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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

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