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KC Sinha Solution Class 12 Chapter 19 Indefinite Integrals Exercise 19.11

Exercise 19.11

Question 1

\int e^{x}\left(\frac{1+x \log x}{x}\right) d x
Sol :
I=\int e^{x}\left(\frac{1+x \log x}{x}\right) d x=\int e^{x}\left(\frac{1}{x}+\frac{x \log x}{x}\right) d x

I=\int e^{x}\left(\frac{1}{x}+\log x\right) d x

=\int e^{x}\left(\log x+\frac{1}{x}\right) d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=logx

I=e^{x} F(x)+c=e^{x} \log x+c


Question 2

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

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=\tan ^{-1} x

I=e^{x} F(x)+c=e^{x} \tan ^{-1} x+c


Question 3

\int e^{x}(\sin x+\cos x) d x
Sol :
I=\int e^{x}(\sin x+\cos x) d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=sinx

I=e^{x} F(x)+c=e^{x} \sin x+c


Question 4

\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x
Sol :
I=\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x

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

I=\int e^{x}\left(\frac{1}{2 \sin ^{2} \frac{x}{2}}-\frac{2 \frac{\sin x}{2} \cdot \cos \frac{x}{2}}{2 \sin \frac{x}{2}}\right) d x

I=\int e^{x}\left(\frac{1}{2} \operatorname{cosec}^{2} \frac{x}{2}-\frac{\cot x}{2}\right) d x

=\int e^{x}\left(-\cot \frac{x}{2}+\frac{1}{2} cosec ^{2} x\right)dx

I=\int e^{x}\left(F(x)+F^{\prime}(x)\right) d x

where F(x)=-\cot \frac{x}{2}

I=e^{x} F(x)+c=e^{x}\left(-\cot \frac{x}{2}\right)+c

=-e^{x} \cot \frac{x}{2}+c


Question 5

\int e^{x}\left(\frac{1+x}{(2+x)^{2}}\right) d x
Sol :
I=\int e^{x} \frac{(1+x)}{(2+x)^{2}} d x=\int e^{x} \frac{(2+x-1)}{(2+x)^{2}} d x

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

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

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=\frac{1}{2+x}

I=e^{x} F(x)+c=\frac{e^{x}}{2+x}+c


Question 6

\int e^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x
Sol :
I=\int e^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x

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

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

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=\frac{1}{1+x^{2}}

I=e^{x} F(x)+c

=e^{x} \times \frac{1}{1+x^{2}}+c=\frac{e^{x}}{1+x^{2}}+c


Question 7

e^{x} \frac{\left(x^{3}+x+1\right)}{\left(1+x^{2}\right)^{3 / 2}} d x
Sol :
I=\int e^{x} \frac{\left(x^{3}+x+1\right)}{\left(1+x^{2}\right)^{3 / 2}} d x

=\int e^{x} \frac{\left(x\left(x^{2}+1\right)+1\right)}{\left(1+x^{2}\right)^{1 / 2}} d x

I=\int e^{x}\left(\frac{x\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{3 / 2}}+\frac{1}{\left(1+x^{2}\right)^{-3/ 2}}\right) d x

I=\int e^{x}\left(\frac{x}{\left(1+x^{2}\right)^{\frac{1}{2}}}+\frac{1}{\left(1+x^{2}\right)^{3 / 2}}\right) d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x\

where F(x)=\frac{x}{\left(1+x^{2}\right)^{\frac{1}{2}}} d x

I=e^{x} F(x)+c

=e^{x} \frac{ x}{\left(1+x^{2}\right) \frac{1}{2}}+c

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


Question 8

(i) \int e^{x}(\tan x-\log \cos x) d x
Sol :
I \cdot \int e^{x}(\tan x-\log \cos x) d x

I=\int e^{x}(-\log \cos x+\tan x) d x

I=\int e^{x}\left(F(x)+F^{\prime}(x)\right) d x

where F(x)=-\log \cos x

I=e^{x} F(x)+c=-e^{x} \log \cos x+c


(ii) \int e^{x}\left(\cot x-\operatorname{cosec}^{2} x\right) d x
Sol :
I=\int e^{x}\left(\cot x-\operatorname{cosec}^{2} x\right) d x

I=\int e^{x}\left[\cot x +\left(-cosec ^{2} x\right)\right] d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=cotx

I=e^{x} F(x)+c=e^{x} \cot x+c


(iii) \int e^{x}(\tan x+\log \sec x) d x
Sol :
I=\int e^{x}(\tan x+\log \sec x) d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where

F(x)=logsecx

I=e^{x} F(x)+c=e^{x} \cdot x \log \sec x+c


(iv) \int e^{x}\left(\frac{1+\sin x \cdot \cos x}{\cos ^{2} x}\right) d x
Sol :

I=\int e^{x}\left(\frac{1+\sin x \cdot \cos x}{\cos ^{2} x}\right) d x

=\int e^{x}\left(\frac{1}{\cos ^{2} x}+\frac{\sin x \cdot \cos x}{\cos ^{2} x}\right)dx

I=\int e^{x}\left(\sec ^{2} x+\tan x\right) d x

I=\int e^{x}\left(F(x)+F^{\prime}(x)\right] d x

where F(x)=tanx

I=e^{x} F(x)+c=e^{x} \tan x+c


(v) \int e^{x}\left(\frac{\sin x \cdot \cos x-1}{\sin ^{2} x}\right) d x
Sol :
I=\int e^{x}\left(\frac{\sin x \cdot \cos x}{\sin ^{2} x}-\frac{1}{\sin ^{2} x}\right) d x

I=\int e^{x}\left(\cot x-\operatorname{cosec}^{2} x\right) d x

I=e^{x} \cot x+c


(vi) \int e^{x}(\cot x+\log \sin x) d x
Sol :
I=\int e^{x}(\cot x+\log \sin x) d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=logsinx

I=e^{x} F(x)+c=e^{x} \log \sin x+c


(vii) \int e^{x} \sec x(1+\tan x) d x
Sol :
I=\int e^{x} \sec x(1+\tan x) d x

=\int e^{x}(\sec x+\sec x \cdot \tan x) d x

I=e^{x} \sec x+c


Question 9

\int e^{x} \frac{\left[1+\sqrt{1-x^{2}} \sin ^{-1} x\right]}{\sqrt{1-x^{2}}} d x
Sol :
I=\int e^{x} \frac{\left[1+\sqrt{1-x^{2}} \sin ^{-1} x\right]}{\sqrt{1-x^{2}}} d x

I=\int e^{x}\left(\frac{1}{\sqrt{1-x^{2}}}+\frac{\sqrt{1-x^{2}} \cdot \sin ^{-1} x}{\sqrt{1-x^{2}}}\right) d x

I=\int e^{x}\left(\sin ^{-1} x+\frac{1}{\sqrt{1-x^{2}}}\right) d x

I=e^{x} \sin ^{-1} x+c


Question 10

(i) \int e^{x}\left(\frac{1}{x^{2}}-\frac{2}{x^{3}}\right) d x
Sol :
I=\int e^{x}\left(\frac{1}{x^{2}}-\frac{2}{x^{3}}\right) d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=\frac{1}{x^{2}}

I=e^{x} F(x)+c=e^{x} \times \frac{1}{x^{2}}+c

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


(ii) \int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x
Sol :
I=\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right) d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=\frac{1}{x}

I=e^{x} F(x)+c=e^{x} x \frac{1}{x}+c

=\frac{e^{x}}{x}+c


(iii) \int e^{2x}\left(\frac{2 x-1}{4 x^{2}}\right) d x
Sol :
I=\int e^{2 x}\left(\frac{2 x-1}{4 x^{2}}\right) d x

putting z=2x , z^{2}=4x^{2}then dz=2dx \Rightarrow \frac{d z}{2}=d x

Now , I=\int e^{2}\left(\frac{z-1}{z^{2}}\right) \frac{d z}{2}

=\frac{1}{2} \int e^{z}\left(\frac{z}{z^{2}}-\frac{1}{z^{2}}\right) d z

I=\frac{1}{2} \int e^{2}\left(\frac{1}{z}-\frac{1}{z^{2}}\right) d z

I=\frac{1}{2} \cdot e^{z} \times \frac{1}{2}+c

=\frac{1}{2} \frac{e^{2 x}}{2 x}

=\frac{e^{2 x}}{4 x}+c


(iv) \int \frac{x-3}{(x+1)^{3}} \cdot e^{x} d x
Sol :
I=\int \frac{(x-3)}{(x-1)^{3}} e^{x} d x

=\int e^{x}\left(\frac{x-3}{(x-1)^{3}}\right) d x

I=\int e^{x} \frac{(x-1-2)}{(x-1)^{3}} d x

=\int e^{x}\left[\frac{(x-1)}{(x-1)^{2}}-\frac{2}{(x-1)^{3}}\right] d x

I=\int e^{x}\left(\frac{1}{(x-1)^{2}}-\frac{2}{(x-1)^{3}}\right) d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=\frac{1}{(x-1)^{2}}

I=e^{x} F(x)+c

=e^{x} \times \frac{1}{(x-1)^{2}}+c

=\frac{e^{x}}{(x-1)^{2}}+c


(v) \int \frac{(2-x) e^{x}}{(1-x)^{2}} d x
Sol :
I=\int \frac{2-x \cdot e^{x}}{(1-x)^{2}} d x

=\int \frac{(1-x+1) \cdot e^{x}}{(1-x)^{2}} d x

I=\int\left(\frac{(1-x)}{(1-x)^{2}}+\frac{1}{(1-x)^{2}}\right) \cdot e^{x} d x

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

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=\frac{1}{1-x}

I=e^{x} F(x)+c=e^{x} \frac{1}{1-x}+c

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


Question 11

\int e^{x}\left(\frac{2-\sin 2 x}{1-\cos 2 x}\right) d x
Sol :
I=\int e^{x}\left(\frac{2-2 \sin x \cdot \cos x}{2 \sin ^{2} x}\right) d x

I=\int e^{x}\left(\frac{2}{2 \sin ^{2} x}-\frac{2 \sin x \cdot \cos x}{2 \sin ^{2} x}\right) d x

I=\int e^{x}\left(\cos e c^{2} x-\cot x\right) d x

I=\int e^{x}\left(-\cot x+\operatorname{cosec}^{2} x\right) d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=-cotx

I=e^{x} F(x)+c=e^{x}(-\cot x)+c

=-e^{x} \cot x+c


Question 12

\int e^{x}\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) d x
Sol :
I=\int e^{x}\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) d x

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

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

=\int e^{x}\left(\sec ^{2} + \tan x\right) d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=tanx

I=e^{x} F(x)+c=e^{x} \tan x+c


Question 13

\int e^{2 x}(-\sin x+2 \cos x) d x
Sol :
putting 2x=z then dz=2dx \Rightarrow \frac{d z}{2}=d x

Now , I=\int e^{z}\left(-\sin \frac{z}{2}+2 \cos \frac{z}{2}\right) \frac{d z}{2}

I=\frac{1}{2} \int e^{2}\left(2 \cos \frac{z}{2}-\sin \frac{z}{2}\right) d z

I=\int e^{2}\left(\cos \frac{z}{2}-\frac{1}{2} \sin \frac{z}{2}\right) d z

I=\int e^{z}\left[F(z)+F^{\prime}(z)\right] d z

where F(z)=\cos \frac{z}{2}

I=e^{z} F(z)+c

=\frac{e^{2} \cdot \cos z}{2 }+c

=e^{2 x} \cdot \cos \frac{2 x}{2}+c

=e^{2 x} \cos x+c


Question 14

(i) \left[\tan (\log x)+\sec ^{2}(\log x)\right] d x
Sol :
putting z=logx  \Rightarrow e^{2}=x \Rightarrow d x=e^{2} d z

Now , I=\int\left(\tan z+\sec ^{2} z\right) \cdot e^{z} d z

I=\int e^{2}\left[F(z)+F^{\prime}(z)\right] d z

where F(z)=tanz

I=e^{z} F(z)+c=e^{2} \tan z+c

=e^{\log x} \tan (\log x)+c \left(\because e^{\log x}=x\right)

=x.tan(log x)+c



(ii) \int e^{x}[\sec x+\log (\sec x+\tan x)] d x
Sol :
I=\int e^{x} \left[ \sec x+\log (\sec x+\tan x)\right] d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=log(secx +tanx)

I=e^{x} F(x)+c

=e^{x} \log (\sec x+\tan x)+c


Question 15

(i) \int \frac{e^{x}(1-x)^{2}}{(1+x)^{2}} d x
Sol :
I=\int \frac{e^{x}(1-x)^{2}}{(1+x)^{2}} d x

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

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

=\int e^{x}\left(\frac{x^{2}-2 x-3+4}{(1+x)^{2}}\right) d x

I=\int e^{x}\left[\frac{x^{2}-2 x-3}{(1+x)^{2}}+\frac{4}{(1+x)^{2}}\right] d x

I=\int e^{x}\left[\frac{x^{2}-3 x+x-3}{(1+x)^{2}}+\frac{4}{(1+x)^{2}}\right] d x

I=\int e^{x}\left[\frac{x(x-3)+(x-3)}{(1+x)^{2}}+\frac{4}{(1+x)^{2}}\right] d x

I=\int e^{x}\left[\frac{(x-3)(x+1)}{(1+x)^{2}}+\frac{4}{(1+x)^{2}}\right] d x

I=\int e^{x}\left[\left(\frac{x-3}{1+x}\right)+\frac{4}{(1+x)^{2}}\right] d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=\left(\frac{x-3}{1+x}\right)

I=e^{x} F(x)+c=e^{x} \times \frac{(x-3)}{1+x}+c


(ii) \int \frac{e^{x}\left(x^{3}-x+2\right)}{\left(1+x^{2}\right)^{2}} d x
Sol :
I=\int e^{x} \frac{\left(x^{3}-x+2\right)}{\left(1+x^{2}\right)^{2}} d x

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

I=\int e^{x}\left[\frac{\left(1+x^{2}\right)(1+x)}{\left(1+x^{2}\right)^{2}}+\frac{1-2 x-x^{2}}{\left(1+x^{2}\right)^{2}}\right] d x

I=\int e^{x}\left[\left(\frac{1+x}{1+x^{2}}\right)+\left(\frac{1-2 x-x^{2}}{\left(1+x^{2}\right)^{2}}\right)\right] d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=\frac{1+x}{1+x^{2}}

I=e^{x} F(x)+c=e^{x}\left(\frac{1+x}{x^{2}+1}\right)+c


(iii) \int \frac{x \cdot e^{2 x}}{(1+2 x)^{2}} d x
Sol :
putting z=2x then dz=2dx \Rightarrow d x=\frac{d z}{2}

Now , I=\int \frac{z}{2(1+z)^{2}} \frac{d z}{2}

=\frac{1}{4} \int e^{2} \frac{z}{(1+z)^{2}} d z

=\frac{1}{4} \int e^{2}\left(\frac{1+z-1}{(1+z)^{2}}\right) d z

I=\frac{1}{4} \int e^{2}\left(\frac{1+z}{(1+z)^{2}}-\frac{1}{(1+z)^{2}}\right) d z

=\frac{1}{4} \int e^{2}\left(\frac{1}{1+z}-\frac{1}{(1+z)^{2}}\right) d z

I=\frac{1}{4} \cdot e^{2} \times \frac{1}{1+z}+c

=\frac{1}{4} \cdot \frac{e^{2 x}}{1+2 x}+c

=\frac{e^{2 x}}{4(1+2 x)}+c

Question 16

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

putting z=\tan ^{-1} x
⇒x=tanz

then d z=\frac{1}{1+x^{2}} d x

Now  , I=\int e^{\tan ^{-1} x}\left(\frac{1+x^{2}+x}{1+x^{2}}\right) d x

=\int e^{z}\left(1+\tan ^{2} z+\tan z\right) d z

I=\int e^{z}\left(\sec ^{2} z+\tan z\right) d z

=e^{z} \tan z+c

I=e^{\tan ^{-1} x} \tan \tan ^{-1} x+c

=x \cdot e^{\tan ^{-1} x}+c


Question 17

\left.\int e^{x} \frac{1}{x} \int x(\log x)^{2}+2 \log x\right] d x
Sol :
I=\int e^{x} \frac{1}{x}\left[x(\log x)^{2}+2 \log x\right] d x

I=\int e^{x}\left[(\log x)^{2}+\frac{2 \log x}{x}\right] d x

I=\int e^{x}\left[F(x)+F^{\prime}(x)\right] d x

where F(x)=(\log x)^{2}

I=e^{x} F(x)+c=e^{x}(\log x)^{2}+c


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