Class 12: Definite Integrals

$\displaystyle \text{Evaluate the following definite integrals (1 - 58):}$

$\displaystyle \textbf{Question 1: }~\int_{4}^{9}\frac{1}{\sqrt{x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{4}^{9} x^{-1/2}\,dx = \left[2\sqrt{x}\right]_{4}^{9}$
$\displaystyle = 2(\sqrt{9}-\sqrt{4})$
$\displaystyle = 2(3-2)$
$\displaystyle = 2$
$\\$

$\displaystyle \textbf{Question 2: }~\int_{-2}^{3}\frac{1}{x+7}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{-2}^{3}\frac{1}{x+7}\,dx = \left[\log|x+7|\right]_{-2}^{3}$
$\displaystyle = \log(10)-\log(5)$
$\displaystyle = \log\!\left(\frac{10}{5}\right)$
$\displaystyle = \log 2$
$\\$

$\displaystyle \textbf{Question 3: }~\int_{0}^{1/2}\frac{1}{\sqrt{1-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{1/2}\frac{1}{\sqrt{1-x^2}}\,dx = \left[\sin^{-1}x\right]_{0}^{1/2}$
$\displaystyle = \sin^{-1}\!\left(\frac{1}{2}\right)-\sin^{-1}(0)$
$\displaystyle = \frac{\pi}{6}$
$\\$

$\displaystyle \textbf{Question 4: }~\int_{0}^{1}\frac{1}{1+x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{1}\frac{1}{1+x^2}\,dx=\left[\tan^{-1}x\right]_{0}^{1}$
$\displaystyle =\tan^{-1}(1)-\tan^{-1}(0)$
$\displaystyle =\frac{\pi}{4}$
$\\$
$\displaystyle \textbf{Question 5: }~\int_{2}^{3}\frac{x}{x^2+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{2}^{3}\frac{x}{x^2+1}\,dx=\frac12\int_{2}^{3}\frac{2x}{x^2+1}\,dx$
$\displaystyle =\frac12\left[\log(x^2+1)\right]_{2}^{3}$
$\displaystyle =\frac12\left(\log(10)-\log(5)\right)$
$\displaystyle =\frac12\log 2$
$\\$
$\displaystyle \textbf{Question 6: }~\int_{0}^{\infty}\frac{1}{a^2+b^2x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\infty}\frac{1}{a^2+b^2x^2}\,dx=\frac{1}{a^2}\int_{0}^{\infty}\frac{1}{1+\left(\frac{b}{a}x\right)^2}\,dx$
$\displaystyle \text{Let }u=\frac{b}{a}x\Rightarrow dx=\frac{a}{b}\,du$
$\displaystyle =\frac{1}{a^2}\cdot\frac{a}{b}\int_{0}^{\infty}\frac{1}{1+u^2}\,du$
$\displaystyle =\frac{1}{ab}\left[\tan^{-1}u\right]_{0}^{\infty}$
$\displaystyle =\frac{1}{ab}\left(\frac{\pi}{2}-0\right)$
$\displaystyle =\frac{\pi}{2ab}$
$\\$
$\displaystyle \textbf{Question 7: }~\int_{-1}^{1}\frac{1}{1+x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{-1}^{1}\frac{1}{1+x^2}\,dx=\left[\tan^{-1}x\right]_{-1}^{1}$
$\displaystyle =\tan^{-1}(1)-\tan^{-1}(-1)$
$\displaystyle =\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)$
$\displaystyle =\frac{\pi}{2}$
$\\$
$\displaystyle \textbf{Question 8: }~\int_{0}^{\infty}e^{-x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\infty}e^{-x}\,dx=\left[-e^{-x}\right]_{0}^{\infty}$
$\displaystyle =\left(0\right)-\left(-1\right)$
$\displaystyle =1$
$\\$
$\displaystyle \textbf{Question 9: }~\int_{0}^{1}\frac{x}{x+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{1}\frac{x}{x+1}\,dx=\int_{0}^{1}\left(1-\frac{1}{x+1}\right)\,dx$
$\displaystyle =\left[x-\log(x+1)\right]_{0}^{1}$
$\displaystyle =(1-\log 2)-(0-\log 1)$
$\displaystyle =1-\log 2$
$\\$
$\displaystyle \textbf{Question 10: }~\int_{0}^{\pi/2}(\sin x+\cos x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}(\sin x+\cos x)\,dx=\left[-\cos x+\sin x\right]_{0}^{\pi/2}$
$\displaystyle =\left(-\cos\frac{\pi}{2}+\sin\frac{\pi}{2}\right)-\left(-\cos 0+\sin 0\right)$
$\displaystyle =(0+1)-(-1+0)$
$\displaystyle =2$
$\\$

$\displaystyle \textbf{Question 11: }~\int_{\pi/4}^{\pi/2}\cot x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{\pi/4}^{\pi/2}\cot x\,dx=\int_{\pi/4}^{\pi/2}\frac{\cos x}{\sin x}\,dx=\left[\log(\sin x)\right]_{\pi/4}^{\pi/2}$
$\displaystyle =\log\left(\sin\frac{\pi}{2}\right)-\log\left(\sin\frac{\pi}{4}\right)$
$\displaystyle =\log(1)-\log\left(\frac{1}{\sqrt{2}}\right)$
$\displaystyle =\log\sqrt{2}$
$\\$
$\displaystyle \textbf{Question 12: }~\int_{0}^{\pi/4}\sec x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/4}\sec x\,dx=\left[\log|\sec x+\tan x|\right]_{0}^{\pi/4}$
$\displaystyle =\log\left(\sec\frac{\pi}{4}+\tan\frac{\pi}{4}\right)-\log(\sec 0+\tan 0)$
$\displaystyle =\log(\sqrt{2}+1)-\log(1)$
$\displaystyle =\log(\sqrt{2}+1)$
$\\$
$\displaystyle \textbf{Question 13: }~\int_{\pi/6}^{\pi/4}\mathrm{cosec}\,x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{\pi/6}^{\pi/4}\mathrm{cosec}\,x\,dx=\left[\log\left|\tan\frac{x}{2}\right|\right]_{\pi/6}^{\pi/4}$
$\displaystyle =\log\left(\tan\frac{\pi}{8}\right)-\log\left(\tan\frac{\pi}{12}\right)$
$\displaystyle =\log\left(\frac{\tan(\pi/8)}{\tan(\pi/12)}\right)$
$\\$
$\displaystyle \textbf{Question 14: }~\int_{0}^{1}\frac{1-x}{1+x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{1}\frac{1-x}{1+x}\,dx=\int_{0}^{1}\left(1-\frac{2x}{1+x}\right)\,dx=\int_{0}^{1}\left(-1+\frac{2}{1+x}\right)\,dx$
$\displaystyle =\left[-x+2\log(1+x)\right]_{0}^{1}$
$\displaystyle =\left(-1+2\log 2\right)-\left(0+2\log 1\right)$
$\displaystyle =2\log 2-1$
$\\$
$\displaystyle \textbf{Question 15: }~\int_{0}^{\pi}\frac{1}{1+\sin x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi}\frac{1}{1+\sin x}\,dx=\int_{0}^{\pi}\frac{1-\sin x}{1-\sin^2 x}\,dx=\int_{0}^{\pi}\frac{1-\sin x}{\cos^2 x}\,dx$
$\displaystyle =\int_{0}^{\pi}\left(\sec^2 x-\sec x\tan x\right)\,dx$
$\displaystyle =\left[\tan x-\sec x\right]_{0}^{\pi}$
$\displaystyle =\left(0-(-1)\right)-\left(0-1\right)$
$\displaystyle =2$
$\\$
$\displaystyle \textbf{Question 16: }~\int_{-\pi/4}^{\pi/4}\frac{1}{1+\sin x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{-\pi/4}^{\pi/4}\frac{1}{1+\sin x}\,dx=\int_{-\pi/4}^{\pi/4}\frac{1-\sin x}{\cos^2 x}\,dx=\int_{-\pi/4}^{\pi/4}\left(\sec^2 x-\sec x\tan x\right)\,dx$
$\displaystyle =\left[\tan x-\sec x\right]_{-\pi/4}^{\pi/4}$
$\displaystyle =\left(1-\sqrt{2}\right)-\left(-1-\sqrt{2}\right)$
$\displaystyle =2$
$\\$
$\displaystyle \textbf{Question 17: }~\int_{0}^{\pi/2}\cos^2 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}\cos^2 x\,dx=\int_{0}^{\pi/2}\frac{1+\cos 2x}{2}\,dx$
$\displaystyle =\left[\frac{x}{2}+\frac{\sin 2x}{4}\right]_{0}^{\pi/2}$
$\displaystyle =\frac{\pi}{4}$
$\\$
$\displaystyle \textbf{Question 18: }~\int_{0}^{\pi/2}\cos^3 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}\cos^3 x\,dx=\int_{0}^{\pi/2}\cos x\,(1-\sin^2 x)\,dx$
$\displaystyle \text{Let }u=\sin x\Rightarrow du=\cos x\,dx$
$\displaystyle =\int_{0}^{1}(1-u^2)\,du=\left[u-\frac{u^3}{3}\right]_{0}^{1}$
$\displaystyle =1-\frac{1}{3}$
$\displaystyle =\frac{2}{3}$
$\\$
$\displaystyle \textbf{Question 19: }~\int_{0}^{\pi/6}\cos x\cos 2x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/6}\cos x\cos 2x\,dx=\int_{0}^{\pi/6}\frac{\cos(x-2x)+\cos(x+2x)}{2}\,dx$
$\displaystyle =\frac12\int_{0}^{\pi/6}\left(\cos x+\cos 3x\right)\,dx$
$\displaystyle =\frac12\left[\sin x+\frac{\sin 3x}{3}\right]_{0}^{\pi/6}$
$\displaystyle =\frac12\left(\sin\frac{\pi}{6}+\frac{\sin\frac{\pi}{2}}{3}-0\right)$
$\displaystyle =\frac12\left(\frac12+\frac13\right)$
$\displaystyle =\frac{5}{12}$
$\\$
$\displaystyle \textbf{Question 20: }~\int_{0}^{\pi/2}\sin x\sin 2x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}\sin x\sin 2x\,dx=\int_{0}^{\pi/2}\frac{\cos(x-2x)-\cos(x+2x)}{2}\,dx$
$\displaystyle =\frac12\int_{0}^{\pi/2}\left(\cos x-\cos 3x\right)\,dx$
$\displaystyle =\frac12\left[\sin x-\frac{\sin 3x}{3}\right]_{0}^{\pi/2}$
$\displaystyle =\frac12\left(\sin\frac{\pi}{2}-\frac{\sin\frac{3\pi}{2}}{3}-0\right)$
$\displaystyle =\frac12\left(1-\frac{-1}{3}\right)$
$\displaystyle =\frac12\cdot\frac{4}{3}$
$\displaystyle =\frac{2}{3}$
$\\$

$\displaystyle \textbf{Question 21: }~\int_{\pi/3}^{\pi/4}(\tan x+\cot x)^2\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle (\tan x+\cot x)^2=\tan^2 x+\cot^2 x+2=(\sec^2 x-1)+(\mathrm{cosec}^2 x-1)+2=\sec^2 x+\mathrm{cosec}^2 x$
$\displaystyle \int_{\pi/3}^{\pi/4}(\tan x+\cot x)^2\,dx=\int_{\pi/3}^{\pi/4}\left(\sec^2 x+\mathrm{cosec}^2 x\right)\,dx$
$\displaystyle =\left[\tan x-\cot x\right]_{\pi/3}^{\pi/4}$
$\displaystyle =\left(\tan\frac{\pi}{4}-\cot\frac{\pi}{4}\right)-\left(\tan\frac{\pi}{3}-\cot\frac{\pi}{3}\right)$
$\displaystyle =(1-1)-\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)$
$\displaystyle =-\left(\frac{3-1}{\sqrt{3}}\right)=-\frac{2}{\sqrt{3}}$
$\\$
$\displaystyle \textbf{Question 22: }~\int_{0}^{\pi/2}\cos^4 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}\cos^4 x\,dx=\int_{0}^{\pi/2}\left(\frac{1+\cos 2x}{2}\right)^2 dx=\int_{0}^{\pi/2}\frac{3+4\cos 2x+\cos 4x}{8}\,dx$
$\displaystyle =\frac{1}{8}\left[3x+2\sin 2x+\frac{\sin 4x}{4}\right]_{0}^{\pi/2}=\frac{3\pi}{16}$
$\\$
$\displaystyle \textbf{Question 23: }~\int_{0}^{\pi/2}(a^2\cos^2 x+b^2\sin^2 x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}(a^2\cos^2 x+b^2\sin^2 x)\,dx=a^2\int_{0}^{\pi/2}\cos^2 x\,dx+b^2\int_{0}^{\pi/2}\sin^2 x\,dx$
$\displaystyle =a^2\cdot\frac{\pi}{4}+b^2\cdot\frac{\pi}{4}=\frac{\pi}{4}(a^2+b^2)$
$\\$
$\displaystyle \textbf{Question 24: }~\int_{0}^{\pi/2}\sqrt{1+\sin x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle 1+\sin x=\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)^2\Rightarrow \sqrt{1+\sin x}=\sin\frac{x}{2}+\cos\frac{x}{2}$
$\displaystyle \int_{0}^{\pi/2}\sqrt{1+\sin x}\,dx=\int_{0}^{\pi/2}\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right)\,dx$
$\displaystyle =\left[-2\cos\frac{x}{2}+2\sin\frac{x}{2}\right]_{0}^{\pi/2}=2$
$\\$
$\displaystyle \textbf{Question 25: }~\int_{0}^{\pi/2}\sqrt{1+\cos x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle 1+\cos x=2\cos^2\frac{x}{2}\Rightarrow \sqrt{1+\cos x}=\sqrt{2}\cos\frac{x}{2}$
$\displaystyle \int_{0}^{\pi/2}\sqrt{1+\cos x}\,dx=\sqrt{2}\int_{0}^{\pi/2}\cos\frac{x}{2}\,dx=\sqrt{2}\left[2\sin\frac{x}{2}\right]_{0}^{\pi/2}=2$
$\\$
$\displaystyle \textbf{Question 26: }~\int_{0}^{\pi/2}x^2\sin x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}x^2\sin x\,dx=\left[-x^2\cos x\right]_{0}^{\pi/2}+\int_{0}^{\pi/2}2x\cos x\,dx=2\int_{0}^{\pi/2}x\cos x\,dx$
$\displaystyle =2\left(\frac{\pi}{2}-1\right)=\pi-2$
$\\$
$\displaystyle \textbf{Question 27: }~\int_{0}^{\pi/2}x\cos x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}x\cos x\,dx=\left[x\sin x\right]_{0}^{\pi/2}-\int_{0}^{\pi/2}\sin x\,dx=\frac{\pi}{2}-\left[-\cos x\right]_{0}^{\pi/2}=\frac{\pi}{2}-1$
$\\$
$\displaystyle \textbf{Question 28: }~\int_{0}^{\pi/2}x^2\cos x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}x^2\cos x\,dx=\left[x^2\sin x\right]_{0}^{\pi/2}-\int_{0}^{\pi/2}2x\sin x\,dx=\frac{\pi^2}{4}-2\int_{0}^{\pi/2}x\sin x\,dx$
$\displaystyle \int_{0}^{\pi/2}x\sin x\,dx=\left[-x\cos x\right]_{0}^{\pi/2}+\int_{0}^{\pi/2}\cos x\,dx=\left[\sin x\right]_{0}^{\pi/2}=1$
$\displaystyle \Rightarrow \int_{0}^{\pi/2}x^2\cos x\,dx=\frac{\pi^2}{4}-2$
$\\$
$\displaystyle \textbf{Question 29: }~\int_{0}^{\pi/4}x^2\sin x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int x^2\sin x\,dx=-x^2\cos x+\int 2x\cos x\,dx=-x^2\cos x+2(x\sin x+\cos x)$
$\displaystyle \int_{0}^{\pi/4}x^2\sin x\,dx=\left[-x^2\cos x+2x\sin x+2\cos x\right]_{0}^{\pi/4}$
$\displaystyle =\frac{\sqrt{2}}{2}\left(2+\frac{\pi}{2}-\frac{\pi^2}{16}\right)-2$
$\\$
$\displaystyle \textbf{Question 30: }~\int_{0}^{\pi/2}x^2\cos 2x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}x^2\cos 2x\,dx=\left[\frac{x^2}{2}\sin 2x\right]_{0}^{\pi/2}-\int_{0}^{\pi/2}x\sin 2x\,dx=-\int_{0}^{\pi/2}x\sin 2x\,dx$
$\displaystyle \int_{0}^{\pi/2}x\sin 2x\,dx=\left[-\frac{x}{2}\cos 2x\right]_{0}^{\pi/2}+\int_{0}^{\pi/2}\frac{1}{2}\cos 2x\,dx=\frac{\pi}{4}$
$\displaystyle \Rightarrow \int_{0}^{\pi/2}x^2\cos 2x\,dx=-\frac{\pi}{4}$
$\\$

$\displaystyle \textbf{Question 31: }~\int_{0}^{\pi/2}x^2\cos^2 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}x^2\cos^2 x\,dx=\frac12\int_{0}^{\pi/2}x^2(1+\cos 2x)\,dx$
$\displaystyle =\frac12\int_{0}^{\pi/2}x^2\,dx+\frac12\int_{0}^{\pi/2}x^2\cos 2x\,dx$
$\displaystyle =\frac12\left[\frac{x^3}{3}\right]_{0}^{\pi/2}+\frac12\left(-\frac{\pi}{4}\right)$
$\displaystyle =\frac{\pi^3}{48}-\frac{\pi}{8}=\frac{\pi(\pi^2-6)}{48}$
$\\$
$\displaystyle \textbf{Question 32: }~\int_{1}^{2}\log x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int \log x\,dx=x\log x-x$
$\displaystyle \int_{1}^{2}\log x\,dx=\left[x\log x-x\right]_{1}^{2}$
$\displaystyle =(2\log 2-2)-(0-1)$
$\displaystyle =2\log 2-1$
$\\$
$\displaystyle \textbf{Question 33: }~\int_{1}^{3}\frac{\log x}{(x+1)^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }u=\log x,\; dv=\frac{1}{(x+1)^2}\,dx\Rightarrow du=\frac{1}{x}\,dx,\; v=-\frac{1}{x+1}$
$\displaystyle \int_{1}^{3}\frac{\log x}{(x+1)^2}\,dx=\left[-\frac{\log x}{x+1}\right]_{1}^{3}+\int_{1}^{3}\frac{1}{x(x+1)}\,dx$
$\displaystyle =-\frac{\log 3}{4}+\int_{1}^{3}\left(\frac{1}{x}-\frac{1}{x+1}\right)\,dx$
$\displaystyle =-\frac{\log 3}{4}+\left[\log x-\log(x+1)\right]_{1}^{3}$
$\displaystyle =-\frac{\log 3}{4}+\left(\log 3-\log 4\right)-\left(0-\log 2\right)$
$\displaystyle =\frac{3}{4}\log 3-\log 2$
$\\$
$\displaystyle \textbf{Question 34: }~\int_{1}^{e} \frac{e^{x}}{x}(1+x\log x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{e^{x}}{x}(1+x\log x)=\frac{e^{x}}{x}+e^{x}\log x$
$\displaystyle \frac{d}{dx}\left(e^{x}\log x\right)=e^{x}\log x+\frac{e^{x}}{x}$
$\displaystyle \Rightarrow \int_{1}^{e} \frac{e^{x}}{x}(1+x\log x)\,dx=\left[e^{x}\log x\right]_{1}^{e}$
$\displaystyle =e^{e}\log(e)-e^{1}\log 1$
$\displaystyle =e^{e}$
$\\$
$\displaystyle \textbf{Question 35: }~\int_{1}^{e}\frac{\log x}{x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{1}^{e}\frac{\log x}{x}\,dx=\left[\frac{(\log x)^2}{2}\right]_{1}^{e}=\frac{1}{2}$
$\\$
$\displaystyle \textbf{Question 36: }~\int_{e}^{e^2}\left(\frac{1}{\log x}-\frac{1}{(\log x)^2}\right)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }t=\log x\Rightarrow x=e^{t},\; dx=e^{t}\,dt;\; x=e\to t=1,\; x=e^2\to t=2$
$\displaystyle \int_{e}^{e^2}\left(\frac{1}{\log x}-\frac{1}{(\log x)^2}\right)\,dx=\int_{1}^{2}e^{t}\left(\frac{1}{t}-\frac{1}{t^2}\right)\,dt$
$\displaystyle \frac{d}{dt}\left(\frac{e^{t}}{t}\right)=e^{t}\left(\frac{1}{t}-\frac{1}{t^2}\right)$
$\displaystyle \Rightarrow \int_{e}^{e^2}\left(\frac{1}{\log x}-\frac{1}{(\log x)^2}\right)\,dx=\left[\frac{e^{t}}{t}\right]_{1}^{2}=\frac{e^{2}}{2}-e$
$\\$
$\displaystyle \textbf{Question 37: }~\int_{1}^{2}\frac{x+3}{x(x+2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+3}{x(x+2)}=\frac{A}{x}+\frac{B}{x+2}\Rightarrow x+3=A(x+2)+Bx$
$\displaystyle A=\frac{3}{2},\; B=-\frac{1}{2}$
$\displaystyle \int_{1}^{2}\frac{x+3}{x(x+2)}\,dx=\left[\frac{3}{2}\log x-\frac{1}{2}\log(x+2)\right]_{1}^{2}=\frac{1}{2}\log 6$
$\\$
$\displaystyle \textbf{Question 38: }~\int_{0}^{1}\frac{2x+3}{5x^2+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{1}\frac{2x+3}{5x^2+1}\,dx=\int_{0}^{1}\frac{2x}{5x^2+1}\,dx+3\int_{0}^{1}\frac{1}{5x^2+1}\,dx$
$\displaystyle \int \frac{2x}{5x^2+1}\,dx=\frac{1}{5}\log(5x^2+1)$
$\displaystyle \int \frac{1}{5x^2+1}\,dx=\frac{1}{\sqrt{5}}\tan^{-1}(\sqrt{5}\,x)$
$\displaystyle \int_{0}^{1}\frac{2x+3}{5x^2+1}\,dx=\left[\frac{1}{5}\log(5x^2+1)+\frac{3}{\sqrt{5}}\tan^{-1}(\sqrt{5}\,x)\right]_{0}^{1}$
$\displaystyle =\frac{1}{5}\log 6+\frac{3}{\sqrt{5}}\tan^{-1}(\sqrt{5})$
$\\$
$\displaystyle \textbf{Question 39: }~\int_{0}^{4}\frac{1}{4+x-x^2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle 4+x-x^2=0\Rightarrow x=\frac{1\pm\sqrt{17}}{2};\;\;\frac{1+\sqrt{17}}{2}\in(0,4)$
$\displaystyle \Rightarrow \int_{0}^{4}\frac{1}{4+x-x^2}\,dx=\int_{0}^{\frac{1+\sqrt{17}}{2}}\frac{1}{4+x-x^2}\,dx+\int_{\frac{1+\sqrt{17}}{2}}^{4}\frac{1}{4+x-x^2}\,dx$
$\displaystyle \text{Both integrals are improper and diverge (log blow-up at }x=\tfrac{1+\sqrt{17}}{2}\text{), hence the integral diverges.}$
$\\$
$\displaystyle \textbf{Question 40: }~\int_{0}^{1}\frac{1}{2x^2+x+1}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle 2x^2+x+1=2\left[\left(x+\frac14\right)^2+\left(\frac{\sqrt7}{4}\right)^2\right]$
$\displaystyle \int_{0}^{1}\frac{1}{2x^2+x+1}\,dx=\frac12\int_{0}^{1}\frac{dx}{\left(x+\frac14\right)^2+\left(\frac{\sqrt7}{4}\right)^2}$
$\displaystyle =\frac{2}{\sqrt7}\left[\tan^{-1}\left(\frac{4x+1}{\sqrt7}\right)\right]_{0}^{1}$
$\displaystyle =\frac{2}{\sqrt7}\left(\tan^{-1}\frac{5}{\sqrt7}-\tan^{-1}\frac{1}{\sqrt7}\right)$
$\\$

$\displaystyle \textbf{Question 41: }~\int_{0}^{1}\sqrt{x(1-x)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }x=\sin^2 t\Rightarrow dx=2\sin t\cos t\,dt,\;\sqrt{x(1-x)}=\sin t\cos t$
$\displaystyle \int_{0}^{1}\sqrt{x(1-x)}\,dx=2\int_{0}^{\pi/2}\sin^2 t\cos^2 t\,dt=2\int_{0}^{\pi/2}\frac{1}{4}\sin^2 2t\,dt=\frac12\int_{0}^{\pi/2}\sin^2 2t\,dt$
$\displaystyle =\frac12\int_{0}^{\pi/2}\frac{1-\cos 4t}{2}\,dt=\frac14\left[t-\frac{\sin 4t}{4}\right]_{0}^{\pi/2}=\frac14\cdot\frac{\pi}{2}=\frac{\pi}{8}$
$\\$
$\displaystyle \textbf{Question 42: }~\int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle 3+2x-x^2=4-(x-1)^2$
$\displaystyle \int_{0}^{2}\frac{1}{\sqrt{3+2x-x^2}}\,dx=\int_{0}^{2}\frac{1}{\sqrt{4-(x-1)^2}}\,dx=\left[\sin^{-1}\left(\frac{x-1}{2}\right)\right]_{0}^{2}$
$\displaystyle =\sin^{-1}\left(\frac12\right)-\sin^{-1}\left(-\frac12\right)=\frac{\pi}{6}-\left(-\frac{\pi}{6}\right)=\frac{\pi}{3}$
$\\$
$\displaystyle \textbf{Question 43: }~\int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle 4x-x^2=4-(x-2)^2$
$\displaystyle \int_{0}^{4}\frac{1}{\sqrt{4x-x^2}}\,dx=\int_{0}^{4}\frac{1}{\sqrt{4-(x-2)^2}}\,dx=\left[\sin^{-1}\left(\frac{x-2}{2}\right)\right]_{0}^{4}$
$\displaystyle =\sin^{-1}(1)-\sin^{-1}(-1)=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi$
$\\$
$\displaystyle \textbf{Question 44: }~\int_{-1}^{1}\frac{1}{x^2+2x+5}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+2x+5=(x+1)^2+4$
$\displaystyle \int_{-1}^{1}\frac{1}{x^2+2x+5}\,dx=\int_{-1}^{1}\frac{1}{(x+1)^2+2^2}\,dx=\left[\frac12\tan^{-1}\left(\frac{x+1}{2}\right)\right]_{-1}^{1}$
$\displaystyle =\frac12\left(\tan^{-1}(1)-\tan^{-1}(0)\right)=\frac12\cdot\frac{\pi}{4}=\frac{\pi}{8}$
$\\$
$\displaystyle \textbf{Question 45: }~\int_{1}^{4}\frac{x^2+x}{\sqrt{2x+1}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }u=2x+1\Rightarrow x=\frac{u-1}{2},\;dx=\frac{du}{2}$
$\displaystyle x^2+x=\frac{u-1}{2}\cdot\frac{u+1}{2}=\frac{u^2-1}{4}$
$\displaystyle \int_{1}^{4}\frac{x^2+x}{\sqrt{2x+1}}\,dx=\int_{3}^{9}\frac{(u^2-1)/4}{\sqrt{u}}\cdot\frac{du}{2}=\frac18\int_{3}^{9}\left(u^{3/2}-u^{-1/2}\right)\,du$
$\displaystyle =\frac18\left[\frac{2}{5}u^{5/2}-2u^{1/2}\right]_{3}^{9}=\frac14\left[\frac{1}{5}u^{5/2}-u^{1/2}\right]_{3}^{9}=\frac{57-\sqrt{3}}{5}$
$\\$
$\displaystyle \textbf{Question 46: }~\int_{0}^{1}x(1-x)^5\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{1}x(1-x)^5\,dx=\int_{0}^{1}x^{2-1}(1-x)^{6-1}\,dx=B(2,6)=\frac{\Gamma(2)\Gamma(6)}{\Gamma(8)}=\frac{1!\,5!}{7!}=\frac{1}{42}$
$\\$
$\displaystyle \textbf{Question 47: }~\int_{1}^{2}\frac{x-1}{x^2}e^x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}$
$\displaystyle \frac{d}{dx}\left(\frac{e^x}{x}\right)=\frac{e^x}{x}-\frac{e^x}{x^2}=e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)$
$\displaystyle \Rightarrow \int_{1}^{2}\frac{x-1}{x^2}e^x\,dx=\left[\frac{e^x}{x}\right]_{1}^{2}=\frac{e^2}{2}-e$
$\\$
$\displaystyle \textbf{Question 48: }~\int_{0}^{1}\!\left(xe^{2x}+\sin\frac{\pi x}{2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int xe^{2x}\,dx=\frac{x e^{2x}}{2}-\int\frac{e^{2x}}{2}\,dx=\frac{x e^{2x}}{2}-\frac{e^{2x}}{4}=\frac{e^{2x}(2x-1)}{4}$
$\displaystyle \int \sin\left(\frac{\pi x}{2}\right)\,dx=-\frac{2}{\pi}\cos\left(\frac{\pi x}{2}\right)$
$\displaystyle \int_{0}^{1}\!\left(xe^{2x}+\sin\frac{\pi x}{2}\right)\!dx=\left[\frac{e^{2x}(2x-1)}{4}-\frac{2}{\pi}\cos\left(\frac{\pi x}{2}\right)\right]_{0}^{1}=\frac{e^2+1}{4}+\frac{2}{\pi}$
$\\$
$\displaystyle \textbf{Question 49: }~\int_{0}^{1}\!\left(xe^x+\cos\frac{\pi x}{4}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{1}\left(xe^x+\cos\frac{\pi x}{4}\right)\,dx=\int_{0}^{1}xe^x\,dx+\int_{0}^{1}\cos\frac{\pi x}{4}\,dx$
$\displaystyle \int xe^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x=e^x(x-1)$
$\displaystyle \int_{0}^{1}xe^x\,dx=\left[e^x(x-1)\right]_{0}^{1}=0-(-1)=1$
$\displaystyle \int \cos\left(\frac{\pi x}{4}\right)\,dx=\frac{4}{\pi}\sin\left(\frac{\pi x}{4}\right)$
$\displaystyle \int_{0}^{1}\cos\frac{\pi x}{4}\,dx=\left[\frac{4}{\pi}\sin\left(\frac{\pi x}{4}\right)\right]_{0}^{1}=\frac{4}{\pi}\sin\frac{\pi}{4}=\frac{4}{\pi}\cdot\frac{1}{\sqrt{2}}=\frac{2\sqrt{2}}{\pi}$
$\displaystyle \Rightarrow \int_{0}^{1}\left(xe^x+\cos\frac{\pi x}{4}\right)\,dx=1+\frac{2\sqrt{2}}{\pi}$
$\\$
$\displaystyle \textbf{Question 50: }~\int_{\pi/2}^{\pi}e^x\!\left(\frac{1-\sin x}{1-\cos x}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1-\sin x}{1-\cos x}=\frac{1-2\sin\frac{x}{2}\cos\frac{x}{2}}{2\sin^2\frac{x}{2}}=\frac{(\sin\frac{x}{2}-\cos\frac{x}{2})^2}{2\sin^2\frac{x}{2}}=\frac12\,\mathrm{cosec}^2\frac{x}{2}-\cot\frac{x}{2}$
$\displaystyle \frac{d}{dx}\left(\cot\frac{x}{2}\right)=-\frac12\,\mathrm{cosec}^2\frac{x}{2}\Rightarrow \frac{d}{dx}\left(e^x\cot\frac{x}{2}\right)=e^x\left(\cot\frac{x}{2}-\frac12\,\mathrm{cosec}^2\frac{x}{2}\right)$
$\displaystyle \Rightarrow e^x\left(\frac12\,\mathrm{cosec}^2\frac{x}{2}-\cot\frac{x}{2}\right)=-\frac{d}{dx}\left(e^x\cot\frac{x}{2}\right)$
$\displaystyle \int_{\pi/2}^{\pi}e^x\!\left(\frac{1-\sin x}{1-\cos x}\right)\!dx=\left[-e^x\cot\frac{x}{2}\right]_{\pi/2}^{\pi}=-e^{\pi}\cot\frac{\pi}{2}+e^{\pi/2}\cot\frac{\pi}{4}=e^{\pi/2}$
$\\$

$\displaystyle \textbf{Question 51: }~\int_{0}^{2\pi}e^{x/2}\sin\!\left(\frac{x}{2}+\frac{\pi}{4}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }u=\frac{x}{2}\Rightarrow x=2u,\;dx=2\,du,\;x:0\to2\pi\Rightarrow u:0\to\pi$
$\displaystyle \int_{0}^{2\pi}e^{x/2}\sin\!\left(\frac{x}{2}+\frac{\pi}{4}\right)\!dx=2\int_{0}^{\pi}e^{u}\sin\!\left(u+\frac{\pi}{4}\right)\,du$
$\displaystyle \int e^{u}\sin(u+\alpha)\,du=\frac{e^{u}}{2}\left(\sin(u+\alpha)-\cos(u+\alpha)\right)$
$\displaystyle \Rightarrow 2\int_{0}^{\pi}e^{u}\sin\!\left(u+\frac{\pi}{4}\right)\,du=\left[e^{u}\left(\sin\!\left(u+\frac{\pi}{4}\right)-\cos\!\left(u+\frac{\pi}{4}\right)\right)\right]_{0}^{\pi}=0$
$\\$
$\displaystyle \textbf{Question 52: }~\int_{0}^{2\pi}e^x\cos\!\left(\frac{\pi}{4}+\frac{x}{2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\theta=\frac{x}{2}+\frac{\pi}{4}\;(b=\tfrac12),\;a=1,\;a^2+b^2=1+\frac14=\frac54$
$\displaystyle \int e^{x}\cos\theta\,dx=\frac{e^{x}}{a^2+b^2}\left(a\cos\theta+b\sin\theta\right)=\frac{4}{5}e^{x}\left(\cos\theta+\frac12\sin\theta\right)$
$\displaystyle \int_{0}^{2\pi}e^x\cos\!\left(\frac{\pi}{4}+\frac{x}{2}\right)\!dx=\left[\frac{4}{5}e^{x}\left(\cos\!\left(\frac{x}{2}+\frac{\pi}{4}\right)+\frac12\sin\!\left(\frac{x}{2}+\frac{\pi}{4}\right)\right)\right]_{0}^{2\pi}$
$\displaystyle =-\frac{3\sqrt{2}}{5}\left(e^{2\pi}+1\right)$
$\\$
$\displaystyle \textbf{Question 53: }~\int_{0}^{\pi}e^{2x}\sin\!\left(\frac{\pi}{4}+x\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\theta=x+\frac{\pi}{4}\;(b=1),\;a=2,\;a^2+b^2=4+1=5$
$\displaystyle \int e^{2x}\sin\theta\,dx=\frac{e^{2x}}{a^2+b^2}\left(a\sin\theta-b\cos\theta\right)=\frac{e^{2x}}{5}\left(2\sin\theta-\cos\theta\right)$
$\displaystyle \int_{0}^{\pi}e^{2x}\sin\!\left(\frac{\pi}{4}+x\right)\!dx=\left[\frac{e^{2x}}{5}\left(2\sin\!\left(x+\frac{\pi}{4}\right)-\cos\!\left(x+\frac{\pi}{4}\right)\right)\right]_{0}^{\pi}$
$\displaystyle =-\frac{\sqrt{2}}{10}\left(e^{2\pi}+1\right)$
$\\$
$\displaystyle \textbf{Question 54: }~\int_{0}^{1}\frac{1}{\sqrt{1+x}-\sqrt{x}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+x}-\sqrt{x}}\,dx=\int_{0}^{1}\frac{\sqrt{1+x}+\sqrt{x}}{(1+x)-x}\,dx=\int_{0}^{1}\left(\sqrt{1+x}+\sqrt{x}\right)\,dx$
$\displaystyle =\int_{0}^{1}\sqrt{1+x}\,dx+\int_{0}^{1}\sqrt{x}\,dx=\left[\frac{2}{3}(1+x)^{3/2}\right]_{0}^{1}+\left[\frac{2}{3}x^{3/2}\right]_{0}^{1}$
$\displaystyle =\frac{2}{3}\left(2^{3/2}-1\right)+\frac{2}{3}=\frac{2}{3}\cdot 2^{3/2}=\frac{4\sqrt{2}}{3}$
$\\$
$\displaystyle \textbf{Question 55: }~\int_{1}^{2}\frac{x}{(x+1)(x+2)}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\Rightarrow x=A(x+2)+B(x+1)$
$\displaystyle A+B=1,\;2A+B=0\Rightarrow A=-1,\;B=2$
$\displaystyle \int_{1}^{2}\frac{x}{(x+1)(x+2)}\,dx=\int_{1}^{2}\left(-\frac{1}{x+1}+\frac{2}{x+2}\right)\,dx=\left[-\log(x+1)+2\log(x+2)\right]_{1}^{2}$
$\displaystyle =\log\!\left(\frac{32}{27}\right)$
$\\$
$\displaystyle \textbf{Question 56: }~\int_{0}^{\pi/2}\sin^3 x\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/2}\sin^3 x\,dx=\int_{0}^{\pi/2}\sin x(1-\cos^2 x)\,dx$
$\displaystyle \text{Let }u=\cos x\Rightarrow du=-\sin x\,dx,\;x:0\to\frac{\pi}{2}\Rightarrow u:1\to 0$
$\displaystyle =\int_{1}^{0}(1-u^2)(-du)=\int_{0}^{1}(1-u^2)\,du=\left[u-\frac{u^3}{3}\right]_{0}^{1}=\frac{2}{3}$
$\\$
$\displaystyle \textbf{Question 57: }~\int_{0}^{\pi}\!\left(\sin^2\frac{x}{2}-\cos^2\frac{x}{2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin^2\frac{x}{2}-\cos^2\frac{x}{2}=-(\cos^2\frac{x}{2}-\sin^2\frac{x}{2})=-\cos x$
$\displaystyle \int_{0}^{\pi}\left(\sin^2\frac{x}{2}-\cos^2\frac{x}{2}\right)\,dx=\int_{0}^{\pi}(-\cos x)\,dx=\left[-\sin x\right]_{0}^{\pi}=0$
$\\$
$\displaystyle \textbf{Question 58: }~\int_{1}^{2}e^{2x}\!\left(\frac{1}{x}-\frac{1}{2x^2}\right)\!dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{d}{dx}\left(\frac{e^{2x}}{2x}\right)=\frac{2e^{2x}\cdot 2x-e^{2x}\cdot 2}{4x^2}=e^{2x}\left(\frac{1}{x}-\frac{1}{2x^2}\right)$
$\displaystyle \Rightarrow \int_{1}^{2}e^{2x}\left(\frac{1}{x}-\frac{1}{2x^2}\right)\,dx=\left[\frac{e^{2x}}{2x}\right]_{1}^{2}=\frac{e^{4}}{4}-\frac{e^{2}}{2}$
$\\$

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

$\displaystyle \textbf{Question 59: }~\int_{1}^{2}\frac{1}{\sqrt{(x-1)(2-x)}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle (x-1)(2-x)=-(x^2-3x+2)=\frac14-\left(x-\frac32\right)^2$
$\displaystyle \int_{1}^{2}\frac{1}{\sqrt{(x-1)(2-x)}}\,dx=\int_{1}^{2}\frac{1}{\sqrt{\frac14-\left(x-\frac32\right)^2}}\,dx$
$\displaystyle \text{Let }u=2x-3\Rightarrow dx=\frac{du}{2},\;x:1\to2\Rightarrow u:-1\to 1$
$\displaystyle =\int_{-1}^{1}\frac{1}{\sqrt{1-u^2}}\,du=\left[\sin^{-1}u\right]_{-1}^{1}=\pi$
$\\$
$\displaystyle \textbf{Question 60: }~\text{If }\int_{0}^{k}\frac{1}{2+8x^2}\,dx=\frac{\pi}{16},\ \text{find the value of }k.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{k}\frac{1}{2+8x^2}\,dx=\int_{0}^{k}\frac{1}{2(1+4x^2)}\,dx=\frac12\int_{0}^{k}\frac{1}{1+(2x)^2}\,dx$
$\displaystyle =\frac12\cdot\frac12\left[\tan^{-1}(2x)\right]_{0}^{k}=\frac14\tan^{-1}(2k)$
$\displaystyle \frac14\tan^{-1}(2k)=\frac{\pi}{16}\Rightarrow \tan^{-1}(2k)=\frac{\pi}{4}\Rightarrow 2k=1\Rightarrow k=\frac12$
$\\$
$\displaystyle \textbf{Question 61: }~\text{If }\int_{0}^{a}3x^2\,dx=8,\ \text{find the value of }a.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{a}3x^2\,dx=\left[x^3\right]_{0}^{a}=a^3$
$\displaystyle a^3=8\Rightarrow a=2$
$\\$
$\displaystyle \textbf{Question 62: }~\int_{\pi}^{3\pi/2}\sqrt{1-\cos 2x}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle 1-\cos 2x=2\sin^2 x\Rightarrow \sqrt{1-\cos 2x}=\sqrt{2}\,|\sin x|$
$\displaystyle \text{On }[\pi,3\pi/2],\;\sin x\le 0\Rightarrow |\sin x|=-\sin x$
$\displaystyle \int_{\pi}^{3\pi/2}\sqrt{1-\cos 2x}\,dx=\sqrt{2}\int_{\pi}^{3\pi/2}(-\sin x)\,dx=\sqrt{2}\left[\cos x\right]_{\pi}^{3\pi/2}$
$\displaystyle =\sqrt{2}\left(\cos\frac{3\pi}{2}-\cos\pi\right)=\sqrt{2}(0-(-1))=\sqrt{2}$
$\\$
$\displaystyle \textbf{Question 63: }~\int_{0}^{2\pi}\sqrt{1+\sin\frac{x}{2}}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }t=\frac{x}{2}\Rightarrow x=2t,\;dx=2\,dt,\;x:0\to2\pi\Rightarrow t:0\to\pi$
$\displaystyle \int_{0}^{2\pi}\sqrt{1+\sin\frac{x}{2}}\,dx=2\int_{0}^{\pi}\sqrt{1+\sin t}\,dt$
$\displaystyle 1+\sin t=\left(\sin\frac{t}{2}+\cos\frac{t}{2}\right)^2\Rightarrow \sqrt{1+\sin t}=\sin\frac{t}{2}+\cos\frac{t}{2}$
$\displaystyle =2\int_{0}^{\pi}\left(\sin\frac{t}{2}+\cos\frac{t}{2}\right)\,dt=2\left[-2\cos\frac{t}{2}+2\sin\frac{t}{2}\right]_{0}^{\pi}=8$
$\\$
$\displaystyle \textbf{Question 64: }~\int_{0}^{\pi/4}(\tan x+\cot x)^{-2}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \tan x+\cot x=\frac{\sin^2 x+\cos^2 x}{\sin x\cos x}=\frac{1}{\sin x\cos x}=2\,\mathrm{cosec}\,2x$
$\displaystyle (\tan x+\cot x)^{-2}=\sin^2 x\cos^2 x=\frac14\sin^2 2x$
$\displaystyle \int_{0}^{\pi/4}(\tan x+\cot x)^{-2}\,dx=\frac14\int_{0}^{\pi/4}\sin^2 2x\,dx=\frac14\int_{0}^{\pi/4}\frac{1-\cos 4x}{2}\,dx$
$\displaystyle =\frac18\left[x-\frac{\sin 4x}{4}\right]_{0}^{\pi/4}=\frac18\cdot\frac{\pi}{4}=\frac{\pi}{32}$
$\\$
$\displaystyle \textbf{Question 65: }~\int_{0}^{1}x\log(1+2x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }u=\log(1+2x),\;dv=x\,dx\Rightarrow du=\frac{2}{1+2x}\,dx,\;v=\frac{x^2}{2}$
$\displaystyle \int_{0}^{1}x\log(1+2x)\,dx=\left[\frac{x^2}{2}\log(1+2x)\right]_{0}^{1}-\int_{0}^{1}\frac{x^2}{2}\cdot\frac{2}{1+2x}\,dx$
$\displaystyle =\frac12\log 3-\int_{0}^{1}\frac{x^2}{1+2x}\,dx$
$\displaystyle \frac{x^2}{1+2x}=\frac12x-\frac14+\frac{1}{4(1+2x)}$
$\displaystyle \int_{0}^{1}\frac{x^2}{1+2x}\,dx=\left[\frac14x^2-\frac14x+\frac18\log(1+2x)\right]_{0}^{1}=\frac18\log 3$
$\displaystyle \Rightarrow \int_{0}^{1}x\log(1+2x)\,dx=\frac12\log 3-\frac18\log 3=\frac{3}{8}\log 3$
$\\$
$\displaystyle \textbf{Question 66: }~\int_{\pi/6}^{\pi/3}(\tan x+\cot x)^2\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle (\tan x+\cot x)^2=\tan^2 x+\cot^2 x+2=\sec^2 x+\mathrm{cosec}^2 x$
$\displaystyle \int_{\pi/6}^{\pi/3}(\tan x+\cot x)^2\,dx=\left[\tan x-\cot x\right]_{\pi/6}^{\pi/3}$
$\displaystyle =\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)-\left(\frac{1}{\sqrt{3}}-\sqrt{3}\right)=\frac{4}{\sqrt{3}}$
$\\$
$\displaystyle \textbf{Question 67: }~\int_{0}^{\pi/4}(a^2\cos^2 x+b^2\sin^2 x)\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_{0}^{\pi/4}(a^2\cos^2 x+b^2\sin^2 x)\,dx=a^2\int_{0}^{\pi/4}\cos^2 x\,dx+b^2\int_{0}^{\pi/4}\sin^2 x\,dx$
$\displaystyle \int_{0}^{\pi/4}\cos^2 x\,dx=\left[\frac{x}{2}+\frac{\sin 2x}{4}\right]_{0}^{\pi/4}=\frac{\pi}{8}+\frac14$
$\displaystyle \int_{0}^{\pi/4}\sin^2 x\,dx=\left[\frac{x}{2}-\frac{\sin 2x}{4}\right]_{0}^{\pi/4}=\frac{\pi}{8}-\frac14$
$\displaystyle \Rightarrow \int_{0}^{\pi/4}(a^2\cos^2 x+b^2\sin^2 x)\,dx=\frac{\pi}{8}(a^2+b^2)+\frac14(a^2-b^2)$
$\\$
$\displaystyle \textbf{Question 68: }~\int_{0}^{1}\frac{1}{1+2x+2x^2+2x^3+x^4}\,dx.$
$\displaystyle \text{Answer:}$
$\displaystyle \int_0^1\frac{1}{1+2x+2x^2+2x^3+x^4}\,dx$
$\displaystyle =\int_0^1\frac{1}{(x^2+1)^2+2x(x^2+1)}\,dx$
$\displaystyle =\int_0^1\frac{1}{(x^2+1)(x^2+1+2x)}\,dx$
$\displaystyle =\int_0^1\frac{1}{(x^2+1)(x+1)^2}\,dx$
$\displaystyle \text{Let }\frac{1}{(x+1)^2(x^2+1)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{Cx+D}{x^2+1}$
$\displaystyle 1=A(x+1)(x^2+1)+B(x^2+1)+(Cx+D)(x+1)^2$
$\displaystyle \text{Putting }x=-1$
$\displaystyle 1=2B$
$\displaystyle B=\frac12$
$\displaystyle \text{Putting }x=0$
$\displaystyle A+B+D=1$
$\displaystyle \text{Equating coefficients of }x^3$
$\displaystyle A+C=0$
$\displaystyle \text{Equating coefficients of }x^2$
$\displaystyle A+B+2C+D=0$
$\displaystyle C=-\frac12$
$\displaystyle A=\frac12$
$\displaystyle D=0$
$\displaystyle \int_0^1\frac{1}{(x+1)^2(x^2+1)}\,dx$
$\displaystyle =\int_0^1\frac{1}{2(x+1)}\,dx+\int_0^1\frac{1}{2(x+1)^2}\,dx-\int_0^1\frac{x}{2(x^2+1)}\,dx$
$\displaystyle =\frac12[\log(x+1)]_0^1+\frac12\left[-\frac{1}{x+1}\right]_0^1-\frac14[\log(x^2+1)]_0^1$
$\displaystyle =\frac12(\log2-\log1)-\frac12\left(\frac12-1\right)-\frac14(\log2-\log1)$
$\displaystyle =\frac12\log2+\frac14-\frac14\log2$
$\displaystyle =\frac14\log2+\frac14$
$\displaystyle =\frac14(\log2+\log e)$
$\displaystyle =\frac14\log(2e)$