(\sin^3 x+\cos^3 x)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{4}}\frac{\sin^{2}x\cos^{2}x}{(\sin^{3}x+\cos^{3}x)^{2}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{\sin^{2}x\cos^{2}x}{\cos^{6}x(\tan^{3}x+1)^{2}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{4}}\frac{\tan^{2}x\sec^{2}x}{(\tan^{3}x+1)^{2}}\,dx.$
$\displaystyle \text{Put } \tan^{3}x+1=z.$
$\displaystyle \text{Then } dz=3\tan^{2}x\sec^{2}x\,dx.$
$\displaystyle \Rightarrow \tan^{2}x\sec^{2}x\,dx=\frac{dz}{3}.$
$\displaystyle \text{When } x=0,\ z=1.$
$\displaystyle \text{When } x=\frac{\pi}{4},\ z=2.$
$\displaystyle \therefore I=\frac{1}{3}\int_{1}^{2}\frac{dz}{z^{2}}.$
$\displaystyle I=\frac{1}{3}\left[-\frac{1}{z}\right]_{1}^{2}.$
$\displaystyle I=-\frac{1}{3}\left(\frac{1}{2}-1\right).$
$\displaystyle I=\frac{1}{6}.$

$\displaystyle \textbf{Question 61: }~\int_{0}^{\pi/2}\sqrt{\cos x-\cos^3 x}\,(\sec^2 x-1)\cos^2 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\sqrt{\cos x-\cos^{3}x}\,(\sec^{2}x-1)\cos^{2}x\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\sqrt{\cos x(1-\cos^{2}x)}\,(\tan^{2}x)\cos^{2}x\,dx.$
$\displaystyle I=-\int_{0}^{\frac{\pi}{2}}\sqrt{\cos x}\,(\sin^{2}x)\sin x\,dx.$
$\displaystyle I=-\int_{0}^{\frac{\pi}{2}}\sqrt{\cos x}(1-\cos^{2}x)\sin x\,dx.$
$\displaystyle \text{Put } \cos x=z^{2}.$
$\displaystyle \text{Then } -\sin x\,dx=2z\,dz.$
$\displaystyle \text{When } x=0,\ z=1.$
$\displaystyle \text{When } x=\frac{\pi}{2},\ z=0.$
$\displaystyle \therefore I=-\int_{1}^{0}z(1-z^{4})\,2z\,dz.$
$\displaystyle I=-2\int_{1}^{0}(z^{2}-z^{6})\,dz.$
$\displaystyle I=-2\left[\frac{z^{3}}{3}\right]_{1}^{0}+2\left[\frac{z^{7}}{7}\right]_{1}^{0}.$
$\displaystyle I=\frac{2}{3}-\frac{2}{7}.$
$\displaystyle I=\frac{8}{21}.$

$\displaystyle \textbf{Question 62: }~\int_{0}^{\pi/2}\frac{\cos x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{n}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\frac{\cos x}{(\cos\frac{x}{2}+\sin\frac{x}{2})^{n}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}}{(\cos\frac{x}{2}+\sin\frac{x}{2})^{n}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{(\cos\frac{x}{2}+\sin\frac{x}{2})(\cos\frac{x}{2}-\sin\frac{x}{2})}{(\cos\frac{x}{2}+\sin\frac{x}{2})^{n}}\,dx.$
$\displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{(\cos\frac{x}{2}+\sin\frac{x}{2})^{\,n-1}}\,dx.$
$\displaystyle \text{Put } \cos\frac{x}{2}+\sin\frac{x}{2}=z.$
$\displaystyle \text{Then } dz=\left(-\frac{1}{2}\sin\frac{x}{2}+\frac{1}{2}\cos\frac{x}{2}\right)dx.$
$\displaystyle \Rightarrow (\cos\frac{x}{2}-\sin\frac{x}{2})\,dx=2\,dz.$
$\displaystyle \text{When } x=0,\ z=1.$
$\displaystyle \text{When } x=\frac{\pi}{2},\ z=\sqrt{2}.$
$\displaystyle \therefore I=2\int_{1}^{\sqrt{2}}\frac{dz}{z^{\,n-1}}.$
$\displaystyle I=2\int_{1}^{\sqrt{2}}z^{\,1-n}\,dz.$
$\displaystyle I=2\left[\frac{z^{\,2-n}}{2-n}\right]_{1}^{\sqrt{2}}.$
$\displaystyle I=\frac{2}{2-n}\left[(\sqrt{2})^{\,2-n}-1\right].$
$\displaystyle I=\frac{2}{2-n}\left(2^{\frac{2-n}{2}}-1\right).$