Exercise 20.1
Question 1
$=\int_{0}^{\pi /2} \frac{d x}{1+\sin x}$
Sol :
$=\int_{0}^{\pi /2} \frac{d x}{1+\sin x}\times \frac{1-\sin x}{1-\sin x}$
$=\int_{0}^{\pi/2} \frac{(1-\sin x)dx}{1-\sin^2 x}$
$=\int_{0}^{\pi/2} \frac{1-\sin x}{\cos^2 x}dx$
$=\int_{0}^{\pi/2} \left(\frac{1}{1-\cos x}-\frac{\sin x}{\cos^2 x}\right)dx$
$=\int_{0}^{\pi/2} \left(\sec^2 x-\tan x \sec x\right)dx$
$=\bigg[\tan x-\sec x\bigg]_{0}^{\frac{\pi}{2}}$
$=\lim_{x\rightarrow \frac{\pi}{2}}(\tan x-\sec x)-(0-1)$
$=\lim_{x\rightarrow \frac{\pi}{2}}\left(\frac{\sin x}{\cos x}-\frac{1}{\cos x}\right)+1$
$=\lim_{x\rightarrow \frac{\pi}{2}}\left(\frac{\sin x-1}{\cos x}\right)+1$
By L Hospital Rule
$=\lim _{x \rightarrow \frac{\pi}{2}^{-}} \frac{\cos x-0}{-\sin x}+1$
$=\frac{0}{1}+1$
=0+1=1
(i) $\int_{2}^{3} x^{2} d x$
Sol :
$=\left[\frac{x^{3}}{3}\right]_{2}^{3}$
$=\frac{3^{3}}{3}-\frac{2^{3}}{3}$
$=9-\frac{8}{2}$
$=\frac{27-8}{3}=\frac{19}{3}$
(iv) $\int_{1}^2\left(4 x^{3}-5 x^{2}+6 x+9\right) d x$
Sol :
$=\left[4 \frac{x^{4}}{x}-5 \frac{x^{3}}{3}+6 \frac{x^{2}}{2}+9 x\right]_{1}^{2}$
$=\left[x^{4}-\frac{5}{3} x^{3}+3 x^{2}+9x\right]_{1}^{2}$
$=2^{4}-\frac{5}{3}(2)^{2}+3(2)^{2}+9(2)-\left(1^{4}-\frac{5}{3}(1)^{3}+3(1)^{2}+9(1)\right)$
(v) $\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}$
Sol :
$=\left[\tan ^{-1} x\right]_{1}^{\sqrt{3}}$
$=\tan ^{-1} \sqrt{3}-\tan ^{-1}1$
$=\frac{x}{3}-\frac{\pi}{4}$
$=\frac{4 \pi-3 \pi}{12}=\frac{\pi}{12}$
(xii) $\int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}$
Sol :
$=\int_{-1}^{1} \frac{d x}{(x+1)^{2}+5-1}$
$=\int_{-1}^{1} \frac{d x}{(x-1)^{2}+2^{2}}$
$=\quad \frac{1}{2}\left[\tan ^{-1}\left(\frac{x+1}{2}\right)\right]_{-1}^{1}$
$=\frac{1}{2}\left[\operatorname{tan}^{-1}\left(\frac{1+1}{2}\right)-\tan ^{-1}\left(\frac{-1+1}{2}\right)\right]$
$=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right]$
$=\frac{1}{2}\left[\frac{\pi}{4}-0\right]$
$=\frac{\pi}{8}$
Question 2
(i) $\int_{0}^{\pi / 4}\left(2 \sec ^{2} x+x^{3}+2\right) d x$
Sol :
$=\left[2 \tan x+\frac{x^{4}}{4}+2 x\right]_{0}^{\pi / 4}$
$=2\left(\operatorname{tan} \frac{\pi}{4}\right)+\frac{\left(\pi/4\right)^{4}}{4}+2 \frac{\pi}{4}-(0)$
$= 2+\frac{\pi^{4}}{4^{5}}+\frac{\pi}{2}$
$=2+\frac{\pi^{4}}{1024}+\frac{\pi}{2}$
(ii) $\int_{0}^{\pi}\left(\sin ^{2} x / 2-\cos ^{2}\frac{x}{2}\right) d x$
Sol :
$=-\int_{0}^{\pi}\left(\cos^{2} x/2-\sin ^{2} x/2\right) d x$
$=-\int_{0}^{\pi} \cos x=-[\sin x]_{0}^{\pi}$
=-[0-0]=0
Question 3
(i) $\int_{0}^{1}\left(x^{1 / 3}+1\right)^{2} d x$
Sol :
$=\int_{0}^{1}\left(x^{2 / 3}+1+2 x^{1 / 3}\right) d x$
$=\left[\frac{x^{2}+1}{\frac{2}{3}+1}+x+2 \frac{x^{\frac{1}{3}}+1}{\frac{1}{3}+1}\right]_{0}^{1}$
$=\left[\frac{x^{5 / 3}}{5/3}+x+2 \frac{x^{4 / 3}}{4 / 3}\right]_{0}^{1}$
$=\left[\frac{3}{5} x^{5 / 3}+x+\frac{3}{2} x^{\frac{4}{3}}\right]_{0}^{1}$
$=\frac{3}{5}+1+\frac{3}{2}-(0)$
$=\frac{6+10+15}{10}=\frac{31}{10}$
(ii) $\int_{0}^{7} \sqrt{9+x} d x$
Sol :
$=\int_{0}^{7}(9+x)^{1/2} d x$
$=\frac{2}{3}\left[(9+x)^{3 / 2}\right]_{0}^{7}$
$=\frac{2}{3}\left[(9+7)^{3}-(9+0)^{3 / 2}\right]$
$=\frac{2}{3}\left[16^{3/ 2}-9^{3/ 2}\right]$
$=\frac{2}{3}\left[\left(4^{2}\right)^{3 / 2}-(3^2)^{3/2}\right]$
$=\frac{2}{3}[64-27]$
$=\frac{37 x^{2}}{3}=\frac{74}{3}$
Question 4
$\frac{8 B}{\sqrt{H}} \int_{0}^{H}(H-h) \sqrt{h} d b$
Sol :
$=\frac{8 B}{\sqrt{H}} \int_{0}^{H}\left(H h^{\frac{1}{2}}-h^{3 / 2}\right) d h$
$=\frac{8 B}{\sqrt{H}}\left[\frac{2 H}{3} h^{3 / 2}-\frac{2 }{5} h^{5/2}\right]_{0}^{H}$
$=\frac{8 B}{\sqrt{H}}\left[\frac{2 H}{3} H^{3/2}-\frac{2}{5} H^{5/2}-(0)\right]$
$=\frac{8 B}{\sqrt{H}}\left[\frac{2}{3} H^{5 / 2}-\frac{2}{5} H^{5 / 2}\right]$
$=\frac{8B \times 2 H^{5 / 2}}{\sqrt{H}}\left[\frac{1}{3}-\frac{1}{5}\right]$
$=168 \mathrm{H}^{2}\left(\frac{5-3}{15}\right)=\frac{32}{15} \mathrm{BH}^{2}$
Question 5
$\int_{0}^{\pi / 4} \frac{1+\sin 2 x}{\cos x+\sin x} d x$
Sol :
$=\int_{0}^{\pi / 4} \frac{(\cos x+\sin x)^{2}}{\cos +\sin x} d x$
$=\int_{0}^{\pi / 4}(\cos x+\sin x) d x$
$=[\sin x-\cos x]_{0}^{\pi / 4}$
$=\sin \frac{\pi}{4}-\cos \frac{\pi}{4}-(0-1)$
$=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+1$
=1
Question 6
$\int_{0}^{\pi /6} \cos x \cdot \cos d x dx$
Sol :
$=\frac{1}{2} \int_{0}^{\pi / 6}(\cos (3 x)+\cos (-x)) d x$
$=\frac{1}{2} \int_{0}^{\pi / 6}(\cos 3 x+\cos x) d \theta$
$=\frac{1}{2}\left[\frac{\sin 3 x}{3}+\sin x\right]_{0}^{\pi / 6}$
$=\frac{1}{2}\left[\frac{1}{3}+\frac{1}{2}-(0+0)\right]$
$=\frac{1}{2}\left[\frac{2+3}{6}\right]$
$=\frac{5}{12}$
Question 7
$\int_{0}^{\pi / 4} \cos ^{2} 3 x d x$
Sol :
$=\int_{0}^{\pi / 4} \frac{1+\cos 6x}{2} d x$
$=\frac{1}{2}\left[x+\frac{\sin 6 x}{6}\right]_{0}^{\pi / 4}$
$=\frac{1}{2}\left[\frac{\pi}{4}+\frac{\sin \frac{3 \pi}{2}}{6}-(0+0)\right]$
$=\frac{1}{2}\left[\frac{\pi}{4}-\frac{1}{6}\right]$
$=\frac{\pi}{8}-\frac{1}{12}$
Question 8
$\int_{\pi / 4}^{\pi / 2} \operatorname{cosec} ^{2} \theta \cdot \cos \theta d \theta$
Sol :
$=\int_{\pi / 4}^{\pi /2} \frac{\cos \theta}{\sin ^{2} \theta} d \theta$
$=\int_{\pi / 4}^{\pi /4} \cot \theta \cdot \operatorname{cosec}\theta d \theta$
$=-[\operatorname{cosec} \theta]_{\pi / 4}^{\pi / 2}$
=-[cosec π/4-cosec π/4]
=-[1-√2]
$=\sqrt{2}-1$
Question 20
$\int_{0}^{\pi/2}(1+\sin x)^{3} \cos x d x$
Sol :
$\left[\int f(x)^{n} f^{\prime}(x) d x=\frac{f(x)^{n+1}}{n+1}+c\right]$
$I=\int(1+\sin x)^{3} \cos d x$
$=-\frac{(1+\sin x)^{4}}{4}$
Now
$\int_{0}^{\pi / 2}(1+\sin n)^{3} \cos d x=\left[\frac{(1+\sin x)^4}{4}\right]_{0}^{\pi / 2}$
$=\frac{(1+1)^4}{4}-\left(\frac{(1+0)^4}{4}\right)$
$=\frac{16}{4}-\frac{1}{4}$
$=\frac{16-1}{4}=\frac{15}{4}$
ALTERNATE METHOD
$I=\int_{0}^{\pi/2}(1+\sin x)^{3} \cos x d x$
$0+\cos x=\frac{d t}{d x}$
$d x=\frac{d t}{a x}$
Now
$=\int_{1}^{2} t^{3} \cos x \frac{d t}{\operatorname{cos} x}$
$=\left[\frac{t^{4}}{4}\right]_{1}^{2}$
$=\frac{2^{4}}{4}-\frac{1^{4}}{4}$
$=\frac{16}{4}-\frac{1}{4}$
$=\frac{15}{4}$
Question 21
$\int_{0}^{\pi/2}(1+\cos x)^{2} \sin x d x$
Question 22
$\int_{0}^{\pi/2}(a+b \sin \theta)^{3} \cos \theta d \theta$
Question 23
$\int_{0}^{\pi / 2} \frac{\cos x}{1+\sin ^{4}} d x$
Sol :
$I=\int \frac{\cos x}{1+\sin x} d x$
Put
1+sin x=t
$0+\cos x=\frac{d t}{d x}$
$d x=\frac{d t}{\cos x}$
$I=\int \frac{\cot x}{t} \frac{d t}{\cos x}$
=log|t|
=log|1+sin x|
Now
$=\int_{0}^{\pi/2} \frac{\cos x}{1+\sin x} d x=[\log \mid 1+\sin x|]_{0}^{\pi / 2}$
=log|1+1|-log|1+0|
=log|2|-log1
=log2-0
=log2
ALTERNATE METHOD
Put 1+sin x=t
$\cos x=\frac{d t}{d x}$
$dx=\frac{d t}{\cos x}$
$=\int_{0}^{2} \frac{\cos x}{t} \frac{d t}{\cos x}$
$=[\log |t|]_{1}^{2}$
=log|2|-log1
$=\log _{e} 2$
Question 24
(iv) $\int_{0}^{\pi /3}{ \frac{\sec x \cdot \tan x}{1+\operatorname{sec}^{2} x}} d x$
Sol :
Put sec x=t
$\sec x.\tan x=\frac{d t}{d x}$
$d x=\frac{d t}{\sec x \cdot \tan x}$
Now
$=\int_{1}^{2} \frac{\sec x \cdot \tan x}{1+t^{2}} \frac{d t}{\sec x+\tan x}$
$=\left[\tan^{-1} t\right]_{1}^{2}$
=tan-1 2-tan-1 1
$=\tan ^{-1}\left(\frac{2-1}{1+2}\right)$
$=\tan^{-1}\left(\frac{1}{3}\right)$
Question 28
$\int_{0}^{\pi / 2} \sin ^{2} x \cdot \cos x d x$
Sol :
$=\int_{0}^{\pi/2}\left(\frac{2 \sin x \cdot \cos x}{2}\right)^{2} dx$
$=\int_{0}^{\pi /2}{\frac{\sin ^{2} 2x}{4}} dx$
$=\frac{1}{4} \int_{0}^{\pi /2}{\dfrac{(1-\cos 4x)}{2}} d x$
$=\frac{1}{8}\left(x-\frac{\sin 4 x}{4}\right)_{0}^{\pi /4}$
$=\frac{1}{8}\left[\frac{\pi}{2}-0-(0-0)\right]$
$=\frac{\pi}{16}$
Question 29
$\int_{0}^{\pi / 4} \tan ^{3} x \cdot \sec ^{2} x dx$
Sol :
Put tan x=t
$=\left(\frac{\tan^{4} x}{4}\right)_{0}^{\pi / 4}$
$=\frac{1}{4}-0$
$=\frac{1}{4}$
Question 30
$\int_{0}^{\pi /4} \tan x\cdot \sec^{4}x dx$
Sol :
Put tan x=t
$\sec ^{2} x=\frac{d t}{d x}$
$d x=\frac{d t}{\sec ^{4} x}$
Now
$=\int_{0}^{1} t \sec ^{4} x \frac{d t}{\sec^2 t}$
$=\int_{0}^{1} t \sec^{2} x d t$
$=\int_{0}^{1} t\left(1+\operatorname{tan}^{2} x\right) d t$
$=\int_{0}^{1} t\left(1+t^{2}\right) d t$
Question 31
$\int_{0}^{\pi / 4} \sec ^{4} x d x$
Sol :
Put tan x=t
$\sec ^{2} x=\frac{d t}{d x}$
$d x=\frac{d t}{\sec ^{2} x}$
Now
$=\int_{0}^{1} \operatorname{sec}^{4} x \frac{d t}{\sec ^{2} x}$
$=\int_{0}^{1}\left(\sec ^{2} x\right) d t$
$=\int_{0}^{1}\left(1+\tan ^{2} x\right) d t$
$=\int_{0}^{0}\left(1+t^{2}\right) d t$
$=\left(t+\frac{t^{3}}{3}\right)_{0}^{1}$
$=1+\frac{1}{3}-(0+0)$
$=\frac{4}{3}$
Question 33
$\int_{0}^{\pi / 4} \operatorname{tan}^{3} x d x$
Sol :
$=\int_{0}^{\pi / 4} \tan x \tan ^{2} x d x$
$=\int_{0}^{\pi / 4} \tan x \left(\sec ^{2} x-1\right) d x$
$=\int_{0}^{\pi / 4}\left(\tan x \cdot \sec ^{2} x-\tan x\right) d x$
$=\left[\frac{\tan ^{2} x}{2}-\log |\sec x|\right]_{0}^{\pi / 4}$
$=\frac{1}{2}-\log \sqrt{2}-(0-0)$
$=\frac{1}{2}-\log \sqrt{2}$
$=\frac{1}{2}-\log 2^{\frac{1}{2}}$
$=\frac{1}{2}-\frac{1}{2}\log 2$
$=\frac{1}{2}(1-\log 2)$
Question 34
$\int_{0}^{\pi / 4}(\tan x-x) \tan ^{2} x d x$
Sol :
Put tan x-x=t
$\sec ^{2} x-1=\frac{d t}{d x}$
$\operatorname{tan}^{2} x=\frac{d t}{d x}$
$d x=\frac{d t}{\tan ^{2} x}$
Now
$=\int_{0}^{1-\frac{\pi}{4}} t \tan ^{2} x \cdot \frac{d t}{\tan ^{2} x }$
$\left(\frac{t^{2}}{2}\right)_{0}^{1-\pi / 4}=\frac{1}{2}\left[\left(1-\frac{\pi}{4}\right)^{2}-0\right]$
$=\frac{1}{2}\left(\frac{4- \pi}{4}\right)^{2}=\frac{1}{2} \frac{(4-\pi)^{2}}{16}=\frac{(4-\pi)^{2}}{32}$
Question 35
$\int_{0}^{\pi / 4} \tan ^{4} x d x$
Sol :
$=\int_{0}^{\pi / 4} tan ^{2} x.\tan^2 x dx$
$=\int_{0}^{\pi / 4} \tan ^{2} x\left(\sec ^{2} x-1\right) d x$
$=\int_{0}^{\pi / 4}\left(\sin ^{2} x \cdot \sec ^{2} x-\tan ^{2} x\right) d x$
$\int_{0}^{4}\left(\tan ^{2} x \cdot \sec ^{2} x-\left(\sec ^{2} x-1\right)\right) d x$
$=\left[\frac{\operatorname{tan}^{3} x}{3}-\tan x+x\right]_{0}^{\pi / 4}$
$=\left(\frac{1}{3}-1+\frac{\pi}{4}\right)-(0)$
$=-\frac{2}{3}+\frac{\pi}{4}=\frac{\pi}{4}-\frac{2}{3}$
Question 36
$\int_{0}^{\pi} \frac{\tan x}{\sec x+\tan x} d x$
Sol :
$=\int_{0}^{\pi} \frac{\tan x}{\sec x+\tan x} \times \frac{\sec x-\tan x}{\sec x-\tan x} d x$
$=\int_{0}^{\pi} \frac{\sec x \tan x-\tan ^{2} x}{\sec ^{2} x-\tan ^{2} x} d x$
$=\int_{0}^{\pi}\left(\sec x \cdot \tan x-\tan ^{2} x\right) d x$
$=\int_{0}^{\pi}\left(\sec x \cdot \tan x-\left(\sec ^{2} x-1\right)\right) d x$
$=\int_{0}^{\pi}\left(\sec x \cdot \tan x-\sec ^{2} x+1\right) d x$
$=[\sec x-\tan x+x]_{0}^{\pi}$
=(-1-0+π)-(1-0+0)
=-1+π-1
=π-2
Question 37
$\int_{0}^{1} \frac{6 x-5}{3 x^{2}-5 x+7} d x$
Sol :
$=\left[\log \left|3 x^{2}-5 x+7\right|\right]_{0}^{1}$
$\left[\int \frac{f^{ \prime}(x)}{f(x)} d x=\log |f(x)|+c\right]$
=[log|5|-log|7|]
$=\log \frac{5}{7}$
Question 38
$\int_{2}^{4} \frac{6 x^{2}-1}{\sqrt{2 x^{2}-x-2}} d x$
Sol :
$=\int_{2}^{4}\left(2 x^{3}-x-2\right)^{-1 / 2}\left(6 x^{2}-1\right) d x$
$=\frac{\left[\left(2 x^{3}-x-2\right)^{\frac{1}{2}}\right]^{4}}{1/2}$
$=2\left[(128-4-2)^{\frac{1}{2}}-(16-2-2)^{1 / 2}\right]$
$=2[\sqrt{122}-\sqrt{12}]$
Question 39
$\int_{0}^{1} \frac{x^{3}}{\sqrt{1+8 x^{4}}} d x$
Sol :
$=\frac{1}{32} \int_{0}^{1} \frac{32 x^{3}}{\left(1+8 x^{4}\right)^{1 / 2}} d x$
$=\frac{1}{32} \int_{0}^{1}\left(1+8 x^{4}\right)^{-\frac{1}{2}} 32 x^{3} d x$
$\left.=\frac{1}{32} \times {2}\bigg[(1+8 x^{4}\right)^{1 / 2}\bigg]_{0}^{1}$
$=\frac{1}{16}[3-1]=\frac{2}{16}=\frac{1}{8}$
Question 40
$\int_{1}^{2} \frac{x}{\sqrt{2 x^{2}+1}} d x$
Sol :
$=\frac{1}{4} \int_{1}^{2} \frac{4 x}{\left(2 x^{2}+1\right)^{\frac{1}{2}}} d x$
$=\frac{1}{4} \int_{1}^{2}\left(2 x^{2}+1\right)^{-1 / 2} 4 x d x$
$=\frac{1}{4} \times {2}\left[\left(2 x^{2}+1\right)^{1/2}\right]_{1}^{2}$
$=\frac{1}{2}\big[3-\sqrt{3}\big]$
$=\frac{3-\sqrt{3}}{2}$
Question 41
$\int_{0}^{1} \frac{d x}{\sqrt{1+x^{2}}}$
Sol :
$=\left[\log \left|x+\sqrt{1+a^{2}}\right|\right]_{0}^{1}$
$=\log |1+\sqrt{2}|-\log |1|$
$=\log |1+\sqrt{2}|+2$
Question 42
$\int_{0}^{a} \frac{x}{\sqrt{a^{2}+x^{2}}} d x$
Sol :
$=\frac{1}{2} \int_{0}^{a} \frac{2 x}{\left(a^{2}+x^{2}\right)^{2}} d x$
$=\frac{1}{2} \int\left(a^{2}+1^{2}\right)^{1/2}(2x) d x$
$=\frac{1}{2} \times 2\left[\left(a^{2}+x^{2}\right)^{1/ 2}\right]_{0}^{a}$
$=\left(2 a^{2}\right)^{\frac{1}{2}}-a$
$=\sqrt{2} a-a$
$=a(\sqrt{2}-1)$
Question 43
$\int_{0}^{2} x \sqrt{x+2} d x$
Sol :
$=\int_{0}^{2}\left((x+2-2)(x+2)^{1 / 2}d x\right.$
$=\int_{0}^{2}\left[(x+2)^{2 / 2}-2(x+2)^{1 / 2}\right] d x$
$=\left[\frac{2}{5}(x+2)^{5 / 2}-\frac{2 \times 2}{3}(x+2)^{3 / 2}\right]_{0}^{2}$
$=\frac{2}{5}(4)^{5 / 2}-\frac{4}{3}(4)^{3 / 2}-\left[\frac{2}{5}(2)^{5 / 2}-\frac{4}{3}(2)^{3 / 2}\right]$
$=\frac{64}{5}-\frac{32}{3}-\left[\frac{8 \sqrt{2}}{5}-\frac{8 \sqrt{2}}{3}\right]$
$=\frac{192-160}{15}-\left[\frac{24 \sqrt{2}-40 / 2}{15}\right]$
$=\frac{32}{15}+\frac{16 \sqrt{2}}{15}$
Question 44
(i) $\int_{2}^{3} \frac{x}{x^{2}+1} d x$
Sol :
$=\frac{1}{2} \int_{2}^{3} \frac{2 x}{x^{2}+1} d x$
$=\frac{1}{2}\left[\log \left|x^{2}+1\right|\right]_{2}^{3}$
$=\frac{1}{2}[\log 10-\log 5]$
$=\frac{1}{2} \log \frac{10}{3}$
$=\frac{1}{2} \log 2$
(ii) $\int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x$
Sol :
$=\int_{0}^{2}\left(\frac{6 x}{x^{2}+4}+\frac{3}{x^{2}+4}\right) d x$
$=\int_{0}^{2}\left(3 \cdot \frac{2 x}{x^{2}+4}+3 \cdot \frac{1}{x^{2}+2^{2}}\right) d x$
$=\left[3\log \left|x^{2}+4\right|+3 \times \frac{1}{2} \tan ^{-1} \frac{x}{2}\right]_{0}^{2}$
$=3 \log 8+\frac{3}{2} \frac{\pi}{4}-(3 \log 4+0)$
$=3 \log {8}+\frac{3 \pi}{8}-3 \log 4$
$=3(\log 8-\log 4)+\frac{3 \pi}{8}$
$=3 \log {\frac{8}{4}}+\frac{3 \pi}{8}$
$=3 \log 2+\frac{3 \pi}{8}$
(iii) $\int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3} d x$
Sol :
$=5 \int_{1}^{2} \frac{x^{2}}{x^{2}+4 x+3} d x$
$=5 \int_{1}^{2} \frac{x^{2}+4 x+3-(4 x+3)}{x^{2}+4 x+3} d x$
$=5 \int_{1}^{2}\left(1-\frac{4 x+3}{x^{2}+4 x+3}\right) d x$
$=5\left[(x)_{1}^{2}-\int_{1}^{2} \frac{4 x+5}{x^{2}+4 x+3} d x\right]$
$5\left[1-\int_{1}^{2} \frac{4{x+3}}{(x+1)(x+3)} d x\right]$
$=5\left[1-\int_{1}^{2}\left(\frac{\frac{-1}{2}}{x+1}+\frac{\frac{9}{2}}{x+3}\right) d x\right]$
$=5\left[1-\left[\left(-\frac{1}{2} \log (x+1)+\frac{9}{2} \log (x+3)\right)\right]_{1}^{2}\right]$
$=5\left[1-\left\{-\frac{1}{2} \log 3+\frac{9}{2} \log 5-\left(-\frac{1}{2} \lg 2+\frac{9}{2} \log 4\right)\right\}\right]$
$=5\left[1-\left\{-\frac{1}{2} \log 3+\frac{9}{2} \log 5+\frac{1}{2} \log 2-\frac{9}{2} \log 4\right]\right.$
$=5-\frac{5}{2}(-\log 3+9 \log 5+\log 2-9 \log 4)$
$=5-\frac{5}{2}\left(9 \log \frac{5}{4}+\log \frac{2}{3}\right)$
Question 45
(i) $\int_{0}^{1} x e^{x^{2}} d x$
Sol :
Put x2=t
$2 x=\frac{d t}{d x}$
$d x=\frac{d t}{2 x}$
Now
$=\int_{0}^{1} x e^{t} \frac{d t}{2 x}$
$=\frac{1}{2} \int_{0}^{1} e^{t} d t$
$=\frac{1}{2}\left[e^{t}\right]_{0}^{1}$
$=\frac{1}{2}\left(e^{1}-e^{0}\right)$
$=\frac{2-1}{2}$
(ii) $\int_{0}^{1} x^{2} e^{x^{3}} d x$
Sol :
Question 46
$\int_{0}^{1}\left(2 x^{2}-3\right)^{n} x d x$
Sol :
$=\frac{1}{4} \int_{0}^{1}\left(2 x^{2}-3\right)^{n} 4 x$
$\left[\int f(x)^{n}. f^{\prime}(x) d x=\frac{f(x)^{n+1}}{n+1}+c\right]$
$=\frac{1}{4}\left[\dfrac{(2 x-3)}{n+1}^{n+1}\right]_{0}^{1}$
$=\frac{1}{4(n-1)}\left((-1)^{n+1}-(-3)^{n+1}\right)$
$=\frac{1}{4(n+1)}\left((-1)^{n+1}-(-1 \times 3)^{n+1}\right)$
$=\frac{1}{4(n+1)}\left((-1)^{n+1}-(-1)^{n+1}(3)^{n+1}\right)$
$=\frac{(-1)^{n+1}\left(1-3^{n+1}\right)}{4(n+1)}$
Question 47
$\int_{0}^{1} x^{3} \sqrt{\left(1+3 x^{4}\right)} d x$
Sol :
$=\frac{1}{12} \int\left[2 x^{3}\left(1+3 x^{4}\right)^{1 / 2} d x\right.$
$=\frac{1}{12} \times \frac{2}{3}\left[\left(1+3 x^{4}\right)^{3 / 2}\right]_{0}^{1}$
$=\frac{1}{18}\left[4^{3 / 2}-1\right]$
$=\frac{1}{18}\left(\left(2^{2}\right)^{3 / 2}-1\right)$
$=\frac{1}{18} \times 7=\frac{7}{18}$
Question 48
$\int_{-2}^{2} \frac{d x}{4+x^{2}}$
Sol :
$=\int_{-2}^{2} \frac{d x}{2^{2}+x^{2}}$
$=\frac{1}{2}\left[\tan ^{-1} \frac{x}{2}\right]_{-2}^{2}$
$=\frac{1}{2}\left[\tan^{-1} 1-\tan^{-1}(-1)\right]$
$=\frac{1}{2}\left[\frac{\pi}{4}+\tan ^{-1} 1\right]$
$=\frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right]$
$=\frac{1}{2} \frac{\pi}{2}=\frac{\pi}{4}$
Question 49
$\int_{0}^{a} \frac{d x}{\left(a^{2}+x^{2}\right)^{3 / 2}}$
Sol :
Putting x=atanθ
$1=a \sec ^{2} \theta \frac{d \theta}{d x}$
dx=asec2θdθ
Now
$=\int_{0}^{\pi / 4} \frac{\operatorname{asec}^{2} \theta d \theta}{\left(a^{2}+a^{2}+\operatorname{tan}^{2} \theta\right)^{3 / 2}}$
$=\int_{0}^{\pi / 4} \frac{d \sec ^{2} \theta d \theta}{a^{3}\left(1+\tan ^{2} \theta\right)^{3 / 2}}$
$=\frac{1}{a^{2}} \int_{0}^{\pi / 4} \frac{\sec^2 \theta d \theta}{\left(\sec ^{2} \theta\right)^{3 / 2}}$
$=\frac{1}{a^{2}} \int_{0}^{\pi / 4} \frac{\sec ^2 \theta d \theta}{\sec ^{3} \theta}$
$=\frac{1}{a^{2}} \int_{0}^{\pi / 4} \cos \theta d \theta$
$=\frac{1}{a^{2}}[\sin \theta]_{0}^{\pi / 4}$
$=\frac{1}{a^{2}}\left[\frac{1}{12}-0\right]$
$=\frac{1}{\sqrt{2} a^{2}} $
Question 50
$\int_{\frac{1}{2}}^{1} \frac{d x}{x^{2} \sqrt{1-x^{2}}}$
Sol :
$=\int_{1 / 2}^{1} \frac{d x}{x^{3} \sqrt{\frac{1}{x^{2}}-1}}$
$=-\frac{1}{2} \int_{1 / 2}^{1}\left(\frac{1}{x^{2}}-1\right)^{-1 / 2}\left(\frac{-2}{x^{3}}\right) d x$
$=-\frac{1}{2} \times 2\left[\left(\frac{1}{x^{2}}-1\right)^{1 / 2}\right]_{1 / 2}^{1}$
$=-\left[0-(4-1)^{1/2}\right]$
=√3
Question 51
$\int_{0}^{a} x y d x$ where $y=\sqrt{a^{2}-x^{2}}$
Sol :
$=\int_{0}^{4} x \sqrt{a^{2}-x^{2}} d x$
$=-\frac{1}{2} \int_{0}^{a}(-2 x)\left(a^{2}-x^{2}\right)^{1/2} d x$
$=-\frac{1}{2}\times \frac{x}{3}\left[\left(a^{2}-x^{2}\right)^{3 / 2}\right]_{0}^{a}$
$=-\frac{1}{3}\left[0-a^{3}\right]$
$=\frac{a^{3}}{3}$
Question 52
$\int_{0}^{1} \frac{\tan ^{-1}x}{1+x^{2}} d x$
Sol :
$=\int_{0}^{1} \tan ^{-1} x \cdot\left(\frac{1}{1+x^{2}}\right) d x$
$=\frac{\left[\left(\tan ^{-1} x\right)^{2}\right]_{0}^{1}}{2}$
$=\frac{1}{2}\left[\frac{\pi^{2}}{16}-0\right]$
$=\frac{\pi^{2}}{32}$
Question 53
$\int_{0}^{1} \frac{\left(\tan ^{-1} x\right)^{2}}{1+x^{2}} dx$
Sol :
Question 54
$\int_{0}^{1} \frac{e^{-x}}{1+e^{x}} d x$
Sol :
$=\int_{0}^{1} \frac{1}{e^{x}\left(1+e^{x}\right)} d x$
Put ex=t
$e^{x}=\frac{d t}{dx}$
$d x=\frac{d t}{e^{x}}$
Now
$=\int_{1}^{e} \frac{1}{t(1+t)} \frac{a t}{e^{t}}$
$=\int_{1}^{e} \frac{d t}{t^{2}(1+t)}$
Let $\frac{1}{t^{2}(1+t)}=\frac{A}{t}+\frac{B}{t^{2}}+\frac{c}{1+t}$
$\frac{1}{t^{2} (1+t)}=\frac{A+(1+t)+B(1+t)+c t^{2}}{t^{2}(1+t)}$
1=(A+C)t2+(A+B)t+B
On Compounding
A+C=0..(1)
A+B=0..(2)
B=1..(3) put in (2)
A=-1 put in (1)
C=1
Now
$=\int_{1}^{e}\left(\frac{-1}{t}+\frac{1}{t^{2}}+\frac{1}{1+t}\right) d t$
$=\left[-\log t-\frac{1}{t}+\log |1+t|\right]_{1}^{e}$
$=\left[\begin{array}{c}-1-\frac{1}{e}+\log|1+e|-(-0-1+\log 2)]\end{array}\right.$
$=-1-\frac{1}{e}+\log (1+e)+1-\log 2$
$=\log \left(\frac{1+e}{2}\right)-\frac{1}{e}$
Question 55
$\int_{2}^{3} \frac{d x}{x \log x}$
Sol :
$=\int_{2}^{3} \frac{\frac{1}{x}}{\log x} d x$
$=\big[\log \mid \log x|\big]_{2}^{3}$
=log(log 3)-log(log 2)
$=\log \left(\frac{\log 3}{\log 2}\right)$
$\left[\int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]$
$=\log \left(\log\frac{3}{2}\right)$
Question 56
$\int_{0}^{1} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x$
Sol :
Putting √x=t
$\frac{1}{2 \sqrt{x}}=\frac{d t}{d x}$
dx=2√xdt
Now
$=\int_{0}^{1} \frac{e^{t}}{\sqrt{x} }2 \sqrt{x} d t$
$=2 \int_{0}^{1} e^{t} d t$
$=2\left[e^{t}\right]_{0}^{1}$
=2(e1-e0)
=2(e-1)
Question 57
$\int_{1}^{2} \frac{(\log x)^{2}}{x} d x$
Sol :
$=\int_{1}^{2}(\log x)^{2} \cdot \frac{1}{x} d x$
$=\frac{\left[(\log x)^{3}\right]_{1}^{2}}{3}$
$=\frac{1}{3}\left[(\log 2)^{3}-0\right]$
$=\frac{(\log 2)^{3}}{3}$
Question 58
$\int_{1}^{2} \frac{\sqrt{\log x}}{x} d x$
Sol :
$=\int_{1}^{2}(\log x)^{1/2} \cdot \frac{1}{x} d x$
$=\frac{2}{3}\left[(\log x)^{3 / 2}\right]_{1}^{2}$
$=\frac{2}{3}\left[( \log 2)^{3 / 2}-0\right]$
$=\frac{2}{3}(\log 2)^{3 / 2}$
Question 59
$\int_{0}^{1 / 2} \frac{e^{\sin ^{-1}x}}{\sqrt{1-x^{2}}} d x$
Sol :
Putting sin-1x=t
$\frac{1}{\sqrt{1-x^{2}}}=\frac{d t}{d x}$
$d x=\sqrt{1-x^{2}} d t$
Now
$=\int_{0}^{\pi / 6} \frac{e^{t}}{\sqrt{1-x^{2}}} \sqrt{1-x^2} d t$
$=\int_{0}^{\pi / 6} e^{t} d t$
$=\left(e^{t}\right)_{0}^{\pi / 6}$
$=e^{\pi/{6}}-e^{0}$
$=e^{\pi / 6}-1$
Question 60
$\int_{0}^{\pi} x \cot ^{2} x d x$
Sol :
$I=\int x \cos ^{2} x d x$
$=\int x\left(\frac{1+\cos 2 x}{2}\right) d x$
$=\frac{1}{2} \int x(1+\cos x) d x$
$=\frac{1}{2} \int(x+x \cos x) d x$
$=\frac{1}{2}\left[\frac{x^{2}}{2}+\int x \cos x d x\right]$
$=\frac{1}{2}\left[\frac{x^{2}}{2}+x \int \operatorname{cos2x} d x-\int\left(1-\int(cos2xdx) d x\right]\right.$
$=\frac{1}{2}\left[\frac{x^{2}}{2}+\frac{x \sin x}{2}-\int \frac{\operatorname{sin2x} }{2} d x\right]$
$=\frac{1}{2}\left[\frac{x^{2}}{2}+\frac{x \sin 2 x}{2}+\frac{\cos x}{4}\right]$
Now
$\int_{0}^{\pi} x \cos ^{2} x d x=\frac{1}{2}\left[\frac{x^{2}}{2}+\frac{x \cdot \sin 2x}{2}+\frac{\cos 2x}{4}\right]_{0}^{\pi}$
$=\frac{1}{2}\left[\frac{\pi^{2}}{2}+\frac{\pi}{2} \times 0+\frac{1}{4}-\left(\frac{1}{4}\right)\right]$
$=\frac{1}{2}\left[\frac{\pi^{2}}{2}+\frac{1}{4}-\frac{1}{4}\right]$
$=\frac{\pi^{2}}{4}$
Question 62
$\int_{0}^{\pi/2} x \sin x \cdot \sin 2 x d x$
Sol :
$=\frac{1}{2} \int_{0}^{\pi} x \cdot 2\sin x \cdot \sin 2 x d x$
$1=\frac{1}{2} \int_{0}^{\pi / 2} x(\cos x-\cos 3x) d x$...(i)
Let $I_{1}=\int{x}({\cos x-\cos 3x}) d x$
$=x \int(\cos x-\cos 3x) d x-\int\left(1 \cdot \int(\cos x-\cos 3x) d x\right) d x$
$=x[\sin x-\frac{\sin 3x}{3}]-\int\left(\sin x-\frac{\sin 3 x}{3}\right) d x$
$I_{1}=x\left[\sin x-\frac{\sin 3x}{3}\right]-\left(-\cos x+\frac{\cos 3x}{9}\right)$
Now from (1)
$I=\frac{1}{2}\left[x\left(\operatorname{sin} x-\frac{\sin 3 x}{3}\right)+\cos x-\frac{\cos 3x}{9}\right]_{0}^{\pi/2}$
$=\frac{1}{2}\left[\frac{\pi}{2}\left(1+\frac{1}{3}\right)+0-0-\left(0+1-\frac{1}{9}\right)\right]$
$=\frac{1}{2}\left[\frac{4 \pi}{-6}-\frac{8}{9}\right]$
$=\frac{\pi}{3}-\frac{4}{9}$
Question 63
$\int_{0}^{\pi / 4} x^{2} \sin 2 \pi d x$
Sol :
$I=\int {x^{2}} \sin 2 x d x$
$=x^{2} \int \sin 2 x d x-\int\left(2 x \cdot \int \sin 2 x d x\right) d x$
$=-x^{2} \frac{\cos 2 x}{2}+\int\left(2 \times \frac{\cos 2 x}{2}\right) d x$
$=-x^{2} \frac{\cos 2x}{2}+\int {x \cos 2x}dx$
$=-x^{2} \frac{\cos 2x}{2}+x \int \cos 2x d x-\int\left(1-\int \cos 2x d x\right) d x$
$=-x^{2} \frac{\cos 2x}{2}+x \frac{\sin 2 x}{2}-\int \frac{\sin 2x}{2} d x$
$I=-x^2 \frac{\cos 2x}{2}+x \frac{\sin 2 x}{2}+\frac{\cos 2x}{4}$
Now
$\int_{0}^{\pi /4} x^{2} \sin 2 x d x=\left[-x^2\frac{\cos 2x}{2}+\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}\right]_{0}^{\pi/4}$
$=\left[-\frac{\pi^{2}}{16} \cdot 0+\frac{\pi}{4} \cdot \frac{1}{2}+0-\left(\frac{1}{4}\right)\right]$
$=\frac{\pi}{8}-\frac{1}{4}$
Question 66
$\int_{1}^{\sqrt{e}} x \log x d x$
Sol :
$I=\int x\log xdx$
$=\log x \int xd x-\int\left(\frac{1}{x} \int x d x\right) d x$
$=\frac{x^{2}}{2} \log x-\int \frac{1}{x} \frac{x^{2}}{2} d x$
$=\frac{x^{2}}{2} \log x-\frac{1}{2} \int x d x$
$=\frac{x^{2}}{2} \log x-\frac{1}{2} \frac{x^{2}}{2}$
$I=\frac{x^2}{2} \log x-\frac{x^{2}}{4}$
Now
$\int_{1}^{\sqrt{e}}x \log x dx=\left[\frac{x^{2}}{2} \log x-\frac{x^{2}}{4}\right]_{1}^{\sqrt{e}}$
$=\frac{e}{2} \log \sqrt{e}-\frac{e^{2}}{4}-\left(0-\frac{1}{4}\right)$
$=\frac{e}{2} \log e^{1 / 2}-\frac{e^{2}}{4}+\frac{1}{4}$
$=\frac{e}{2} \cdot \frac{1}{2} \log e-\frac{e^{2}}{4}+\frac{1}{4}$
$=\frac{e}{4}-\frac{e^{2}}{4}+\frac{1}{4}$
$=\frac{1}{4}$
Question 67
$\int_{0}^{1}1.\tan ^{-1} x d x$
Sol :
$=\operatorname{tan}^{-1} x \int d x-\int\left(\frac{1}{1+x^{2}} \int d x\right) d x$
$=x+\tan^{-1} x-\int \frac{1}{1+x^{2}} \cdot x d x$
$=x \tan ^{-1} x-\frac{1}{2} \int \frac{2x}{1+x^2}dx$
$I=x \operatorname{\tan}^{-1} x-\frac{1}{2} \log \left|1+x^{2}\right|$
Now
$\int_{0}^{1} \tan^{-1}x d x=\left[x \tan ^{1} x-\frac{1}{2} \log \mid 1+x^2|\right]_{0}^{1}$
$=\frac{\pi}{4}-\frac{1}{2} \log 2-(0-0)$
$=\frac{\pi}{4}-\frac{1}{2} \log 2$
Question 68
$\int_{0}^{1} {x}^{2} {e^{x}} d x$
Sol :
$=x^{2} \int e^{x} d x-\int\left(2 x \cdot \int e^{x} d x\right) d x$
$=x^{2} e^{x}-\int\left(2 x e^{x}\right) d x$
$=x^{2} e^{x}-2\left[x \int e^{x} d x-\int\left(1 \cdot \int e^{x} d x\right) d x\right]$
$=x^{2} e^{x}-2\left[x e^{x}-\int e^{x} dx\right]$
$=x^{2} e^{x}-2\left[x e^{x}-e^{x}\right]$
$I=x^{2} e^{x}-2 x e^{x}+2 e^{x}$
Now
$\int_{0}^{1} x^{2} e^{x} d x=\left[x^{2} e^{x}-2 x e^{x}+2 e^{x}\right]_{0}^{1}$
=e-2e+2e-(0-0+2)
=e-2
Question 69
$\int_{0}^{\pi/2} e^{-x} \cos xd x$
Sol :
$=\cos x \int e^{-x} d x+\int\left(\sin x \int e^{-x} d x\right) d x$
$=-\cos e^{-x}+\int\left(-\sin e ^{-x}\right) d x$
$I=-\cos x \cdot e^{-x}-\int {e}^{-x} \sin x d x$
$=-e^{-x} \cos x -\left[\sin x \int e^{-x} d x-\int\left(\cos x \cdot \int e^{-x} d x\right) d x\right]$
$=-e^{-x} \cos x-\left[-{e}^{-x} \sin x+\int \cos {e}^{-x} d x\right]$
$=-e^{-x} \cos x+e^{-x} \sin x -\int e^{-x} \cos d x$
$I=-e^{-x} \cos x+e^{-x} \sin x-I$
$2I=-e^{-x} \cos x+e^{-x} \sin x$
$I=\frac{1}{2}\left[e^{-x} \sin x-e^{-x} \cos x\right]$
Now
$\int_{0}^{\pi / 2} e^{-x} \cos d x=\frac{1}{2}\left[{e}^{-x} \sin x-e^{-x} \cos x\right]_{0}^{\pi / 2}$
$=\frac{1}{2}\left[e^{-\pi / 2}-0-(0-1)\right]$
$=\frac{e^{-\pi/{2}}+1}{2}$
$=\frac{1+e^{-\pi / 2}}{2}$
Question 71
$\int_{a}^{b} \frac{\log x}{x^{2}}dx$
Sol :
Put $log_e x=t$⇒x=et
$\frac{1}{x}=\frac{dt}{d x}$
dx=xdt
Now
$=\int_{\log a}^{\log {b}} \frac{t}{x^{2}} \times xdt$
$=\int_{\log {a}}^{\log b} \frac{t}{e^{t}} d t$
$=\int_{\log a}^{\log b} e^{-t} t d$
$I=\int t e^{-t} d t$
$=t \int e^{-t} d t-\int\left(1 \int e^{-t} d t\right) d t$
$=-t {e}^{-t}+\int e^{-t} dt$
$I=-t e^{-t}-e^{t}$
Now
$\int_{\log a}^{\log b}+e^{-t} d t=\left[-t e^{-t}-e^{-t}\right]_{\log a}^{\log b}$
$=\left[-\log b e^{-\log b}-e^{-log b}-(-\log a e^{-log a}-e^{-\log a})\right]$
$=-\log b e^{\log \frac{1}{b}}-e^{\log \frac{1}{b}}+\log a^{\log \frac{1}{a}}+e^{\log \frac{1}{a}}$
$=-\frac{\log b}{b}-\frac{1}{b}+\frac{\log a}{a}+\frac{1}{a}$
$=-\frac{\log b+1}{b}+\frac{\log a+1}{a}$
Question 72
(i) $\int_{0}^{1} \sin ^{-1} x d x$
Sol :
$=\sin ^{-1} x \int 1 d x-\int\left(\frac{1}{\sqrt{1-x^{2}}} \int1 d x\right) d x$
$=x \sin ^{-1} x-\int \frac{x}{\sqrt{(1-x^2)}} d x$
$=x \sin ^{-1} x+\frac{1}{2} \int \frac{-2 x}{\left(1-x^{2}\right)^{1 / 2}} d x$
$=x \sin ^{-1} x+\frac{1}{2} \int\left(1-x^{2}\right)^{-1/2}(-2 x) d x$
$=x \sin ^{-1} x+\frac{1}{2} \frac{\left(1-x^{2}\right)^{1/2}}{\frac{1}{2}}$
$=x \sin ^{-1} x+\left(1-x^{2}\right)^{1 / 2}$
Now
$\int_{0}^{1} \sin ^{1} x d x=\left[x \sin^{-1} x+\left(1-x^{2}\right)^{1 / 2}\right]_{0}^{1}$
$=\frac{\pi}{2}+(1-1)^{1/2}-(1)$
$=\frac{\pi}{2}-1$
(ii) $\int_{0}^{\pi / 2} e^{x}\left(\frac{1+\sin x}{1+\cos x}\right) d x$
Sol :
$\int_{0}^{\pi / 2} e^{x}\left(\frac{1}{2 \cos ^{2} x/2}+\frac{\sin x}{2 \cos^2 x/2}\right) d x$
$\int_{0}^{\pi/2} e^{x}\left(\frac{1}{2} \sec ^{2} \frac{x}{2}+\frac{2\sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right) d x$
$\int_{0}^{\pi/2} e^{x}\left(\frac{1}{2} \sec ^{2} x+\tan \frac{x}{2}\right) d x$
$=\left[e^{x} \tan \frac{x}{2}\right]_{0}^{\pi/{2}}$
$=e^{\pi/2}-0=e^{\pi/2}$
(ii) $\int_{0}^{\pi /2} \frac{x+\sin x}{1+\cos x} d x$
Sol :
$\int_{0}^{\pi /2} \frac{x+\sin x}{2\cos^2 \frac{x}{2}} d x$
$=\int_{0}^{\pi/2}\left(\frac{x}{2 \cos^2 \frac{x}{2}}+\frac{2 \cos \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right) d x$
$=\int_{0}^{\pi/2}\left(\frac{x}{2} \sec^2 \frac{x}{2}+\tan \frac{x}{2}\right) d x$...(i)
Let $\int\left(\frac{x}{2} \sec ^{2} x\right)+\tan \frac{x}{2} d x$
$=\int \frac{x}{2} \sec ^{2} \frac{x}{2} d x+\int \tan \frac{x}{2} d x$
$=\frac{1}{2}\left[x \int \sec ^{2} \frac{x}{2} d x-\int\left(1 \int \sec ^{2} \frac{x}{2} d x\right) d x\right]+\int \tan \frac{x}{2} d x$
$=\frac{1}{2}\left[2 x \tan \frac{x}{2}-2 \int \tan \frac{x}{2} d x\right]+\int \tan \frac{x}{2} d x$
$=x \tan \frac{x}{2}-\int \tan \frac{x}{2}+\int \tan \frac{x}{2} d x$
=x tan x/2
From (i)
$\int_{0}^{\pi /2}\left(\frac{x}{2}\sec^2 \frac{x}{2}+\tan \frac{x}{2}\right)dx=\left[x \tan \frac{x}{2}\right]_{0}^{\pi /2}$
$=\frac{\pi}{2}-0$
$=\frac{\pi}{2}$
(iv) $\int_{0}^{a} \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$
Sol :
$I=\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x$
Putting x=atan2θ
1=2atanθ sec2θ$\frac{d \theta}{d x}$
dx=2atanθsec2θdθ
Now
$=\int \sin ^{2} \sqrt{\frac{a \operatorname{tan}^{2} \theta}{a+a\tan^2 \theta}} 2a\tan \theta \sec ^{2} \theta d \theta$
$=2 a \int \sin^{-1} \sqrt{\frac{\tan ^2 \theta}{1+\tan^2 \theta}} \tan \theta \sec ^{2} \theta d \theta$
$=2 a \int \sin^{-1} \sqrt{\frac{\tan ^2 \theta}{\sec^2 \theta}} \tan \theta \sec ^{2} \theta d \theta$
$=2 a \int \sin^{-1} \frac{\tan \theta}{\sec \theta} \tan \theta \sec ^{2} \theta d \theta$
$=2 a \int \sin^{-1} \frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta} } \tan \theta \sec ^{2} \theta d \theta$
$=2 a \int \sin ^{-1}(\sin \theta) \tan \sec^{2} \theta d \theta$
$=2a \int \theta \tan \sec ^{2} \theta d \theta$
$=2a\left[\theta \int \tan \theta \sec ^{2} \theta d \theta-\int\left(1 \cdot \int \tan \theta \sec^2 \theta d \theta\right) d \theta\right]$
$\left.=2a\left[\theta \frac{\operatorname{tan}^{2} \theta}{2}-\frac{1}{2}\right. \int \tan ^{2} \theta d \theta\right]$
$=2 a\left[\theta \frac{\tan^2 \theta}{2}-\frac{1}{2} \int(\sec^2 \theta-1) d \theta\right]$
$=2a\left[\frac{\theta \tan ^2 \theta}{2}-\frac{1}{2}(\tan \theta-\theta)\right]$
=aθtan2θ-tanθ+θ
$\int_{0}^{a} \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=a \theta \tan ^{2} \theta-a\tan \theta+a\theta$
$\left[x=a\tan^2 \theta\\\sqrt{\frac{x}{a}}=\tan^2 \theta \\ \tan^{-1} \sqrt{\frac{x}{a}}=\theta\right]$
$=\left[x \tan ^{-1} \sqrt{\frac{x}{a}}-a\sqrt{\frac{x}{a}}+a\tan ^{-1} \sqrt{\frac{x}{a}}~\right]_{0}^{a}$
$=\frac{a \pi}{4}-a+0 \frac{\pi}{4}-(0)$
$=0 \frac{\pi}{4}+\frac{0 \pi}{4}-a$
$=2 \frac{a\pi}{4}-a$
$=a\left(\frac{\pi}{2}-1\right)$
(vi) $\int_{0}^{1} x\left(\tan^{-1} x\right)^{2} d x$
Sol :
$I=\int_{0}^{1} x\left(\tan^{-1} x\right)^{2} d x$
Putting tan-1x=t⇒x=tan t
$\frac{1}{1+x^{2}}=\frac{dt}{dx}$
dx=(1+x2)dt
Now
$=\int_{0}^{\pi /4} x t^2\left(1+x^{2}\right) d x$
$=\int_{0}^{\pi /4} \tan t~ t^{2} \cdot\left(1+\operatorname{tan}^{2} t\right) d t$
$=\int_{0}^{\pi /4} t^{2}(\tan t \sec^2 t) d t$
$=t^2 \int \tan t \sec^2 t dt-\int (2t \int \tan t \sec^2 t dt)dt$
$=t^{2} \frac{\operatorname{tan}^{2} t}{2}-\int\left(2 t \frac{\operatorname{tan}^{2} t}{2}\right) d t$
$=t^{2} \frac{\operatorname{tan}^{2} t}{2}-\left[t \int \tan ^{2} t d t-\int\left(1 \cdot \int \tan ^{2} t d t\right) d t\right.$
$=t^{2} \frac{\tan ^{2} t}{2}-\left[t(\tan t-t)-\int(\tan t-t) d t\right.$
$=\frac{t^{2} \tan^{2} t}{2}-\left[t (\tan t-t)-\int (\tan t-t)dt\right]$
$=\frac{t^{2} \tan^{2} t}{2}-\left[t\tan t-t^{2}-\operatorname{log}|\sec t|+\frac{t^{2}}{2}\right]$
$=\frac{t^{2}\tan^{2} t}{2}-t \tan t+t^{2}+\log|\sec t|-\frac{t^{2}}{2}$
$=\frac{t^{2}\tan ^2 t}{2}-t \tan t+\frac{t^{2}}{2}+\log | \sec t|$
$\int_{0}^{\pi / 4} t^{2} \tan t \cdot \sec^2 t d t=\left[\frac{t^2 \tan^2 t}{2}-t \tan t+\frac{t^2}{2}+\log|\sec t|\right]_{0}^{\pi /2}$
$=\left[\frac{\pi^{2}}{16\times 2}-\frac{\pi}{4}+\frac{\pi^{2}}{32}+\log \sqrt{2}-10\right]$
$=\frac{n^{2}}{32}-\frac{\pi}{4}+\frac{\pi^{2}}{32}+\frac{1}{2} \log ^{2}$
$=\frac{\pi^{2}}{16}-\frac{\pi}{4}+\frac{1}{2} \log 2$
(vi) $\int_{0}^{\pi / 2} 2 \sin x \cdot \cos x \tan ^{-1}(\sin x) d x$
Sol :
Putting
sinx=t
$\cos x=\frac{d t}{d x}$
$dx=\frac{d t}{\cos x}$
Now
$=\int_{0}^{1} 2 t \cos x \tan ^{-1}(t) \frac{d t}{\cos x}$
$=2 \int_{0}^{1} t\tan^{-1} t dt$
$I=\int t \tan^{-1} t d t$
$=\tan^{-1} t \int t d t-\int\left(\frac{1}{1+t^{2}} \int t d t\right) d t$
$=\frac{t^{2}}{2} \tan ^{-1} t-\frac{1}{2} \int \frac{t^{2}}{1+t^{2}} d t$
$=\frac{t^{2}}{2} \tan ^{-1} t-\frac{1}{2} \int \frac{1+t^{2}-1}{1+t^{2}} d x$
$=\frac{t^{2}}{2} \tan ^{-1} t-\frac{1}{2} \int\left(1-\frac{1}{1+t^{2}}\right) d t$
$=\frac{t^{2}}{2} \tan ^{-1} t-\frac{1}{2}\left[t-\tan ^{-1}t \right]$
Now
$2 \int_{0}^{1} \tan^{2} dt=1\left[\frac{t^2}{2}\tan^{-1}t-\frac{t}{2}+\frac{\tan^{-1 }t}{2}\right]_{0}^{1}$
$=2\left[\frac{1}{2} \cdot \frac{\pi}{4}-\frac{1}{2}+\frac{\pi}{8}-0\right]$
$=\frac{\pi}{4}-1+\frac{\pi}{4}$
$=\frac{2 \pi}{4}-1$
$=\frac{\pi}{2}-1$
Question 73
(i) $\int_{1}^{2} \frac{x}{(x+1)(x+2)} d x$
Sol :
$=\int_{1}^{2}\left(\frac{-1}{x+1}+\frac{2}{x+2}\right) d x$
$=\Big[-\log |x+1|+2 \log |x+2|\Big]_{1}^{2}$
=-log 3+2log 4-(-log 2+2log 3)
=-log3+2log22+log2-2log3
=4log2+log2-3log3
=5log2-3log3
=log25-log33
=log32-log27
$=\log \frac{32}{27}$
(ii) $\int_{1}^{3} \frac{d x}{(x-1)(5-x)}$
Question 74
$\int_{0}^{a}\left(\sqrt{a^{2}-x^{2}}+x e^{x}\right) d x$
Sol :
$I=\int \sqrt{a^{2}+x^{2}}+x e^{n}$
$=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+x \int e^{x} d x-\int\left(1 \int e^{x} dx\right) d x$
$=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin^{-1} \frac{\pi}{4}+x e^{x}-\int e^{x} d x$
$I=\frac{x}{2} \sqrt{a^2-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+x e^{x}-e^{x}$
$\int_{0}^{a}\left(\sqrt{a^2-x^{2}}+x e^{x}\right) d x=\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+x e^{x}-e^{x}\right]_{0}^{a}$
$I=\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+x e^{x}-e^{x}$
$\int_{0}^{a}\left(\sqrt{a^2-x^{2}}+x e^{x}\right) d x=\left[\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}+x e^{x}-e^{x}\right]_{0}^{a}$
$=\left[0+\frac{a^2 \pi}{4}+a e^{a}-e^{a}-(1)\right]$
$=\frac{a^{2} x}{4}+e^{a(a-1)-1}$
Question 75
$\int_{1}^{2} \frac{d x}{x\left(1+x^{2}\right)}$
Sol :
$\frac{1}{x\left(1+x^{2}\right)}=\frac{A}{x}+\frac{Bx+C}{1+x^{2}}$
$\frac{1}{x\left(1+ x^{2}\right)}=\frac{A+A x^{2}+B x^{2}+C x}{x\left(1+x^{2}\right)}$
1=(A+B)x2+Cx+A
On comparing
A+B=0..(1)
C=0..(2)
A=1..(3) putting in (1)
B=-1
$\int_{1}^{2} \frac{d x}{x\left(1+x^{2}\right)}=\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2} \frac{2 x}{1+x^2}\right) d x$
$=\left[\log x-\frac{1}{2} \log \mid 1+x^{4}\right]_{1}^{2}$
$=\log 2-\frac{1}{2} \log 5-\left(0-\frac{1}{2} \log 2\right)$
$=\log 2-\frac{1}{2} \log 5+\frac{1}{2} \log 2$
$=\frac{3}{2} \log 2-\frac{1}{2} \log 5$
$=\frac{1}{2}[3 \log 2-\log 5]$
$=\frac{1}{2}\left[\log ^{3}-\log 5\right]$
$=\frac{1}{2}[\log 8-\log 5)$
$=\frac{1}{2} \operatorname{log} \frac{8}{5}$
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