KC Sinha Solution Class 12 Chapter 19 Indefinite Integrals (अनिश्चित समाकल) Exercise 19.14

 Exercise 19.14

Question 1

$\int \frac{\cos xdx}{1+9\sin^2x}$

Sol :

Given : $I=\int \frac{\cos xdx}{1+9\sin^2 x}$

$=\int \frac{\cos xdx}{(1)^2+(3\sin x )^2}$

3sinx=z then 3cosxdx=dz

$\cos xdx=\frac{1}{3}dz$

Now $I=\int \frac{\cos xdx}{(1)^2+(3\sin x)^2}$

$=\int \frac{1}{3(1+z^2)}dx$

$=\frac{1}{3}\int \frac{1}{1+z^2}dz$

$=\frac{1}{3}\tan^{-1}z+C$

$=\frac{1}{3}\tan^{-1}(3\sin x)+C$

Question 2

$\int \frac{1}{\sin ^4 x+\sin^2 c.\cos ^2 x+\cos^4 x}dx$

Sol :

Given: $I=\int \frac{1}{\sin ^4 x+\sin^2 c.\cos ^2 x+\cos^4 x}dx$

$=\int \dfrac{1}{\frac{\cos ^4 x}{\cos ^4 x}\left(\sin^4 x+\sin^2x.\cos^2 x+\cos ^4 x\right)}dx$

$=\int \dfrac{\sec ^4 x}{\frac{\sin^4 x}{\cos ^4 x}+\frac{\sin^2 x.\cos ^2 x}{\cos ^2 x.\cos ^2 x}+\frac{\cos ^2 x}{\cos ^4 x}}dx$

$=\iint \frac{\sec^2 x.\sec^2 xdx}{\tan ^4 x+\tan^{2}x+1}$

$=\int \frac{(1+\tan^2 x)\sec^2 xdx}{\tan ^4 x+\tan^2 x+1}$

Let tanx=z then sec²xdx=dz

Now $I\int \frac{(1+z^2)dz}{z^4+z^2+1}$

Dividing numerator and denominator by z²

we get $I=\int \dfrac{\frac{1}{z^2}+1}{z^2+1+\frac{1}{z^2}}dz$

$=\int \dfrac{\left(1+\frac{1}{z^2}\right)}{\left(z^2+\frac{1}{z^2}\right)}dz$

$\left[\therefore \left(z^2+\frac{1}{z^2}\right)=\left(z-\frac{1}{z}\right)^2+2z\times \frac{1}{z}\right]$

$=\int \dfrac{\left(1+\frac{1}{z^2}\right)dz}{\left(z-\frac{1}{z}\right)+2z\times \frac{1}{z}+1}$

$=\int \dfrac{\left(1+\frac{1}{z^2}\right)dz}{\left(z-\frac{1}{z}\right)^2+3}$

$=\int \dfrac{\left(1+\frac{1}{z^2}\right)dz}{\left(z-\frac{1}{z}\right)^2+(\sqrt{3})^2}$

Let $\left(z-\frac{1}{z}\right)=y$ then $\left(1+\frac{1}{z^2}\right)dz=dy$

Now $I=\int \frac{dy}{z^2+(\sqrt{3})^2}$

$=\frac{1}{\sqrt{3}}\tan^{-1}\frac{y}{\sqrt{3}}+C$

$=\frac{1}{\sqrt{3}}\tan^{-1}\dfrac{\left(z-\frac{1}{z}\right)}{\sqrt{3}}+C$

$=\frac{1}{\sqrt{3}}\tan^{-1}\dfrac{(z^2-1)}{\sqrt{3}z}+C$

$I=\frac{1}{\sqrt{3}}\tan^{-1}\frac{\tan^{2}x-1}{\sqrt{3}\tan x}+C$

Question 3

(i) $\int \frac{\cos 2x.\sin 2xdx}{\sqrt{9-\cos ^4 2x}}$

Sol :

$=\int \frac{\cos 2x .\sin 2xdx}{\sqrt{(3)^2-(\cos ^2 2x)^2}}$

Let cos²x=z then -2cos2x.sin2x×2dx=dz

-4cos2x.sin2xdx=dz

cos2x.sin2xdx$=-\frac{1}{4}dz$

Now $I=\int \frac{-1dz}{4\sqrt{(3)^2-(z)^2}}$

$=\frac{-1}{4}\int \frac{dz}{\sqrt{(3)^2-z^2}}$

$=\frac{-1}{4}\sin^{-1}\frac{z}{3}+C$

$=\frac{-1}{4}\sin ^{-1} \left(\frac{\cos^2 2x}{3}\right)+C$

(ii) $\int \frac{\cos xdx}{\sqrt{\sin^2 x-2\sin x -3}}$

Sol :

Let sinx=z then cosxdx=dz

Now $I=\int \frac{dz}{\sqrt{z^2-2z-3}}$

$=\int \frac{dz}{\sqrt{z^2-2z+1-1-3}}$

$=\int \frac{dz}{\sqrt{z^2-2z+1-4}}$

$=\int \frac{dz}{\sqrt{(z-1)^2-(2)^2}}$

$\therefore \int \frac{dx}{\sqrt{x^2-a^2}}=\log|x+\sqrt{x^2-a^2}|$

$=\log|(z-1)+\sqrt{(z-1)^2-(2)^2}+C|$

$=log|(\sin x-1)+\sqrt{(\sin x-1)^2-4}|+C$

$=\log|(\sin x-1)+\sqrt{\sin^2 x+1-\sin x-4}|+C$

$I=\log|(\sin x-1)+\sqrt{\sin^2 x-2\sin x-3}|+C$

(iii) $\int \frac{dx}{\sin^4 x+\cos ^4 x}$

Sol :

Given : $I=\int \frac{dx}{\sin^4 x+\cos ^4 x}$

Dividing numerator and denominator by cos²x

We get $I=\int \dfrac{dx}{\frac{\cos ^4 x}{\cos ^4x }(\sin^4 x+\cos ^4 x)}$

$=\int \frac{\sec^4 xdx}{\left(\frac{\sin^4 x}{\cos^4 x}+\frac{\cos ^4x}{\cos ^4 x}\right)}$

$=\int \frac{\sec^2 x.\sec^2x dx}{(\tan^4x+1)}$

$=\int \frac{(1+\tan^2x)\sec^2 xdx}{\tan^4 x+1}$

Let tanx=z then sec²xdx=dz

Now $=\int \frac{(1+z^2)dz}{z^4+1}$

Dividing numerator and denominator by z²

We get $I=\int \dfrac{\left(\frac{1}{z^2}+1\right)dz}{\left(z^2+\frac{1}{z^2}\right)}$

$=\int \dfrac{\left(1+\frac{1}{z^2}\right)dz}{\left(z-\frac{1}{z}\right)^2+2}$

Let $ \left(z-\frac{1}{z}\right)=y$ then $\left(1+\frac{1}{z^2}\right)dz=dy$

Now $I=\int \frac{dy}{y^2+2}$

$=\int \frac{dy}{y^2+(\sqrt{2})^2}$

$=\frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{y}{\sqrt{2}}\right)+C$

$=\frac{1}{\sqrt{2}}\tan^{-1}\dfrac{\left(z-\frac{1}{z}\right)}{\sqrt{2}}+C$

$=\dfrac{1}{\sqrt{2}}\tan^{-1}\left(\dfrac{z^2-1}{\sqrt{2}z}\right)+C$

$=\frac{1}{\sqrt{2}}\tan^{-1}\frac{(\tan^{-1}x-1)}{\sqrt{2}\tan x}+C$

(iv) $\int \frac{x^2+4}{x^4+16}dx$

Sol :

Given : $I=\int \frac{x^2+4}{x^4+16}dx$

Dividing numerator and denominator by x²

we get

$I=\int \dfrac{1+\frac{4}{x^2}}{x^2+\frac{16}{x^2}}dx$

$=\int \dfrac{\left(1+\frac{4}{x^2}\right)dx}{\left(x-\frac{4}{x}\right)^2+8}$

$=\int \dfrac{\left(1+\frac{4}{x^2}\right)dx}{\left(x-\frac{4}{x}\right)^2+\left(2\sqrt{2}^2\right)}$

Let $\left(x-\frac{4}{x}\right)=z$ then $\left(1+\frac{4}{x^2}\right)dx=dz$

Now $I=\int \frac{dz}{z^2+(2\sqrt{2})^2}$

$=\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{z}{2\sqrt{2}}\right)+C$

$=\frac{1}{2\sqrt{2}}\tan^{-1}\dfrac{\left(x-\frac{4}{x}\right)}{2\sqrt{2}}+C$

$=\frac{1}{2\sqrt{2}}\tan^{-1}\left(\frac{x^2-4}{2\sqrt{2}x}\right)+C$

Question 4

$\int \frac{\cos xdx}{\sin x+\cos x}$

Sol :

Let cosx=A(sinx+cosx)+B$\frac{d}{dx}$(sinx+cosx)

=cosx=A(sinx+cosx)+B(cosx-sinx)

=cosx=Asinx+Acosx+Bcosx-Bsinx

=cosx=(A-B)sinx+(A+B)cosx

Equating the coefficient of sinx and cosx we get

A-B=0

A=B

and A+B=1

⇒A+A=1

⇒2A=1

⇒$A=\frac{1}{2}$

⇒$B=\frac{1}{2}$

Now $I=\int \frac{\cos xdx}{\sin x+\cos x}$

$=\int \frac{A(\sin x+\cos x )+B(\cos x-\sin x)}{\sin x+\cos x}$

$=A\int \frac{\sin x+\cos xdx}{\sin x +\cos x}+B\int \frac{\cos x-\sin dx}{\sin x+\cos x}$

$=\frac{1}{2}\int dx+\frac{1}{2}\int \frac{dz}{z}$

$=\frac{1}{2}x+\frac{1}{2}\log|z|+C$

$=\frac{x}{2}+\frac{1}{2}\log|\sin x +\cos x|+C$

Question 5

$\int \frac{\cos xdx}{2\sin x+3\cos x}$

Sol :

Let cosx=A(2sinx+3cosx)+B$\frac{d}{dx}$(2sinx+3cosx)

=cosx=A(2sinx+3cosx)+B(2cosx-3sinx)

=cosx=2Asinx+3Acosx+2Bcosx-2Bsinx

=1.cosx=(2A-3B)sinx+(3A+2B)cosx

Equating the coefficient of sinx and cosx

We get

2A=3B=0 and 3A+2B=1

$\begin{array}{l|l}2A=3B&3.\frac{3}{2}B+2B=1\\\therefore A=\frac{3}{2}B& \frac{9B+4B}{2}=1\\\therefore A=\frac{3}{2}\times \frac{2}{13}& 13B^2=2\\A=\frac{3}{13}&B=\frac{2}{13}\end{array}$

Now $I=\int \frac{\cos xdx}{2\sin x +3\cos x}$

$=\int \frac{A(2\sin x+3\cos x)+B(2\cos x-3\sin x)}{2\sin x+3\cos x}$

$=A\frac{2\sin x+3 \cos xdx}{2\sin x+3\cos x}+B\int {2\cos x-3\sin x dx}{2\sin x+3\cos x}$

$=\frac{3}{13}\int dx+\frac{2}{13}\int \frac{dz}{z}$

$I=\frac{3}{13}x+\frac{2}{13}\log|z|+C$

$=\frac{3}{13}x+\frac{2}{13}\log|2\sin x+3\cos x|+C$

Question 6

$\int \frac{dx}{1+\tan x}$

Sol :

Given : $I=\int \frac{dx}{1+\tan x}$

$=\int \frac{dx}{1+\frac{\sin x }{\cos x}}$

$=\int \frac{\cos xdx}{\cos x+\sin x}$

$I=\frac{x}{2}+\frac{1}{2}\log|\cos x+\sin x|+C$

Question 7

$\int \frac{dx}{1+\cot x}$

Sol :

Given : $I=\int \frac{dx}{1+\cot x}$

$=\int \frac{dx}{1+\frac{\cos x}{\sin x}}$

$=\int \frac{\sin xdx}{\sin x +\cos x}$

Let sinx=A(sinx+cosx)+B$\frac{d}{dx}$(sinx+cosx)

=sinx=A(sinx+cosx)+B(cosx-sinx)

=sinx=Asinx+Acosx+Bcosx-Bsinx

=sinx=(A-B)sinx+(A+B)cosx

Equating the coefficient of sinx and cosx 

we get $\begin{array}{l|l}A-B=1&A+B=0\\-B-b+1&\therefore A=-B\\-2B=1&A=-\left(-\frac{1}{2}\right)=\frac{1}{2}\\B=\frac{-1}{2}&\end{array}$

Now $I=\int \frac{\sin xdx}{\sin x+cos x}$

$=\int \frac{A(\sin x+\cos x)+B(\cos x-\sin x)dx}{\sin x+\cos x}$

$=A\int \frac{\sin x+\cos xdx}{\sin x+\cos x}+B\int \frac{\cos x-\sin xdx}{\sin x+\cos x}$

$=A\int dx+B\frac{dz}{z}$

$I=\frac{1}{2}x-\frac{1}{2}\log|z|+C$

$I=\frac{x}{2}-\frac{1}{2}\log|\sin x+\cos x|+C$

Question 8

(i) $\int \frac{\sin x+2\cos xdx}{2\sin x+\cos x}$

Sol :

Let sinx+2cosx=A(2sinx+cosx)+B$\frac{d}{dx}$(2sinx+cosx)

=sinx+2cosx=A(2sinx+cosx)+B(2cosx-sinx)

=sinx+2cosx=2Asinx+Acosx+2Bcosx-Bsinx

=sinx+2cosx=(2A-B)sinx+(A+2B)cosx

Equating the coefficient of sinx and cosx

We get

$\begin{array}{l|l}2A-B=1&A+2B=2\\2A=1+B&\frac{1+B}{2}+2B=2\\A=\frac{1+B}{2}&\frac{1+B+4B}{2}=2\\ \therefore A=\frac{1+\frac{3}{5}}{2}=\frac{8}{10}&5B=4-1\\ \therefore A=\frac{4}{5}& B=\frac{3}{5}\end{array}$

Now $I=\int \frac{\sin x+2\cos xdx}{2\sin x+\cos x}$

$=\int \frac{A(2\sin x+\cos x)+B(2\cos x-\sin x)}{2\sin x+\cos x}$

$=A\int \frac{2\sin x+\cos xdx}{2\sin x+\cos x}+B\int \frac{2\cos x-\sin xdx}{2\sin x+\cos x }dx$

$=A\int dx+B\int \frac{dz}{z}$

=Ax+Blogz+C

$I=\frac{4}{5}x+\frac{3}{5}\log|2\sin x+\cos x|+C$

(ii) $\int \frac{\sin xdx}{3\sin x+5\cos x}$

Sol :

Let sinx=A(3sinx+5cosx)

Let sinx=A(3sinx+5cosx)+B$\frac{d}{dx}$(3sinx+5cosx)

=sinx=A(3sinx+5cosx)+B(3cosx-5sinx)

=sinx=3Asinx+5Acosx+3Bcosx-5Bsinx

=sinx=(3A-5B)sinx+(5A+3B)cosx

Equating the coefficient of sinx and cosx  

We get

$\begin{array}{l|l}3A-5B=1& 5A+3B=0\\\therefore 3\left(-\frac{3}{5}B\right)-5B=1&5A-3B\\-\frac{9B}{5}-5B=1&A=\frac{-3}{5}B\\\frac{-9B-25B}{5}=1& \therefore A=\frac{-3}{5}\times \frac{-5}{34}\\-34B=5&A=\frac{3}{34}\\\therefore B=-\frac{5}{34}&\end{array}$

Now $I=\int \frac{\sin xdx}{3\sin x+5\cos x}$

$=\int \frac{A(3\sin x+5\cos x)+B(3\cos x-5\sin x)}{3\sin x+5\cos x}$

$=A\int \frac{3\sin x+5\cos x dx}{3\sin x+5\cos x}+B\int \frac{3\cos x-5\cos xdx}{3\sin x+5\cos x}$

$=A\int dx+B\int \frac{dz}{z}$

$=\frac{3}{34}x-\frac{5}{34}\log|z|+C$

$=\frac{3}{34}x-\frac{5}{34}\log|3\sin x+5\cos x|+C$

Question 9

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

Sol :

Given : $I=\int \frac{dx}{1+\cos^2 x}$

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

$=\int \dfrac{\sec^2 xdx}{\frac{1}{\cos^2 x}+\frac{\cos^2 x}{\cos^2 x}}$

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

$=\int \frac{\sec^2 xdx}{1+\tan^2 x+1}$

$=\int \frac{\sec^2 xdx}{2+\tan^2 x}$

$=\int \frac{\sec^2 xdx}{(\sqrt{2})^2+(\tan x)}$

Let tanx=z then sec²xdx=dz

Now $I=\int \frac{\sec^2 xdx}{(\sqrt{2})^2+(\tan x)^2}$

$=\int \frac{dz}{(\sqrt{2})^2+z^2}$

$=\frac{1}{\sqrt{2}}\tan^{-1}+\frac{z}{\sqrt{2}}+C$

$=\frac{1}{\sqrt{2}}\tan^{-1}\frac{(\tan x)}{\sqrt{2}}+C$

Question 10

$\int \frac{dx}{4\sin^2x+9\cos^2 x}$

Sol :

Given : $I=\int \frac{dx}{4\sin^{2}x+9\cos^2 x}$

$=\dfrac{dx}{\frac{\cos^2 x}{\cos^2 x}(4\sin^2 x+9\cos^2 x)}$

$=\int \dfrac{\sec^2 xdx}{\left(\frac{4\sin^2 x}{\cos^2 x}+\frac{9\cos^2 x}{\cos^2 x}\right)}$

$=\int \frac{\sec^2 xdx}{4\tan^{2}x+9}$

Let tanx=z then sec²xdx=dz

Now $I=\int \frac{dz}{4z^2+9}$

$=\int \frac{dz}{(2z)^2+(3)^2}$

$\left[\therefore \int \frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}\right]$

$=\frac{1}{3}\tan^{-1}\left(\frac{2z}{3}\right)+C$

$=\frac{1}{3\times 2}\tan^{-1}\left(\frac{2}{3}\tan x\right)+C$

$I=\frac{1}{6}\tan^{-1}\left(\frac{2}{3}\tan x\right)+C$

Question 11

$\int \frac{dx}{(a\sin x+b\cos x)}$

Sol :

Given : $I=\int \frac{dx}{(a\sin x+b\cos x)}$

$=\int \frac{dx}{a^2\sin^2 x+b^2 \cos ^2 x+2ab\sin x. \cos x}$

$=\int \dfrac{dx}{\frac{\cos ^2 x}{\cos ^2 x}\left(a^2 \sin ^2 x+b^2 \cos ^2 x+2ab \sin x.\cos x\right)}$

$=\int \dfrac{\sec ^2 xdx}{\frac{a^2 \sin ^2 x}{\cos ^2 x}+\frac{b^2 \cos ^2 x}{\cos ^2 x}+\frac{2ab\sin x.\cos x}{\cos ^2 x}}$

$=\int \frac{\sec^2 xdx}{a^2 \tan^2 x+b^2 +2ab\tan x}$

Let tanx=z then sec²xdx=dz

Now $I=\int \frac{dz}{a^2z^2+b^2+2abz}$

$=\int \frac{dz}{(az+b)^2}$

$=\int (az+b)^{-2}dz$

$=\frac{(az+b)^{-2+1}}{(-2+1)a}+C$

$=\frac{(az+b)^{-1}}{-1\times a}+C$

$=\frac{-1}{a(az+b)}+C$

$=\frac{-1}{a(a\tan x+b)}+C$

Question 12

$\int \frac{dx}{5+4\sin x}$

Sol :

Given : $I=\int \frac{dx}{5+4\sin x}$

$\left[\therefore \sin x=\dfrac{2\tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}\right]$

$=\int 5+4\left(\dfrac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)$

$=\int \frac{1+\tan^{2}\frac{x}{2}dx}{5+5\tan^2 \frac{x}{2}+8\tan \frac{x}{2}}$

$=\int \frac{1+\tan^2 \frac{x}{2}dx}{5+5\tan^2 \frac{x}{2}+8\tan \frac{x}{2}}$

$=\int \frac{\sec^2 \frac{x}{2}}{5\tan^2 \frac{x}{2}+8\tan \frac{x}{2}+5}$

Let $\tan \frac{x}{2}=z$ then $\frac{1}{2}\sec^2 \frac{x}{2}=dz$

$\sec^2 \frac{x}{2}dx=2dz$

Now $I=\int \frac{2dz}{5z^2+8z+5}$

$=2\int \frac{dz}{5\left(z^2+\frac{8}{5}z+1\right)}$

$=\frac{2}{5}\int \dfrac{dz}{z^2+2.z.\frac{4}{5}+\frac{16}{25}-\frac{16}{25}+1}$

$=\frac{2}{5}\int \dfrac{dz}{\left(z+\frac{4}{5}\right)^2+\frac{9}{25}}$

$=\frac{2}{5}\int \frac{1dz}{\left(z+\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2}$

$=\frac{2}{5}\times \dfrac{1}{\frac{3}{5}} \tan ^{-1} \dfrac{\left(z+\frac{4}{5}\right)}{\frac{3}{5}}+C$

$=\frac{2}{5}\times \frac{5}{3}\tan^{-1} \dfrac{\left(5z+4\right)}{\frac{3}{5}\times 5}+C$

$=\frac{2}{5}\tan ^{-1}\left(\frac{5z+4}{3}\right)+C$

$=\frac{2}{3}\tan^{-1}\left(\frac{5\tan \frac{x}{2}+4}{3}\right)+C$

Question 13

$\int \frac{dx}{4+5\cos x}$

Sol :

Given : $I=\int \frac{dx}{4+5\cos x}$

$=\int \dfrac{dz}{4+5\left(\dfrac{1-\tan ^2\frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)}$

$=\int \frac{dx \sec^2 \frac{x}{2}}{4+4\tan ^2 \frac{x}{2}+5-5\tan ^2 \frac{x}{2}}$

$=\int \frac{\sec^2 \frac{x}{2}dx}{9-\tan ^2 \frac{x}{2}}$

Let $\tan \frac{x}{2}=z$ then $\frac{1}{2}\sec^2 \frac{x}{2}dx=dz$ 

$\therefore \sec^2 \frac{x}{2}dx=2dz$

Now $I=\int \frac{2dz}{9-z^2}$

$=\int \frac{2dz}{(3)^2-z^2}$

$=2\int \frac{dz}{(3)^2-(z)^2}$

$\left[\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|\right]$

$=2\times \frac{1}{2\times 3} \log\left|\frac{3+z}{3-z}\right|+C$

$=\frac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan \frac{x}{2}}\right|+C$