KC Sinha Solution Class 12 Chapter 19 Indefinite Integrals Exercise 19.2

Exercise 19.2

Question 1

(i) $\displaystyle\int \dfrac{dx}{\sqrt{x+a}+\sqrt{x+b}}$
Sol :
=$\displaystyle\int \dfrac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a}+\sqrt{x+b})((\sqrt{x+a}-\sqrt{x+b})}dx$

=$\displaystyle\int \dfrac{\sqrt{x+a}-\sqrt{x+b}}{(\sqrt{x+a})^2-(\sqrt{x+b})^2}dx$

=$\displaystyle\int \dfrac{\sqrt{x+a}-\sqrt{x+b}}{x+a-x-b}dx$

=$\displaystyle\int \dfrac{\sqrt{x+a}-\sqrt{x+b}}{a-b}dx$

=$\dfrac{1}{a-b}\displaystyle\int \sqrt{x+a}-\sqrt{x+b}dx$

=$\dfrac{1}{a-b}\displaystyle\int \sqrt{x+a}-\dfrac{1}{a-b}\displaystyle\int\sqrt{x+b}dx$

=$\dfrac{1}{a-b}\times \dfrac{(x+a)^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\dfrac{1}{a-b}\times \dfrac{(x+b)^{\frac{1}{2}+1}}{\frac{1}{2}+1}$+c

=$\dfrac{1}{a-b} \left[\dfrac{2}{3}(x+a)^{3/2}-\dfrac{2}{3}(x+b)^{3/2}\right]$+c

=$\dfrac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]$+c


(ii) $\displaystyle\int \dfrac{1}{\sqrt{x+1}+\sqrt{x-1}}dx$

Sol :

=$\displaystyle\int \dfrac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x+1}+\sqrt{x-1})(\sqrt{x+1}-\sqrt{x-1})}dx$

=$\displaystyle\int \dfrac{\sqrt{x+1}-\sqrt{x-1}}{(\sqrt{x+1})^2-(\sqrt{x-1})^2}dx$

=$\displaystyle\int \dfrac{\sqrt{x+1}-\sqrt{x-1}}{x+1-x+1}dx$

=$\displaystyle\int \dfrac{\sqrt{x+1}-\sqrt{x-1}}{2}dx$

=$\dfrac{1}{2}\displaystyle\int \sqrt{x+1} dx - \dfrac{1}{2}\displaystyle\int \sqrt{x-1} dx$

=$\dfrac{1}{2}\times \dfrac{2}{3}(x+1)^{\frac{3}{2}}-\dfrac{1}{2} \times \dfrac{2}{3}(x-1)^{\frac{3}{2}}$+c

=$\dfrac{1}{3}(x+1)^{\frac{3}{2}}-\dfrac{1}{3}(x-1)^{\frac{3}{2}}$+c

=$\dfrac{1}{3} \left[(x+1)^{\frac{3}{2}}-(x-1)^{\frac{3}{2}}\right]$

Question 2
(i) $\displaystyle\int \dfrac{dx}{\sqrt{x+3}-\sqrt{x+2}}$
Sol :
=$\displaystyle\int \dfrac{\sqrt{x+3}+\sqrt{x+2}}{(\sqrt{x+3}-\sqrt{x+2})(\sqrt{x+3}+\sqrt{x+2})}dx$

=$\displaystyle \int \dfrac{\sqrt{x+3}+\sqrt{x+2}}{(\sqrt{x+3})^2-(\sqrt{x+2})^2}dx$

=$\displaystyle \int \dfrac{\sqrt{x+3}+\sqrt{x+2}}{x+3-x-2}dx$

=$\displaystyle \int \sqrt{x+3}+\sqrt{x+2}dx$

=$\displaystyle \int \sqrt{x+3}+\displaystyle \int\sqrt{x+2}dx$

=$\dfrac{2}{3}(x+3)^{\frac{3}{2}}+\dfrac{2}{3}(x+2)^{\frac{3}{2}}$+c

=$\dfrac{2}{3}\left[(x+3)^{\frac{3}{2}}+(x+2)^{\frac{3}{2}}\right]$+c


(ii) $\displaystyle\int \dfrac{dx}{\sqrt{x+1}+\sqrt{x+2}}$
Sol :
=$\displaystyle\int \dfrac{\sqrt{x+1}-\sqrt{x+2}}{(\sqrt{x+1}+\sqrt{x+2})(\sqrt{x+1}-\sqrt{x+2})}dx$

=$\displaystyle\int \dfrac{\sqrt{x+1}-\sqrt{x+2}}{(\sqrt{x+1})^2-(\sqrt{x+2})^2} dx$

=$\displaystyle\int \dfrac{\sqrt{x+1}-\sqrt{x+2}}{x+1-x-2} dx$

=$\displaystyle\int \dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}dx$

=$\displaystyle\int \sqrt{x+2}-\sqrt{x+1}dx$

=$\displaystyle\int \sqrt{x+2}dx-\displaystyle\int \sqrt{x+1}dx$

=$\dfrac{2}{3}(x+2)^{\frac{3}{2}}-\dfrac{2}{3}(x+1)^{\frac{3}{2}}$+c

=$\dfrac{2}{3} \left[(x+2)^{\frac{3}{2}}-(x+1)^{\frac{3}{2}}\right]$+c


Question 3
(i) $\displaystyle\int \dfrac{dx}{\sqrt{5+3x}+\sqrt{4+3x}}$
Sol :
=$\displaystyle\int \dfrac{\sqrt{5+3x}-\sqrt{4+3x}}{(\sqrt{5+3x}+\sqrt{4+3x})(\sqrt{5+3x}-\sqrt{4+3x})}dx$

=$\displaystyle\int \dfrac{\sqrt{5+3x}-\sqrt{4+3x}}{(\sqrt{5+3x})^2-(\sqrt{4+3x})^2}dx$

=$\displaystyle\int \dfrac{\sqrt{5+3x}-\sqrt{4+3x}}{5+3x-4-3x}$

=$\displaystyle\int (\sqrt{5+3x}-\sqrt{4+3x})dx $

=$\displaystyle\int \sqrt{5+3x}dx-\displaystyle\int \sqrt{4+3x}dx $

=$\dfrac{2}{3}.\dfrac{(5+3x)^{\frac{3}{2}}}{3}-\dfrac{2}{3}.\dfrac{(4+3x)^{\frac{3}{2}}}{3}$+c

=$\dfrac{2}{9}(5+3x)^{\frac{3}{2}}-\dfrac{2}{9}(4+3x)^{\frac{3}{2}}$+c

=$\dfrac{2}{9}\left[(5+3x)^{\frac{3}{2}}-(4+3x)^{\frac{3}{2}}\right]$+c


(ii) $\displaystyle\int \dfrac{dx}{\sqrt{1-2x}+\sqrt{3-2x}}$
Sol :
=$\displaystyle\int \dfrac{\sqrt{1-2x}-\sqrt{3-2x}}{(\sqrt{1-2x}+\sqrt{3-2x})(\sqrt{1-2x}-\sqrt{3-2x})}dx$

=$\displaystyle\int \dfrac{\sqrt{1-2x}-\sqrt{3-2x}}{(\sqrt{1-2x})^2-(\sqrt{3-2x})^2}dx$

=$\displaystyle\int \dfrac{\sqrt{1-2x}-\sqrt{3-2x}}{1-2x-3+2x}dx$

=$\displaystyle\int \dfrac{\sqrt{1-2x}-\sqrt{3-2x}}{-2}dx$

=$-\dfrac{1}{2}\displaystyle\int (1-2x)^{\frac{1}{2}} dx+\dfrac{1}{2}\displaystyle\int (3-2x)^{\frac{1}{2}}dx$

=$-\dfrac{1}{2}\times \dfrac{2}{3} \dfrac{(1-2x)^{\frac{3}{2}}}{-2}+\dfrac{1}{2}\times \dfrac{2}{3}\times \dfrac{(3-2x)^{\frac{3}{2}}}{-2}$+c

=$\dfrac{1}{6}(1-2x)^{\frac{3}{2}}-\dfrac{1}{6}(3-2x)^{\frac{3}{2}}$+c

=$\dfrac{1}{6}\left[(1-2x)^{\frac{3}{2}}-(3-2x)^{\frac{3}{2}}\right]$+c


Question 4
(i) $\displaystyle \int \sqrt{ax+b} dx$
Sol :
Given, $I=\displaystyle \int \sqrt{ax+b} dx$

=$\displaystyle\int (ax+b)^{\frac{1}{2}}dx$

=$\dfrac{(ax+b)^{\frac{1}{2}}}{\left(\frac{1}{2}+1\right)a}$+c

=$\dfrac{2}{3a}(ax+b)^{\frac{3}{2}}$+c


(ii) $\displaystyle \int e^{2x+3}dx$
Sol :

$I=\displaystyle \int e^{2x+3}dx$

=$\dfrac{e^{2x+3}}{2}$+c

=$\dfrac{1}{2}e^{2x+3}$+c


Question 5
$\int \frac{x}{\sqrt{1-2 x}} d x$
Sol :
let $x=a(1-2 x)+b \Rightarrow x=a-2 a x+b$

$x=-2 a x, a+b=0$

$a=\frac{-1}{2} \quad b=\frac{1}{2}$

$x=-\frac{1}{2}(1-2 x)+\frac{1}{2}$

$\therefore A / Q, \quad \int \frac{x}{\sqrt{1-2 x}} d x=\int \frac{-\frac{1}{2}(1-2 x)+\frac{1}{2}}{\sqrt{1-2 x}} d x$

$=\frac{-1}{2} \int \frac{(1-2 x)}{\sqrt{1-2 x}} d x+\frac{1}{2} \int \frac{1}{\sqrt{1-2 x}} d x$

$=-\frac{1}{2} \int \sqrt{1-2 x} d x+\frac{1}{2} \int \frac{1}{\sqrt{1-2 x}} d x$

$=-\frac{1}{2} \cdot \frac{2}{3}\left.\frac{1-2 x^{\frac{3}{2}}}{-x^{2}}+\frac{1}{2} \cdot \frac{(1-2 x)^{-\frac{1}{2}+1}}{\left(-\frac{1}{2}+1\right)(-2)}+c\right.$


$=\frac{1}{6}(1-2 x)^{\frac{3}{2}} \cdot-\frac{1}{2}(1-2 x)^{-\frac{1}{2}(-1)(-2)}$

$=\frac{1}{6}(1-2 x) \cdot \sqrt{1-2 x}-\frac{1}{2} \sqrt{1-2 x}+c$

$=\sqrt{1-2 x}\left(\frac{1-2 x}{6}-\frac{1}{2}\right)+c$

$=\sqrt{1-2 x}\left(\frac{1-2 x-3}{6}\right)+c$

$=\sqrt{1-2} x\left(\frac{-2-2 x}{6}\right)$

$=-\frac{2}{6}(1+x) \cdot \sqrt{1-2 x}$

$=-\frac{1}{3} \cdot(1+x) \sqrt{1-2 x}+c$


Question 6
$\int(x+2) \sqrt{x-3} d x$
Sol :
$=\int(x-3) \cdot \sqrt{x-3} d x+5 \int \sqrt{x-3}$

$=\int(x-3)^{\frac{2}{2}} d x+5 \int(x-3)^{\frac{2}{2}} d x$

$=\frac{(x-3)^{\frac{3}{2}+1}}{\frac{3}{2}+1}+5.\frac{(x-3)^{\frac{1}{2}}+c}{\frac{1}{2}+1}+c$

$=\frac{2}{5} \cdot(x-3)^{\frac{5}{2}}+\frac{10}{3} \cdot(x-3)^{\frac{3}{2}}+c$

$=\frac{2}{5} \cdot(x-3)^{\frac{3}{2}} \cdot(x-3)+\frac{10}{3} \cdot(x-3)^{\frac{3}{2}}+c$

$(x-3)^{\frac{3}{2}}\left[\frac{2}{5}(x-3)+\frac{10}{3}\right]+c$

$=(x-3)^{\frac{1}{2}}\left[\frac{2 x-6}{5}+\frac{10}{3}\right]+c$

$=(x-3)^{3 / 2}\left[\frac{6 x-18+5 0}{15}\right]$

$=(x-3)^{\frac{3}{2}}\left[\frac{6 x+38}{15}\right]+c$

$=\frac{2}{15}(3 x+16)(x-3)^{\frac{3}{2}}+c$



Question 7
$\int \frac{x^{2}-1}{x^{2}+1} d x=\int \frac{x^{2}+1-2}{x^{2}+1} d x$
Sol :
$=\int \frac{x^{2}+1}{x^{2}+1} d x-2 \int \frac{1}{x^{2}+1} d x$

$=\int d x-2 \int \frac{1}{x^{2}+1} d x$

$=x-2 \tan ^{-1} x+c$

Question 8
(i) $\int \sin ^{2} x d x=\int \frac{2 \sin ^{2} x}{2} d x=\int \frac{1-\cos 2 x}{2} d x$
Sol :
$\left[2 \sin ^{2} x=1-\cos 2 x\right]$

$=\frac{1}{2} \int d x-\frac{1}{2} \int \cos 2 x d x$

$=\frac{1}{2} \cdot x-\frac{1}{2} \sin 2 x+c$


$=\frac{x}{2}-\frac{\sin 2 x}{2}+c$

(ii) $\int \sin ^{2}(2 x+5) d x=\int \frac{2 \sin ^{2}(2 x+5)}{2} d x$
Sol :
$\int \frac{1-\cos 2(2 x+5)}{2} d x$

$=\frac{1}{2} \int d x-\frac{1}{2} \int \cos (4 x+10)$

$=\frac{1}{2} \cdot x-\frac{1}{2} \frac{\sin (4 x+10)}{4}+c$

$=\frac{x}{2}-\frac{1}{8} \sin (4 x+10)+c$

$=\frac{1}{2}\left[\dfrac{x-\sin (4 x+10)}{4}\right]+0$

(iii) $\int \cos ^{2} x d x$
Sol :
$=\int \frac{2 \cos ^{2} x d x}{2}$

$=\int \frac{1+\cos 2 x}{2} d x$

$=\frac{1}{2} \int d x+\frac{1}{2} \int \cos 2 x d x$

$=\frac{1}{2} \cdot x+\frac{1}{2} \cdot \frac{\sin 2 x}{2}+c$

$=\frac{1}{2}\left[x+\frac{\sin 2 x}{2}\right]+c$


Question 9
(i) $\int \sin ^{3}(2 x+1) d x$
Sol :
$=\int \frac{3 \sin (2 x+1)-\sin (6 x+3)}{4} d x$

$\left[\because \sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4}\right]$

$=\frac{3}{4} \int \sin (2 x+1) d x-\frac{1}{4} \int \sin (6 x+3) d x$

$=\frac{3}{4}\left(\frac{-\cos (2 x+1)}{2}\right)-\frac{1}{4}\left(\frac{-\cos (6 x+3)}{6}\right)+c$

$=-\frac{3}{8} \cdot \cos (2 x+1)+\frac{1}{24} \cos (6 x+3)+c$

$=\frac{1}{24} \cos (6 x+3)-\frac{3}{8} \cos (2 x+1)+c$

$=\frac{1}{24}[\cos (6 x+3)-9 \cos (2 x+1)]+c$

(ii) $\int \sin ^{3} x \cdot \cos ^{3} x d x$
Sol :
$=\int \frac{1}{8} \sin ^{3} 2 x d x$

$=\frac{1}{8} \int \sin ^{3} 2 x d x$

$=\frac{1}{8} \int \frac{3 \sin 2 x-\sin 6 x}{4} d x$

$=\frac{1}{8}\left[\frac{3}{4} \int \sin 2 x d x-\frac{1}{4} \int \sin 6 x d x\right]$

$=\frac{1}{8}\left[\frac{3}{4} \cdot\left(\frac{-\cos 2 x}{2}\right)-\frac{1}{4}\left(\frac{-\cos 6 x}{6}\right)\right]+c$

$\frac{1}{8}\left[-\frac{3 \cos 2 x}{8}+\frac{1}{24} \cos 6 x\right]_{+}^{6}$

$\frac{1}{8}\left[\frac{\cos 6 x}{24}-\frac{3 \cos 2 x}{8}\right]+c$

$=\frac{1}{8}\left[\frac{\cos 6 x-9 \cos 2 x}{24}\right]+c$

$=\frac{1}{192}[-9 \cos 2 x+\cos 6 x]+c$



Question 10

(i)$\int \tan ^{2} x d x=\int \sec ^{2} x-1 d x=\int \sec ^{2} x d x-\int d x$
Sol:
$=\tan x-x+c$

(ii)$\int \frac{\tan x}{\cot x} d x=\int \tan x \cdot \tan x d x=\int \tan ^{2} x d x$
Sol:
$=\int\left(8 e c^{2} x-1\right) d x=\int \sec ^{2} x d x-\int d x$

$=\tan x-x+c$

Question 11
$\int \sec ^{2} x \cdot \operatorname{cosec}^{2} x d x$
Sol :
$=\int \frac{1}{\sin ^{2} x \cdot \cos ^{2} x} d x$

$=4 \int \frac{1}{4 \sin ^{2} x.cos ^{2} x} d x$

$=4 \int \frac{1}{(2 \sin x \cdot \cos x)^{2}} d x$

$=4 \int \frac{1}{(\sin 2 x)^{2}}$

$=4 \int \frac{1}{\sin ^{2} 2 x}$

$=4 \int \operatorname{cosec}^{2} 2 x d x$

$=-4 \frac{\cot 2 x}{2}+c$

$=-2 \cot 2 x+c$

Question 12
(i) $\int\left(\cot ^{2} 2 x+\tan ^{2} 3 x\right) d x$

Sol: $\int\left(\operatorname{cosec}^{2} 2 x-1+\sec ^{2} 3 x-1\right) d x$

$\int \operatorname{cosec}^{2} 2 x d x+\int \sec ^{2} 8 x d x-2 \int d x$

$\frac{\cot 2 x}{2}+\tan \dfrac{3x}{3}-2x+c$

(ii) $\int(\tan x+\cot x)^{2} d x$

Sol: $=\int\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)^{2} d x$

$\int\left(\frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cdot \cos x}\right)^{2} d x$

$=\int \frac{4\times1}{4 \sin ^{2} x \cdot \cos ^{2} x} d x$

$=4 \int \frac{1}{(2 \sin x \cdot \cos x)^{2}} d x$

$=4 \int \frac{1}{(\sin 2 x)^{2}}=4 \int \frac{1}{\sin ^{2} 2 x}$

$=4 \int \operatorname{cosec}^{2} 2 x d x$

$4\left(\frac{-\cot 2 x}{2}\right)+c$

$=-2 \cot 2 x+c$

(ii) $\int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x$

Sol:  $=\int \frac{2 \cos ^{2} x-1+2 \sin ^{2} x}{\cos ^{2} x}$

$=\int \frac{2\left(\sin ^{2} x+\cos ^{2} x\right)-1}{\cos ^{2} x} d x$

$=\int \frac{2-1}{\cos ^{2} x} d x$

$=\int \frac{1dx}{\cos ^{2} x}$

$=\int \sec ^{2} x d x$

$=\tan x+c$

Question 14
$\int \cos ^{3} x d x$

Sol: $=\int \frac{3 \cos x+\cos 3 x}{4} d x$

$=\frac{3}{4} \int \cos x d x+\frac{1}{4} \int \cos 3 x d x$

$=\frac{3}{4} \sin x+\frac{1}{4} \sin 3 x+c$

$=\frac{3}{4} \sin x+\frac{\sin 3 x}{12}+c$


Question 15
$\int \sin ^{4} x d x$
Sol:
$=\int \sin ^{2} x \cdot \sin ^{2} x d x$

$=\int \frac{(1-\cos 2 x) \cdot(1-\cos 2 x)}{4}dx$

$=\frac{1}{4} \int(1-\cos 2 x)^{2}$

$=\frac{1}{4} \int\left(1+\cos ^{2} 2 x-2 \cos 2 x\right) d x$

$=\frac{1}{4} \int 1+\frac{1+\cos 4 x}{2}-2 \cos 2 x d x$

$=\frac{1}{4} \int\left(1+\frac{1}{2}+\frac{\cos 4 x}{2}-2 \cos 2 x\right) d x$

$=\frac{1}{4} \int \frac{3}{2}+\frac{\cos 4 x}{2}-2 \cos 2 x d x$

$=\frac{1}{4} \times \frac{3}{2} \int d x+\frac{1}{4 x} \int \cos 4 x-\frac{2}{4} \int \cos 2 x d x$

$\frac{3}{8} x+\frac{1}{8} \frac{\sin 4 x}{4}-\frac{2}{4} \cdot \frac{\sin 2 x}{2}+c$

$=\frac{3 x}{8}-\frac{\sin 2 x}{4}+\frac{\sin 4 x}{32}+c$


Question 16
$\int \sqrt{1+\cos 2 x} d x$
Sol :
$=\int \sqrt{2 \cos ^{2} x} d x$

$=\int \sqrt{2} \cdot \cos x d x$

Question 17
$\int \sqrt{1-\sin 2 x} d x$
Sol :
$\int \sqrt{1-\cos \left(\frac{x}{2}-2 x\right)} d x$

$=\int \sqrt{2 \sin ^{2}\left(\frac{\pi}{4}-\frac{8 x}{2}\right) d x}$

$=\int \sqrt{2} \cdot \sin \left(\frac{\pi}{4}-x\right) d x$

$=\sqrt{2} \cdot \frac{\left(-\cos \left(\frac{\pi}{4}-x\right)\right.}{-1}+c$

$=\sqrt{2} \cos \left(\frac{\pi}{4}-x\right)+c$

$=\sqrt{2} \cos \left(x-\frac{x}{4}\right)+c$


Question 18
$\int \sqrt{1+\sin \frac{x}{2}} d x$
Sol :
$\int \sqrt{1+\cos \left(\frac{\pi}{2}-\frac{x}{2}\right)} d x$

$=\int \sqrt{2 \cos ^{2}\left(\frac{\pi}{4}-\frac{x}{4}\right)} d x$

$=\int \sqrt{2} \cdot \cos \left(\frac{\pi}{4}-\frac{x}{4}\right) d x$

$=\sqrt{2} \int \cos \left(\frac{x}{4}-\frac{x}{4}\right) d x$

$=\frac{\sqrt{2} \cdot \sin \left(\frac{x}{4}-\frac{x}{4}\right)}{1 / 4}+c$

$=4 \sqrt{2} \cdot \sin \left(\frac{x}{4}-\frac{\pi}{4}\right)+c$


Question 19
$\int \sqrt{1+\sin k x} d x$
Sol :
$\int \sqrt{1+\sin k x} d x$

$=\int \sqrt{1+\cos \left(\frac{\pi}{2}-k x\right)} d x$

$=\int \sqrt{2 \cos ^{2}\left(\frac{\pi}{4}-\frac{k x}{2}\right)} d x$

$=\int \sqrt{2} \cdot \cos \left(\frac{x}{4}-\frac{k x}{2}\right) d x$

$=\sqrt{2} \int \cos \left(\frac{k x}{2}-\frac{\pi}{4}\right) d x=$

$=\sqrt{2} \cdot \frac{\sin \left(\frac{k x}{2}-\frac{x}{4}\right)}{k / 2}+c$

$=\frac{2 \sqrt{2}}{k} \cdot \sin \left(\frac{k x}{2}-\frac{\pi}{4}\right)+c$

Question 20
(i) $\int \frac{2}{1-\sin 2 x} d x$
Sol :
$=\int \frac{2}{1-\cos \left(\frac{x}{2}-2 x\right)} d x$

$=\int \frac{2}{2 \sin ^{2}\left(\frac{\pi}{4}-x\right)} d x$

$=\int \frac{1}{\sin ^{2}\left(\frac{\pi}{4}-x\right)} d x$

$=\int \operatorname{cosec}^{2}\left(\frac{\pi}{4}-x\right) d x$

$=-\frac{\cot \left(\frac{\pi}{4}-x\right)+c}{-1}$

$=\cot \left(\frac{\pi}{4}-x\right)+c$

(ii) $\int \frac{d x}{1-\sin 3 x}$
Sol :
$=\int \frac{1}{1-\cos \left(\frac{\pi}{2}-3 x\right)} d x$

$=\int \frac{1}{2 \sin ^{2}\left(\frac{\pi}{4}-\frac{3 x}{2}\right)} d x$

$=\frac{1}{2} \int \operatorname{cosec}^{2}\left(\frac{\pi}{4}-\frac{3 x}{2}\right) d x$

$=\frac{1}{2} \frac{\left(-\cot \left(\frac{\pi}{4}-\frac{3 x}{2}\right)\right.}{\frac{-3}{8}}+c$

$=\frac{1}{3} \cdot \cot \left(\frac{x}{4}-\frac{3 x}{2}\right)+c$

Question 21
(i) $\int \frac{1-\cos x}{1+\cos x} d x$
Sol :
$=\int \frac{2 \sin ^{2} \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} d x$

$=\int \tan ^{2} \frac{x}{2} d x$

$=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x$

$=\int \sec ^{2} \frac{x}{2} d x-\int d x$

$=\int\left(\sec ^{2} \frac{x}{2}-1\right) d x=\int \sec ^{2} \frac{x}{2} d x-\int d x$

$=2 \tan \frac{x}{2}-x+c$

(ii) $\int \frac{\cos x}{1+\cos x}-d x$
Sol :
$=\int \frac{1+\cos x-1}{1+\cos x} d x$

$=\int \frac{1+\cos x}{1+\cos x} d x-\int \frac{1}{1+\cos x} d x$

$=\int d x-\int \frac{1}{2 \cos \frac{x}{2}} d x$

$=x-\frac{1}{2} \int \sec ^{2} \frac{x}{2}$

$=x-\frac{1}{2} \tan \frac{x}{2} x z+c$

$=x-\tan \frac{x}{2}+c$


Question 22
$\int \frac{\sec x}{\sec x+\tan x} d x$
Sol :
$=\int \frac{1}{1+\sin x} d x$

$=\int \frac{1}{1+\cos \left(\frac{\pi}{2}-x\right)} d x$

$=\int \frac{1}{2 \cos ^{2}\left(\frac{x}{4}-\frac{x}{2}\right)} d x$

$=\frac{1}{2} \int \sec ^{2}\left(\frac{\pi}{4}-\frac{x}{2}\right) d x$

$=\frac{1}{2} \cdot \frac{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)}{\frac{-1}{2}}+c$

$=-\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)+c$

$=\tan \left(\frac{x}{2}- \frac{\pi}{4}\right)+c$


Question 23
$\int \frac{\operatorname{cosec} x}{\operatorname{cosec} x-\cot x} d x$
Sol :
$=\int \frac{1}{\sin x\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)} d x$

$=\int \frac{1}{\sin x\left(\frac{1-\cos x}{\sin x}\right)} d x$

$=\int \frac{1}{1-\cos x} d x$

$=\int \frac{1}{2 \sin ^{2} \frac{x}{2}} d x$

$=\frac{1}{2} \int \operatorname{cosec}^{2} \frac{x}{2} d x$

$=\frac{1}{2}\left(\frac{\left.-\cot \frac{x}{2}\right)}{1 / 2}+c\right.$

$=-\cot \frac{x}{2}+c$


Question 24
If $\int \sqrt{1+\sin 2 x} d x=\sqrt{2} \sin (x+a)+b$ ,  then find the value of a and be
Sol :
$\int \sqrt{1+\sin 2 x} d x=\sqrt{2} \sin (x+a)+b$..(i)

Now , $\int \sqrt{1+\sin 2 x} d x$

$=\int \sqrt{1+\cos \left(\frac{\pi}{2}-2 x\right)} d x$

$=\int \sqrt{2 \cos ^{2}\left(\frac{x}{4}-x\right)} d x$

$=\sqrt{2} \cdot \int \cos \left(\frac{\pi}{4}-x\right) d x$

$=\sqrt{2} \cdot \frac{\sin \left(\frac{\pi}{4}-x\right)}{-1}+c$

$=-\sqrt{2} \sin \left(\frac{\pi}{4}-x\right)+c$

$=\sqrt{2} \sin \left(x-\frac{\pi}{4}\right)+c$

$\int \sqrt{1+\sin 2 x} d x=\sqrt{2} \cdot \sin \left(x-\frac{x}{4}\right)+c$...(ii)
On comparing equations (i) and (ii) , we get
$a=\frac{-\pi}{4}$ , b=c=arbitrary constant


Question 25
(i) $\int \sin 3 x \cdot \sin 5 x d x$
Sol :
$=\int \frac{2}{2} \cdot \sin 5 x \cdot \sin 3 x d x$

$=\frac{1}{2} \int 2 \sin 5 x \cdot \sin 3 x d x$

$\left[\begin{array}{cc}\because 2 \sin A \cdot \sin B= \cos (A-B) -\cos (A+B)\end{array}\right]$

$=\frac{1}{2} \int \cos (5 x-3 x)-\cos (5 x+3 x) d x$

$=\frac{1}{2} \int(\cos 2 x-\cos 8 x) d x$

$=\frac{1}{2} \int \cos 2 x d x-\frac{1}{2} \int \cos 8 x d x$

$=\frac{1}{2} \frac{\sin 2 x}{2}-\frac{1}{2} \frac{\sin 8 x}{8}+c$

$=\frac{1}{4} \sin 2 x-\frac{\sin 8 x}{16}+c$

(ii) $\int \sin 4 x \cdot \sin 8 x d x$
Sol :
$=\int \frac{2}{2} \cdot \sin 8 x \cdot \sin 4 x d x$

$=\frac{1}{2} \int 2 \sin 8 x \cdot \sin 4 x d x$

$=\frac{1}{2} \int \cos (8 x-4 x)-\cos (8 x+4 x) d x$

$=\frac{1}{2} \int(\cos 4 x-\cos 12 x) d x$

$=\frac{1}{2} \int \cos 4 x d x-\frac{1}{2} \int \cos 12 x d x$

$=\frac{1}{2} \frac{\sin 4 x}{4}-\frac{1}{2} \cdot \frac{\sin 12 x}{12}+c$

$=\frac{1}{8} \sin 4 x-\frac{\sin 12 x}{24}+c$


Question 26
(i)$\int \sin 2 x \cdot \cos 3 x d x$
Sol :
$=\int \frac{2}{2} \cdot \cos 3 x \cdot \sin 2 x d x$

$=\frac{1}{2} \int 2 \cos 3 x \cdot \sin 2 x d x$

[∵2cosA.sinB=sin(A+B)-sin(A-B)]

$=\frac{1}{2} \int \sin (3 x+2 x)-\sin (3 x-2 x) d x$

$=\frac{1}{2} \int(\sin 5 x-\sin x) d x$

$=\frac{1}{2} \int \sin 5 x d x-\frac{1}{2} \int \sin x d x$

$=\frac{1}{2} \frac{(-\cos 5 x)}{5}-\frac{1}{2}(-\cos x)+c$

$-\frac{\cos 5 x}{10}+\frac{1}{2} \cos x+c$

(ii) $\int \sin 3 x \cdot \cos 4 x d x$
Sol :
$=\int \frac{2}{2} \cdot \cos 4 x \cdot \sin 3 x d x$

$=\frac{1}{2} \int \sin (4 x+3 x)-\sin (4 x-3 x) d x$

$=\frac{1}{2} \int \sin 7 x-\sin x d x$

$=\frac{1}{2} \int \sin 7 x d x-\frac{1}{2} \int \sin x d x$

$=\frac{1}{2}\left(-\frac{\cos 7 x}{7}\right)-\frac{1}{2} \cdot(-\cos x)+c$

$-\frac{\cos 7 x}{14}+\frac{1}{2} \cos x+c$

(iii) $\int \sin 4 x \cdot \cos 3 x d x$
Sol :
$=\frac{1}{2} \int 2 \sin 4 x \cdot \cos 3 x d x$

$=\frac{1}{2} \int \sin (4 x+3 x)+\sin (4 x-3 x) d x$

[2sinA.cosB=sin(A+B)+sin(A-B)]

$=\frac{1}{2} \int(\sin 7 x+\sin x) d x$

$=\frac{1}{2} \int \sin 7 x d x+\frac{1}{2} \int \sin x d x$

$=\frac{1}{2}\left(\frac{-\cos 7 x}{7}\right)+\frac{1}{2} \cdot(-\cos x)+c$

$=-\frac{\cos 7 x}{14}-\frac{\cos x}{2}+c$

$=\frac{1}{2}\left(-\cos x-\frac{\cos 7 x}{7}\right)+c$


(iv) $\int \cos 2 x \cdot \cos 4 x d x$
Sol :
$=\frac{1}{2} \int 2 \cos 4 x \cdot \cos 2 x d x$

$=\frac{1}{2} \int \cos (4 x-2 x)+\cos (4 x+2 x) d x$

$=\frac{1}{2} \int \cos 2 x+\cos 6 x d x$

$=\frac{1}{2} \int \cos 2 x d x+\frac{1}{2} \int \cos 6 x d x$

$=\frac{1}{2} \cdot \frac{\sin 2 x}{2}+\frac{1}{2} \cdot \frac{\sin 6 x}{6}+c$

$=\frac{1}{4} \sin 2 x+\frac{1}{12} \sin 6 x+c$


Question 27
(i) $\int \sin 3 x \cdot \sin \frac{x}{2} d x$
Sol :
$=\frac{1}{2} \int 2 \sin 3 x \cdot \sin \frac{x}{2} d x$

$=\frac{1}{2} \int \cos \left(3 x-\frac{x}{2}\right)-\cos \left(3 x+\frac{x}{2}\right) d x$

$=\frac{1}{2} \int\left(\cos \frac{5 x}{2}-\cos \frac{7 x}{2}\right) d x$

$=\frac{1}{2} \int \cos \frac{5 x}{2} d x-\frac{1}{2} \int \cos \frac{7 x}{2} d x$

$=\frac{1}{2} \sin \frac{5 x}{2} \cdot \frac{2}{5}-\frac{1}{2} \cdot \sin \frac{7 x}{2} \cdot \frac{2}{7}+c$
$=\frac{1 \sin 5 x}{5}-\frac{1}{7} \cdot \sin \frac{7 x}{2}+c$

(ii) $\int \sin x \cdot \sin 7 x d x$
Sol :
$=\frac{1}{2} \int 2 \sin 7 x \cdot \sin x d x$

$=\frac{1}{2} \int \cos (7 x-x)-\cos (7 x+x) d x$

$=\frac{1}{2} \int \cos 6 x-\cos 8 x d x$

$=\frac{1}{2} \int \cos 6 x d x-\frac{1}{2} \int \cos 8 x d x$

$=\frac{1}{2} \int \cos 6 x d x-\frac{1}{2} \int \cos 8 x d x$

$=\frac{1}{2} \frac{\sin 6 x}{6}-\frac{1}{2} \cdot \frac{\sin 8 x}{8}+c$

$=\frac{\sin 6 x}{12}-\frac{\sin 8 x}{16}+c$

Question 28
(i) $\int \cos x \cdot \cos 2 x \cdot \cos 3 x d x$
Sol :
$=\frac{1}{2} \int \dot{2} \cos 3 x \cdot \cos 2 x \cdot \cos x d x$

$=\frac{1}{2} \int\{\cos (3 x-2 x)+\cos (3 x+2 x)\} \cdot \cos x d x$

$=\frac{1}{2} \int\{\cos x+\cos 5 x) \cdot \cos x d x$

$=\frac{1}{2} \int \cos x \cdot \cos x d x+\frac{1}{2} \int \cos 5 x \cdot \cos x d x$

$\frac{1}{2} \int\{\cos x+\cos 5 x) \cdot \cos x d x$

$\frac{1}{2} \int \cos x \cdot \cos x d x+\frac{1}{2} \int \cos 5 x \cdot \cos x d x$

$\frac{1}{2} \int \cos ^{2} x d x+\frac{1}{2} \cdot \frac{1}{2} \int 2 \cos 5 x \cdot \cos x d x$

$=\frac{1}{2}\times \frac{1}{2} \int 2 \cos ^{2} x d x+\frac{1}{4} \int \cos (5 x-x)+\cos (5 x+x) d x$

$=\frac{1}{4} \int 1+\cos 2 x d x+\frac{1}{4} \int \cos 4 x+\cos 6 x d x$

$\frac{1}{4} \int d x+\frac{1}{4} \int \cos 2 x d x+\frac{1}{4} \int \cos 4 x d x+\frac{1}{4} \int \cos 6 x d x$

$=\frac{x}{4}+\frac{1}{4} \cdot \frac{\sin 2 x}{2}+\frac{1}{4} \frac{\sin 4 x}{4}+\frac{1}{4} \cdot \frac{\sin 6 x}{6}$

$\frac{x}{4}+\frac{1}{8} \sin 2 x+\frac{1}{16} \sin 4 x+\frac{1}{24} \sin 6 x+c$

$
\frac{1}{4}\left(x+\frac{\sin 2 x}{2}+\frac{\sin 4 x}{4}+\frac{\sin 6 x}{6}\right)+c
$

$=\frac{1}{4}\left(x+\frac{\sin 6 x}{6}+\frac{\sin y x}{4}+\frac{\sin 2 x}{2}\right)+c$

(ii) $\int \cos 2 x \cdot \cos 4 x \cdot \cos 6 x d x$\
Sol :
$=\frac{1}{2} \int 2 \cos 6 x \cdot \cos 4 x \cdot \cos 2 x d x$

$\frac{1}{2} \int\{\cos (6 x-4 x)+\cos (6 x+4 x)\} \cdot \cos 2 x d x$

$=\frac{1}{2} \int\{\cos 2 x+\cos 10x \}\cdot \cos 2 x d x$

$\frac{1}{2} \int \cos 2 x \cdot \cos 2 x d x+\frac{1}{2} \int \cos 2 x \cdot \cos 10 x d x$

$=\frac{1}{2 \times 2} \int 2 \cos ^{2} 2 x d x+\frac{1}{2 \times 2} \int 2 \cos 10 x \cdot \cos 2 x d x$

$=\frac{1}{4} \int 1+\cos 2 \cdot 2 x d x+\frac{1}{4} \int \cos (10 x-2 x)+\cos(10x+2x)dx$

$=\frac{1}{4} \int d x+\frac{1}{4} \int \cos 4 x d x+\frac{1}{4} \int \cos 8 x d x+\frac{1}{4} \int \cos 12 x d x$

$=\frac{x}{4}+\frac{\sin 4 x}{4 \times 4}+\frac{1}{4} \cdot \frac{\sin 8 x}{8}+\frac{1}{4} \times \frac{\sin 12 x}{12}+c$

$=\frac{1}{4}\left[x+\frac{\sin 4 x}{4}+\frac{\sin 8 x}{8}+\frac{\sin 12 x}{12}\right]+c$

$\left.=\frac{1}{4} \int x+\frac{\sin 12 x}{12}+\frac{\sin 8 x}{8}+\frac{\sin 4 x}{4}\right]+c$

Question 29
$\int \sin ^{2} x \cdot \cos 2 x d x$
Sol :
$=\int \frac{(1-\cos 2 x) \cdot \cos 2 x}{2} \cdot dx$

$=\frac{1}{2} \int \cos 2 x d x-\frac{1}{2} \int \cos ^{2} 2 x d x$

$=\frac{1}{2} \frac{\sin 2 x}{2}-\frac{1}{2 \times 2} \int 2 \cos ^{2} 2 x d x$

$=\frac{1}{2} \frac{\sin 2 x}{2}-\frac{1}{2 \times 2} \int 2 \cos ^{2} 2 x d x$

$\frac{\sin 2 x}{4}-\frac{1}{4} \int 1+\cos 4 x d x$

$=\frac{\sin 2 x}{4}-\frac{x}{4}-\frac{1}{4} \frac{\sin 4 x}{4}+c$

$=\frac{1}{4}\left(\sin 2 x-x-\frac{\sin 4 x}{4}\right)+c$

Question 30
(i) $
\int \frac{1-\cos 2 x}{1+\cos 2 x} d x
$
Sol:
$=\int \frac{2 \sin ^{2} x}{2 \cos ^{2} x} d x$

$=\int \tan ^{2} x d x$

$=\int\left(\sec ^{2} x-1\right) d x$

$=\int \sec ^{2} x d x-\int d x$

=tanx-x+c

(ii) $\int \frac{\cos 2 x}{\cos ^{2} x \cdot \sin ^{2} x} d x$
Sol :
$=\int \frac{\cos ^{2} x-\sin ^{2} x}{\cos ^{2} x \cdot \sin ^{2} x} d x$

$=\int \frac{\cos ^{2} x}{\cos ^{2} x \cdot \sin ^{2} x}-\int \frac{\sin ^{2} x}{\cos ^{2} x \cdot \sin ^{2} x} d x$

$=\int \operatorname{cosec}^{2} x d x-\int \sec ^{2} x d x$

=-cotx-tanx+c

(iii) $\int \frac{4-5 \cos x}{\sin ^{2} x} d x$
Sol :
$=\int \frac{4}{\sin ^{2} x} d x-5 \int \frac{\cos x}{\sin x \cdot \sin x} d x$

$=4 \int \operatorname{cosec}^{2} x d x-5 \int \cot x \cdot \operatorname{cosec} x d x$

=-4cotx-5(-cosecx)+c
=4cotx+5cosecx+c

(iv) $\int \frac{3 \cos x+4}{\sin ^{2} x} d x$
Sol :
$=3 \int \frac{\cos x}{\sin x \cdot \sin x}+4 \int \frac{1}{\sin ^{2} x} d x$

$=3 \int \cot x \cdot \operatorname{cosec} x d x+4 \int \operatorname{cosec}^{2} x d x$

=3(-cosecx)-4cotx+c

=-3cosecx-4cotx+c

Question 31
$\int \sin (a x+b) \cdot \cos (a x+b) d x$
Sol :
$\frac{1}{2} \int 2 \sin (a x+b) \cdot \cos (a x+b) d x$

$=\frac{1}{2} \int \sin (2 a x+2 b) d x$

$=\frac{1}{2} \frac{(-\cos (2 a x+2 b)}{2 a}$

$=-\frac{1}{4 a}[\cos (2 a x+2 b)]+c$

Question 32
(i) $\int \sec ^{2}(7-4 x) d x$
Sol :
$=\frac{\tan (7-4 x)}{-4}+c$

$=-\frac{1}{4} \tan (7-4 x)+c$

(ii) $\int \tan ^{2}(2 x-3) d x$
Sol :
$=\int \sec ^{2}(2 x-3)-1 d x$

$=\int \sec ^{2}(2 x-3) d x-\int d x$

$=\frac{\tan (2 x-3)}{2}-x+c$

Question 33
(i) $\int \sin ^{-1} \cos x d x$
Sol :
$=\int \sin ^{-1} \cdot \sin \left(\frac{\pi}{2}-x\right) d x$

$=\int \frac{\pi}{2}-x d x$

$=\frac{\pi}{2} x-\frac{x^{2}}{2}+c$

(ii) $\int \tan ^{-1} \sqrt{\frac{1-\cos 2 x}{1+\cos 2 x}} d x$
Sol :
$=\int \tan ^{-1} \sqrt{\frac{2 \sin ^{2} x}{2 \cos ^{2} x}} d x$

$=\int \tan ^{-1} \cdot \tan x d x$

$=\int x d x=\frac{x^{2}}{2}+c$

(iii) $\int \cos ^{-1}\left(\frac{1-\tan ^{2} x}{1+\tan ^{2} x}\right) d x$
Sol :
$=\int \cos ^{-1} \cos 2 x d x$

$=\int 2 x d x$

$\left[ \because \cos 2 x=\dfrac{1-\tan ^{2} x}{1+\tan ^{2} x}\right]$

$=2 \cdot \frac{x^{2}}{2}+c$

=x²+c

(iv) $\int \tan ^{-1}\left(\frac{\sin 2 x}{1+\cos 2 x}\right) d x$
Sol :
$=\int \tan ^{-1}\left(\frac{\frac{2 \tan x}{1+\tan ^{2} x}}{1+\frac{1-\tan ^{2} x}{1+\tan ^{2} x}}\right) d x$

$=\int \tan ^{-1}\left(\frac{2\tan x}{2}\right) d x$

$=\int \tan ^{-1} \tan x d x$

$=\int x d x$

$=\frac{x^{2}}{2}+c$

(v) $\int\tan ^{-1}(\sec x+\tan x) d x$
Sol :
=$\int \tan ^{-1}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right) dx$

$=\int \tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) d x$

$\int \tan ^{-1}\left(\frac{1+\frac{2 \tan x}{1+\tan ^{2} x}}{\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}}\right) d x$

$=\int \tan ^{-1}\dfrac{\left(\frac{1+\tan \frac{x}{2}+2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)}{\left(\dfrac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}\right)} d x$

$=\int \tan ^{-1} \frac{\left(1+\tan \frac{x}{2}\right)^2}{\left(1-\tan \frac{x}{2}\right)(1+\tan \frac{x}{2})} d x$

$=\int \tan ^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right) d x$

$=\int \tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan \frac{x}{2}}{1-\tan \frac{\pi}{4} \cdot \tan x}\right) d x$

$=\int \tan ^{-1} \cdot \tan \left(\frac{\pi}{4}+\frac{x}{2}\right) d x$

$=\int \tan ^{-1} \cdot \tan \left(\frac{\pi}{4}+\frac{x}{2}\right) d x$

$=\int \frac{x}{4}+\frac{x}{2} d x$

$=\int \frac{\pi}{4} d x+\int \frac{x}{2} d x$

$=\frac{\pi}{4} x+\frac{x^{2}}{2 \cdot 2}+c$

$=\frac{\pi}{4} x+\frac{x^{2}}{4}+c$

Question 34
$\int e^{3 x-1} d x$
Sol :
$=\frac{e^{3 x-1}}{3}+c$

$=\frac{1}{3} \cdot e^{3 x-1}+c$

Question 35
$\int \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}} d x$
Sol :
$=\int \frac{e^{\log x^5}-e^{\log x^{4}}}{e^{\log x^{3}}-e^{\log x^{2}}} d x$

$\left(\because m \log x=\log x^{m}\right)$

$=\int \frac{x^{5}-x^{4}}{x^{3}-x^{2}} d x$

$\left(\because e^{\log a}=a\right)$

$=\int \frac{x^{4}(x-1)}{x^{2}(x-1)} d x$

$=\int x^{2} d x=\frac{x^{3}}{3}+c$

Question 36
$\int \frac{e^{\log \sqrt{x}}}{x} d x$
Sol :
$=\int \frac{\sqrt{x}}{x} d x$ $\left(\because e^{\log \sqrt{x}}=\sqrt{x}\right)$

$=\int \frac{\sqrt{x}}{\sqrt{x} \cdot \sqrt{x}} d x$

$=\int \frac{1}{\sqrt{x}} d x$

$=\int x^{-\frac{1}{2}} d x$

$=\frac{x^{-\frac{1}{2}+1}}{\frac{-1}{2}+1}+c$

$=2 \cdot x^{\frac{1}{2}}+c=2 \sqrt{x}+c$