KC Sinha: Exercise 21.1- Mathematics Solution Class 12 Chapter 21 निश्चित समकलनों के गुणधर्म

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Question 1

Evaluate the following:

(i) $\int_{-2}^{2}|x| d x$

Sol :

Put x=0

$=\int_{-2}^{0} |x| d x+\int_{0}^{2}|x| d x$

$=-\int_{-2}^{0} x d x+\int_{0}^{2} x d x$

$=-\frac{\left[x^{2}\right]_{-2}^{0}}{2}+\frac{\left[x^{2}\right]_{0}^{2}}{2}$

$=-\frac{1}{2}\left[0-(-2)^{2} \right]+\frac{1}{2}\left[(2)^{2}-0^{2}\right]$

$=-\frac{1}{2}(-4)=\frac{1}{2} \times 4$

=2+2

=4

Diagram to be added

(ii) $\int_{1}^{5}|x-4| d x$

Sol :

x-4=0⇒x=4

$=\int_{1}^{5}|x-4| d x$

$=\int_{1}^{4}|x-4| d x+\int_{4}^{5}|x-4| d x$

$=-\int_{1}^{4}(x-4) d x+\int_{4}^{5}(x-4) dx$

$=-\frac{\left[(x-4)^{2}\right]_{1}^{4}}{2}+\frac{\left[(x-4)^{2}\right]_{4}^{5}}{2}$

$=-\frac{1}{2}[0-9]+\frac{1}{2}[1-0]$

$=\frac{9}{2}+\frac{1}{2}$

$=\frac{10}{2}=5$

Diagram to be added

(iii) $\int_{0}^{2}|x-3| d x$

Sol :

$=-\int_{0}^{2}(x-3) dx$

$=\frac{-\left[(x-3)^{2}\right]_{0}^{2}}{2}$

$=-\frac{1}{2}[1-9]$

$=\frac{8}{2}=4$

(iv) $\int_{-1}^{2}|2 x-1| d x$

Sol :

2x-1=0

$x=\frac{1}{2}$

Diagram to be added

$\int_{-1}^{2}|2 x-1| dx=\int_{-1}^{1/2}|2 x-1| d x+\int_{1 / 2}^{2}|2 x-1| d x$

$=-\int_{-1}^{1 / 2}(2 x-1) d x+\int_{1/2}^{2}(2 x-1) d x$

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

$=-\frac{1}{4}[0-3]+\frac{1}{4}[9-0]$

$=\frac{9}{4}+\frac{9}{4}$

$=\frac{18}{4}=\frac{9}{2}$

(v) $\int_{-1}^{1}(|x|+|x-1|) dx$

Sol :

$=\int_{-1}^{1}|x| d x+\int_{-1}^{1}|x-1| d x$

$=\int_{-1}^{0}|x| dx+\int_{0}^{1}|x| d x+\int_{-1}^{1}|x-1| d x$

$=-\int_{-1}^{0} x d x+\int_{0}^{1} x d x-\int_{-1}^{1}(x-1) d x$

$\left.=-\frac{1}{2}\left[x^{2}\right]_{-1}^{0}+\frac{1}{2}\left[x^{2}\right]_{0}^{1}-\frac{1}{2}(x-1)^{2}\right]_{-1}^{1}$

$=-\frac{1}{2}[0-1]+\frac{1}{2}[1-0]-\frac{1}{2}[0-4]$

$=\frac{1}{2}+\frac{1}{2}+2$

=1+2

=3

Diagram to be added

(vi) $\int_{0}^{4}|x-1| d x$

Sol :

Diagram to be added

$=\int_{0}^{1}|x-1| d x+\int_{1}^{4}|x-1| dx$

$=-\int_{0}^{1}(x-1)dx+\int_{1}^{4}(x-1) dx$

Question 2

(i) $\int_{-4}^{4}|x+2| d x$

Sol :

x+2=0

x=-2

Diagram to be added

$\int_{-4}^{4}|x+2| d x=\int_{-4}^{-2}|x+2| dx+\int_{-2}^{4}|x+2| d x$

$=-\int_{-4}^{-2}(x+2) dx+\int_{-2}^{4}(x+2) d x$

$=-\frac{1}{2}\left[(x+2)^{2}\right]_{-4}^{-2}+\frac{1}{2}\left[(x+2)^{2}\right]_{-2}^{4}$

$=-\frac{1}{2}[0-4]+\frac{1}{2}[36-0]$

=2+18

=20

Diagram to be added

(iii) $\int_{0}^{3}|3 x-1| d x$

Sol :

3x-1=0

$x=\frac{1}{3}$

Diagram to be added

$\int_{0}^{3}|3 x-1| d x=\int_{0}^{1/3}|3 x-1| d x+\int_{1 / 3}^{3}|3 x-1| d x$

$=-\int_{0}^{1 / 3}(3 x-1) d x+\int_{1 / 3}^{3}(3 x-1) dx$

$=-\frac{1}{2\times 3}\left[(3 x-1)^{2}\right]_{0}^{1 / 3}+\frac{1}{2\times 3}\left[(3 x-1)^{2}\right]_{1 / 3}^{3}$

$=-\frac{1}{2\times 3}[0-1]+\frac{1}{2\times 3}[64-0]$

$=\frac{1}{6}+\frac{64}{6}$

$=\frac{1+64}{6}=\frac{65}{6}$

Diagram to be added

Question 3

(i) Evaluate $\int_{1}^{4} f(n) d x$ where f(x)=|x-1|+|x+2|+|x+3|

Sol :

$\left.\int_{1}^{4}(|x-1+| x-2|+| x-3 \mid)\right|dx=\int_{1}^{4}|x-1| d x+\int_{1}^{4}|x-2| d x+\int_{1}^{4}|x-3| d x$

$=\int_{1}^{4}(x-1) d x+\int_{1}^{2}|x-2| d x+\int_{2}^{4}|x-2| d x+\int_{1}^{3}|x-3| d x+\int_{3}^{4}(x-3) d x$

$=\frac{1}{2}\left[(x-1)^{2}\right]_{1}^{4}-\int_{1}^{2}(x-2) d x+\int_{2}^{4}(x-2) d x-\int_{1}^{3}(x-3) d x+\int_{2}^{4}(x-3) d x$

$=\frac{1}{2}[9-0]-\frac{1}{2}\left[(x-2)^{2}\right]_{1}^{2}+\frac{1}{2}(x-2)^{2} \int_{2}^{4}-\frac{1}{2}\left[(x-3)^{2}\right]_{1}^{3}+\frac{1}{2}\left[(x-3)^{2}\right]_{3}^{4}$

$=\frac{9}{2}-\frac{1}{2}[0-1]+\frac{1}{2}[4-0]-\frac{1}{2}[0-4]+\frac{1}{2}[1-0]$

$=\frac{9}{2}+\frac{1}{2}+2+2+\frac{1}{2}$

$=\frac{9}{2}+5=\frac{19}{2}$

(ii) $\int_{-1}^{2}\left|x^{3}-x\right| d x$

Sol :

x³-x=0

x(x²-1)=0⇒x(x+1)(x-1)=0

x=0,±1

Diagram to be added

$\int_{-1}^{2}\left|x^{3}-x\right| d x=\int_{-1}^{0}\left|x^{3}-x\right| d x+\int_{0}^{1}\left|x^{3}-x\right| d x+\int_{1}^{2}\left|x^{3}-x\right| d x$

$=\int_{-1}^{0}\left(x^{3}-x\right) d x-\int_{0}^{1}\left(x^{3}-x\right) d x+\int_{1}^{2}\left(x^{3}-x\right) d x$

$=\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{-1}^{0}-\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{0}^{1}+\left[\frac{x^{4}}{4}-\frac{x^{2}}{2}\right]_{1}^{2}$

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

$=-\frac{1}{4}+\frac{1}{2}-\frac{1}{4}+\frac{1}{2}+2-\frac{1}{4}+\frac{1}{2}$

$=2+\frac{3}{2}-\frac{3}{4}$

$=\frac{8+6-3}{4}=\frac{11}{4}$

Question 4

$\int_{-\pi / 3}^{\pi / 4}|\tan x| dx$

Sol :

tanx=0

x=0

Diagram to be added

$\int_{-\pi / 4}^{\pi / 4}|\tan x| d x=\int_{-\pi / 4}^{0}|\operatorname{tan}| d x+\int_{0}^{\pi / 4}|\tan | d x$

$=-\int_{-\pi / 4}^{0} \operatorname{tan} dx+\int_{0}^{\pi / 4} \tan dx$

$\left.=-[\log |\sec |]_{-\pi / 4}^{0}+(\log \mid \sec x|\right]_{0}^{\pi / 4}$

$=-[0-\log \sqrt{2}]+[\log \sqrt{2}-0]$

=log√2+log√2

=2log√2

$=2 \log 2^{1/2}$

$=1\times  \frac{1}{2} \log 2$

=log 2

Diagram to be added

Question 5

$\int_{0}^{\pi}|\cos x| d x$

Sol :

cosx=0

$x=\frac{\pi}{2}$

Diagram to be added

$\int_{0}^{\pi}|\cos x| d x=\int_{0}^{\pi/2}|\cos x| d x+\int_{\pi/2}^{\pi}|\cos x| dx$

$=\int_{0}^{\pi / 2} \cos d x-\int_{\pi/2}^{\pi} \cos x dx$

$[\sin x]_{0}^{\pi/2}-[\sin x]_{\pi/2}^{\pi}$

=1-0-[0-1]

=2

$\int_{0}^{\pi/2} \sin x d x=\int_{0}^{\pi /2} \cos x dx=1$

$-[\cos x]_{0}^{\pi /{2}}=[\sin x]_{0}^{\pi/ 2}$

-[0-1]=1-0

1=1

Diagram to be added

Question 6

$\int_{0}^{\pi/2}|\cos x-\sin x| d x$

Sol :

cosx-sinx=0

cosx=sinx

tanx=1

$x=\frac{\pi}{4}$

Diagram to be added

$\int_{0}^{\pi / 2}|\cos x-\sin x| dx=\int_{0}^{\pi / 4}|\cos x-\sin x|d x+\int_{\pi / 4}^{\pi/2}| \cos x-\sin x| d x$

$=\int_{0}^{\pi / 4}(\cos x-\sin x) d n-\int_{\pi /4}^{\pi/2}(\cos x-\sin x) d x$

$=[\sin x+\cos x]_{0}^{\pi / 4}-\left[\sin x +\cos x\right]_{\pi /4}^{\pi /2}$

$=\frac{1}{\sqrt2}+\frac{1}{\sqrt2}-(0+1)-\left[1+0-\left(\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\right)\right]$

$=\frac{2}{\sqrt{2}}-1-\left[1-\frac{2}{\sqrt2}\right]$

$=\frac{2}{\sqrt2}-1-1+\frac{2}{\sqrt{2}}$

=2√2-2

$=2(\sqrt{2}-1)$

Question 7

$\int_{-1}^{1} \frac{|x|}{x} d x$

Sol : 

x=0

Diagram to be added

$\int_{-1}^{1} \frac{|x|}{x} d x=\int_{-1}^{0} \frac{|x|}{x} d x+\int_{0}^{1} \frac{|x|}{x} d x$

$=-\int_{-1}^{0} \frac{x}{x} dx+\int_{0}^{1} \frac{x}{x} d x$

$=-[x]_{-1}^{0}+[x]_{0}^{1}$

=-[0-(1)]+[1-0]

=-1+1=0

Question 8

$\int_{1}^{4} f(x) dx$ when $f(n)=\left\{\begin{array}{ll}2 x+8 & ; 1 \leq x \leq 2 \\ 6 x & ,2 \leq x \leq 4\end{array}\right.$

Sol :

$\int_{1}^{4} f(x) d x=\int_{1}^{2} f(x) d x+\int_{2}^{4} f(x) d x$

$=\int_{1}^{2}(2 x+8) d x+\int_{2}^{4} 6 x d x$

$=\frac{1}{2\times 2}\left[(2x+8)^{2}\right]_{1}^{2}+\frac{6}{2}\left(x^{2}\right)_{2}^{4}$

$=\frac{1}{4}[144-100]+3[16-4]$

$=\frac{1}{4} \times 44 +36=47$

Question 10

$\int_{-\pi / 4}^{\pi / 4}(\sin |x|+\cos |x| )d x$

Sol :

x=0

Diagram to be added

$\int_{-\pi / 2}^{\pi / 2}(\sin |x|+\cos |x|) d x=\int_{-\pi / 2}^{0}(\sin |x|+\cos (x)) dx+\int_{0}^{\pi / 2}(\sin |x|+\cos |x|) d x$

$=\int_{-\pi/2}^{0}(\sin (-x)+\cos (-x)) d x+\int_{0}^{\pi}(\sin x+\cos x) d x$

$=\int_{-\pi/2}^{0}(-\sin x+\cos x) d x+\int_{0}^{\pi/2}[-\sin x+\cos x] dx$

$=\int_{-\pi/2}^{0}(-\sin x+\cos ) d x+[-\cos x+\sin x]_{0}^{\pi /2}$

$=\left[\cos x+\sin x\right]_{-\pi / 2}^{0}+[0+1-(-1+0)]$

=[1+0-(0-1)+1+1]

=1+1+1+1=4

Question 11

$\int_{0.2}^{3.5}[x] dx$

Sol :

$=\int_{0.2}^{1}[x] d x+\int_{1}^{2}[x] d x+\int_{2}^{3}[x] dx+\int_{3}^{3 \cdot 5}[x] d x$

$=\int_{0.2}^{1} 0 d x+\int_{1}^{2} 1 d x+\int_{2}^{3} 2 d x+\int_{3}^{3 \cdot 5} 3 d x$

$=0+[x]_{1}^{2}+2[x]_{2}^{3}+3[x]_{3}^{3.5}$

=2-1+2(3-2)+3(3.5-3)

=1+2+3×0.5

=1+2+1.5

=4.5

Diagram to be added

Question 12

$\int_{0}^{\pi / 2} \cos ^{2} x d x$

Sol :

Let $\int_{0}^{\pi / 2} \cos ^{2} x d x$..(i)

$I=\int_{0}^{\pi / 2} \cos ^{2}\left(\frac{\pi}{2}-x\right) d x$

$I=\int_{0}^{\pi /2} \sin ^{2} x d x-0$..(2)

From (1)+(2)

$2I=\int_{0}^{\pi/{2}} \cos ^{2} x dx+\int_{0}^{\pi /2} \sin ^{2} x d x$

$2 I=\int_{0}^{\pi/2}\left(\cos ^{2} x+\sin^2 x\right) d x$

$2 I=\int_{0}^{\pi/2} d x$

$2 I=[x]_{0}^{\pi/2}$

$2 I=\frac{\pi}{2}=0$

$I=\frac{\pi}{4}$

Question 13

$\int_{0}^{\pi/2} \frac{\cos x}{\cos x+\sin x} d x$

Sol :

Let $\int_{0}^{\pi/2} \frac{\cos x}{\cos x+\sin x} d x$..(i)

$I=\int_{0}^{\pi/2} \dfrac{\cos \left(\frac{\pi}{2}-x\right)}{\cos \left(\frac{\pi}{2}-x\right)+\sin \left(\frac{\pi}{2}-x\right)} dx$

$=\int_{0}^{\pi /2} \frac{\sin x}{\sin x+\cos x} dx$...(2)

From (1)+(2)

$2I=\int_{0}^{\pi /2} \frac{\cos x}{\cos x+\sin x} d x+\int_{0}^{\pi/2} \frac{\sin x}{\sin x+\cos x} d x$

$2 I=\int_{0}^{\pi /2}\left(\frac{\cos x}{\cos x+\sin x}+\frac{\sin x}{\sin x+\cos x }\right) d x$

$=\int_{0}^{\pi /{2}} \frac{\cos x+\sin x}{\sin x+\cos x} d x$

$2 I=[x]_{0}^{\pi /2}$

$2I=\frac{\pi}{2}-0$

$I=\frac{\pi}{4}$

Question 14

$\int_{0}^{\pi /2} \frac{d x}{1+\tan x}$

Sol :

Let $I=\int_{0}^{\pi /2} \frac{d x}{1+\tan x}$

$=\int_{0}^{\pi / 2} \frac{d x}{1+\frac{\sin x}{\cos x}}$

$I=\int_{0}^{\pi/2} \frac{\cos x dx}{\cos x+\sin x}$

Question 15

$\int_{0}^{\pi /2} \log \tan x dx$

Sol :

$1=\int_{0}^{\pi / 2} \log \mid \operatorname{tan x}| dx$

$I=\int_{0}^{\pi/2} \operatorname{log} \tan \left(\frac{\pi}{2}-x\right) dx$

$I=\int_{0}^{\pi /2} \log \cot x dx$..(2)

From (1)+(2)

$2 I=\int_{0}^{\pi /2} \log \tan xdx+\int_{0}^{\pi /2} \log \cot xdx$

$=\int_{0}^{\pi / 2}(\log \tan x+\log \cot x dx$

$=\int_{0}^{\pi/2} \log \left(\operatorname{\tan x-cot x}\right) d x=0$

I=0

(1) $\int_{0}^{\pi/{2}} \log \sin x d x$

Sol :

Let $\int_{0}^{\pi/{2}} \log \sin x d x$...(i)

$1=\int_{0}^{\pi /2} \log \sin (\frac{\pi}{2}-x) dx$

$I=\int_{0}^{\pi /2} \log \cos x dx$..(2)

By (1)+(2)

$2 I=\int_{0}^{\pi /2} \log \sin x dx+\int_{0}^{\pi /2} \log \cos xdx$

$2 I=\int_{0}^{\pi /2}(\log \sin x+\log \cos x) d x$

$2 I=\int_{0}^{\pi /2} \log \left(\frac{2 \sin x-\cos x}{2}\right) d x$

$2I=\int_{0}^{\pi / 2} \log \frac{\operatorname{sin} 2 x}{2}$

$2 I=\int_{0}^{\pi /2}(\log \sin 2 x-\log 2) dx$

$2I=\int_{0}^{\pi /2}(\log \sin 2 x)dx-\int_{0}^{\pi/ 2} \log 2 d x$..(3)

Let $I_{1}=\int_{0}^{\pi /2} \log \sin 2x~dx$

Putting 2x=t

$2=\frac{d x}{d t}$

$dx=\frac{dt}{2}$

Now

$I_{1}=\frac{1}{2} \int_{0}^{\pi} \log \sin t d t$

Let f(t)=log sin t

f(π-t)=log sin(π-t)

=log sin t=f(t)

$I_{1}=\frac{1}{2} \cdot 2 \int_{0}^{\pi /2} \log \sin t d t$

$I_{1}=\int_{0}^{\pi / 2} \log \sin t d t$..(2)...(4)

$I_{1}=\int_{0}^{\pi /2} \log \sin \left(\frac{\pi}{2}-t\right) d t$

$y=\int_{0}^{\pi /2} \log \cot t dt$..(5)

From (4)+(5)

$2I_1=\int_{0}^{\pi /2} \log \frac{(2\sin t \cos t)}{2}dt$

$2 I_{1}=\int_{0}^{\pi /2 } \frac{\sin2t}{2}d t$

$2 I_{1}=\int_{0}^{\pi /2 } \sin2td t-\int_{0}^{\pi /2}\log2 dt$

$2 I_{1}=\int_{0}^{2} \log \sin 2x d x-\int_{0}^{\pi /2}\log 2dt$

$2 I_{1}=I_{1}-\int_{0}^{\pi/{2}} \log {2} dt$

$I_{1}=-\log 2 \int_{0}^{\pi /{2}} d t$

$=-\log 2[t]_{0}^{\pi/{2}}$

$=-\log 2 \frac{\pi}{2}$

$I_{1}=-\frac{\pi}{2} \log 2$

From (3)

$2I=-\frac{\pi}{2} \log 2-\log 2[x]_{0}^{\pi /2}$

$=-\frac{\pi}{2} \log 2-\log 2\left[\frac{\pi}{2}-0\right]$

$=-\frac{\pi}{2} \log 2-\frac{\pi}{2} \log 2$

$=-\frac{2\pi}{2}\log {2}$

$2 I=-\pi \log 2$

$I=-\frac{\pi}{2} \log 2$

(2) $\int_{0}^{\pi} \log \cos xdx=-\frac{\pi}{2} \log 2$

(3) $\int_{0}^{\pi /2} \log \tan x d x=0$

(4) $\int_{0}^{\pi /2} \log \cot xd x=0$

(5) $\int_{0}^{\pi / 2} \log \sec x d x=\frac{\pi}{2} \log 2$

Sol :

$\int_{0}^{\pi /2} \log \frac{1}{\cos x} d x$

$=\int_{0}^{\pi/{2}} \log (\cos x)^{-1} dx$

$-\int_{0}^{\pi /2} \log \cos xdx =\frac{\pi }{2} \log 2$

(6) $\int_{0}^{\pi} \log \operatorname{cosec}xd x=\frac{\pi}{2} \operatorname{log}2$

Sol :

$=\int_{0}^{\pi/{2}} \log \frac{1}{\sin x} d x$

$=\int_{0}^{\pi /2}-\log \sin xdx$

$=\frac{\pi}{2} \log 2$

Question 16

$\int_{0}^{\pi} \frac{x \tan x}{\sec x \cdot \operatorname{cosec x}} d x$

Sol :

Let $I=\int_{0}^{\pi} \frac{x \tan x}{\sec x \cdot \operatorname{cosec x}} d x$..(1)

$I=\int_{0}^{\pi} \frac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x) \operatorname{cosec}(\pi-x)} d x$

$I=\int_{0}^{\pi} \frac{(\pi-x) \operatorname{tan} x}{\sec x \cdot \operatorname{cosec x}} d x$..(2)

From (1)+(2)

$2 I=\int_{0}^{\pi} \frac{x\tan x+(\pi-x)+\tan x}{\operatorname{sec} x\cdot \operatorname{cosec x}} d x$

$2 I=\int_{0}^{\pi} \frac{\operatorname{tan} x(x+\pi-x)}{\sec x \cdot \operatorname{cosec} x} d x$

$2I=\pi \int_{0}^{\pi} \frac{\operatorname{tan} x}{\operatorname{sec} x \cdot \operatorname{cosec x}} d x$

$=\pi \int_{0}^{\pi} \frac{\frac{\sin x}{\cos x}}{\frac{1}{\sin x} \cdot \frac{1}{\sin x}} dx$

$2 t=\pi \int_{0}^{\pi} \sin ^{2} x d x$

$=\pi \int_{0}^{\pi}\left(\frac{1-\cos 2 x}{2}\right) dx$

$2 I=\frac{\pi}{2}\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\pi}$

$I=\frac{\pi}{4}[\pi-0-(0-0)]$

$I=\frac{\pi ^2}{4}$

Question 17

$\int_{0}^{\pi /{2}} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x$

Sol :

Let $I=\int_{0}^{\pi /{2}} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}} d x$

$=\int_{0}^{\pi /2} \frac{\frac{\sqrt{\sin x}}{\sqrt{\cos x}}}{1+\frac{\sqrt{\sin x}}{\sqrt{\cos x}}} d x$

$I=\int_{0}^{\pi /2} \frac{\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x$..(1)

$=\int_{0}^{\pi /2} \frac{\sqrt{\sin (\pi / 2-x)}}{\sqrt{\cos (\pi/2-x)}+\sqrt{\sin (\pi /2-x)}} d x$

Question 18

(i) $\int_{0}^{\pi} x \sin ^{6} x \cos ^{4} x d x=\frac{\pi}{2} \int_{0}^{\pi} \sin ^{6} x \cdot \cos ^{4} x d x$

Sol :

L.H.S

$=\int_{0}^{\pi} x \sin ^{6} x \cdot \cos ^{4} x d x$

$I=\int_{0}^{\pi} x \sin ^{6} x \cdot \cot ^{4} x d x$..(i)

$=\int_{0}^{\pi}(\pi-x) \sin ^{6}(\pi-4) \cdot \cos ^{4}(\pi-4) d x$

$I=\int_{0}^{\pi}(\pi-x) \sin ^{6} x \cos ^{4} x d x$..(2)

From (1)+(2)

$2I=\int_{0}^{\pi} \sin ^{6} x \cdot \cos ^{4}(x+\pi-x) d x$

$2 I=\pi \int_{0}^{\pi} \sin^{6}x \cdot \cos ^{4} x dx$

$I=\frac{\pi}{2} \int_{0}^{\pi} \sin ^6 x \cdot \cos ^{4}xdx$

R.H.S proved

(ii) $\int_{0}^{\pi / 2} \log (\tan x+\cot x) d x=\pi \log 2$

Sol :

L.H.S

$=\int_{0}^{\pi / 2} \log (\tan x+\cot x) d x$

$=\int_{0}^{\pi /{2}} \log \left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right) d x$

$=\int_{0}^{\pi /2} \log \left(\frac{\sin ^{2} x+\cos ^{2}x}{\cos x \sin x}\right) d x$

$=\int_{0}^{\pi / 2} \log \left(\frac{1 \times 2}{2(\cos x\sin x)}\right) d x$

$=\int_{0}^{\pi /2} \log \left(\frac{2}{\sin 2x}\right) d x$

$=\int_{0}^{\pi/{2}}(\log 2-\log \sin x2x) dx$

$I=\log _{2} \int_{0}^{\pi /2} dx-\int_{0}^{\pi / 2} \log \sin 2x dx$

$I=\log {2}[x]_{0}^{\pi/2}-I_{1}$

$I=\frac{\pi}{2} \log 2-I_{1}$...(i)

$I_{1}=\int_{0}^{\pi /2} \log \sin 2 xdx$

Putting 2x=t

$2=\frac{d t}{d x}$

$d x=\frac{d t}{2}$

$I_{1}=\frac{1}{2} \int_{0}^{\pi} \log \sin t d t$

f(t)=log sint

f(π-t)=log sin(π-t)

f(π-t)=log sint

f(π-t)=f(t)

$I_{1}=\frac{1}{2}\times 2 \int_{0}^{\pi /2} \log \sin t d t$

$I_{1}=-\frac{\pi}{2} \log 2$ putting in (i)

$\left[\int_{0}^{\pi} \log sinx d x=-\frac{\pi}{2} \log 2\right]$

$I=\frac{\pi}{2}\log+\frac{\pi}{2} \log 2$

$=\frac{2\pi}{2} \log 2$

2=π log2=R.H.S

Question 29

$\int_{0}^{\pi /2} \frac{\sin ^{2} x}{1+\sin x \cdot \cos x} d x$

Sol :

Let $\int_{0}^{\pi /2} \frac{\sin ^{2} x}{1+\sin x \cdot \cos x} d x$..(1)

$I=\int_{0}^{\pi / 2} \frac{\sin ^{2}\left(\frac{\pi}{2}-x\right)}{1+\sin \left(\frac{\pi}{2}-x\right) \cdot \cos \left(\frac{\pi}{2}-x\right)} dx$

$1=\int_{0}^{\pi /2} \frac{\cos^2 x}{1+\cos x.\sin x} d x$..(2)

From (1)+(2)

$2 I=\int_{0}^{\pi/{2}} \frac{\sin ^{2} x+\cos^2 x}{1+\sin x \cos x} d x$

$2I=\int_{0}^{\pi / 2} \frac{1}{1+\sin x\cos x} dx$

$2 I=\int_{0}^{\pi / 2} \frac{1}{1+\frac{2 \sin x.\cos x}{2}} dx$

$2I=\int_{0}^{\pi /2} \frac{2}{2+\sin 2 x} d x$

$2I=2 \int_{0}^{\pi /2} \frac{1}{2+\frac{2\tan x}{1+\tan^{2} x}} dx$

$2I=2 \int_{0}^{\pi /2} \frac{\left(1+\tan ^{2} x\right)}{2+2\tan ^2 x+2\tan x} d x$

$I=\frac{1}{2} \int_{0}^{\pi /2} \frac{\sec ^{2} x}{\left.\tan ^{2} x+\tan x+1\right)} d x$

Putting tanx=t

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

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

$I=\frac{1}{2} \int_{0}^{\infty} \frac{\sec ^{2}x}{t^{2}+t+1} \frac{dt}{\sec^2 x}$

$I=\frac{1}{2} \int_{0}^{\infty} \frac{d t}{\left(t+\frac{1}{2}\right)^{2}+1-\frac{1}{4}}$

$=\frac{1}{2} \int_{0}^{\infty}\frac{dt}{(t+\frac{1}{2})^{2}+(\frac{\sqrt{3}}{2})^{2}}$

$=\frac{1}{2} \cdot \frac{2}{\sqrt{3}}\left[\tan^{-1} \left(\frac{t+\frac{1}{2}}{\frac{\sqrt3}{2}}\right)\right]_{0}^{\infty}$

$=\frac{1}{\sqrt3}\left[\tan ^{-1}\left(\frac{2 t+1}{\sqrt3}\right)\right]_{0}^{\infty}$

$=\frac{1}{\sqrt3}[\tan^{-1} \infty-\tan^{-1} \frac{1}{\sqrt3}]$

$=\frac{1}{\sqrt3}\left[\frac{\pi}{2}-\frac{\pi}{6}\right]$

$I=\frac{1}{\sqrt3} \frac{\pi}{3}$

$I=\frac{\pi}{3\sqrt3}$

Question 30

$\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x$

Sol :

Let $\int_{1}^{2} \frac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}} d x$..(1)

$=\int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{3-(3-x)}+\sqrt{3-x}} dx$

$=\int_{1}^{2} \frac{\sqrt{3-x}}{\sqrt{x}+\sqrt{3-x}} d x$..(2)

From (1)+(2)

$2 I=\int_{1}^{2} \frac{\sqrt{x}+\sqrt{3-x}}{\sqrt{3-x}+\sqrt{x}} d x$

$2 I=[x]_{1}^{2}$

2I=2-1

$I=\frac{1}{2}$

Question 36

$\int_{-1}^{1} x^{5} e^{x^{8}} d x$

Sol :

Let $f(x)=x^{5} e^{x^{8}}$

$f(-x)=(-x)^{5} e^{(-x)^{8}}$

$=-x^{5} e^{x^{8}}$

f(-x)=-f(x)

f(x) is odd function

$\int_{-1}^{1} x^{5} e^{x^{8}} d x=0$

Question 37

$\int_{-\pi}^{\pi} \sin ^{5} x \cos xd x=0$

Sol :

Let f(x)=sin⁵x.cosx

f(-x)=sin⁵(-x).cos(-x)

=-sin⁵x.cosx

=-f(x)

Question 38

$\int_{-1}^{1} x e^{-x^{2}\tan^2 x} d x=0$

Sol :

$f(x)=x e^{-x^{2}\operatorname{tan}^{2} x}$

$f(-x)=-x e^{-(x)^{2}+\tan^{2}(-x)}$

$=-x e^{-x^{2}+\operatorname{tan}^{2} x}$

f(-x)=-f(x)

f(x) is odd

Question 39

$\int_{-a}^{a} x f\left(x^{4}\right) d x=0$

Sol :

Let f(x)=xf(x⁴)

f(-x)=xf((-x)⁴)

=-xf(x⁴)

f(-x)=-f(x)

f(x) is odd

Question 40

$\int_{-2}^{2} f\left(x^{4}\right) d x=2 \int_{0}^{2} f\left(x^{4}\right) d x$

Sol :

L.H.S

$\int_{-2}^{2} f\left(x^{4}\right) d x$

f(x)=f(x⁴)

f(-x)=f((-x)⁴)

f(-x)=f(x⁴)

f(-x)=f(x)

f(x) is even function

L.H.S$=\int_{-2}^{2} f\left(x^{4}\right) d x=2 \int_{0}^{2} f\left(x^{4}\right) d x$=R.H.S

Question 41

$\int_{-1}^{1} e^{|x|} d x$

Sol :

x=0

Diagram to be added

$\int_{-1}^{1} e^{|x|} d x=\int_{-1}^{0} e^{|x|} d x+\int_{0}^{1} e^{|x|} d x$

$=\int_{-1}^{0} e^{-x} d x+\int_{0}^{1} e^{x} d x$

$=-\left[e^{-x}\right]_{-1}^{0}+\left[e^{x}\right]_{0}^{1}$

=-[1-e]+e-1

=-1+e+e-1

=2e-2

Question 42

(i) $\int_{-1}^{1} \sin ^{5} x \cdot \cos ^{4} x dx=0$

(ii) $\int_{-1}^{1} x^{17} \cos^4 x d x=0$

Question 43

$\int_{-\pi / 2}^{\pi / 2} \sin ^{7} x d x=0$

Question 44

$\int_{-\pi / 2}^{\pi} \sin ^{2} x d x$

Sol :

Let f(x)=sin²x

f(-x)=sin²(-x)

=sin²x

f(-x)=f(x); f(x) is even

$\int_{-\pi / 2}^{\pi} \sin ^{2} xdx=2 \int_{0}^{\pi/2} \sin ^{2} xdx$

$=2\int_{0}^{\pi/2} \frac{1-\cos 2 x}{2} dx$

$=\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\pi/2}$

$=\frac{\pi}{2}-0-(0-0)$

$=\frac{\pi}{2}$

Question 45

$\int_{-\pi /2}^{\pi /2} \sin x d x=0$

Question 46

$\int_{-1}^{1} \log \left(\frac{3-x}{3+x}\right) d x$

Sol :

$f(x)=\log \left(\frac{3-x}{3+x}\right)$

$f(-x)=\log \left(\frac{3+x}{3-x}\right)$

$f(-x)=-\log \left(\frac{3-x}{3+x}\right)$

f(-x)=-f(x)

f(x) is odd

Question 47

$\int_{-\pi /2}^{\pi /2} \log \left(\frac{2-\sin x}{2+\sin x}\right) d x$

Sol :

Let $f(x)=\log \left(\frac{2-\sin x}{2+\sin x }\right) d x$

$f(-x)=\log \left(\frac{2-\sin (-x)}{2+\sin (-x)}\right)$

$=\log \left(\frac{2+\sin x}{2-\sin x }\right)$

$f(-x)=-\log \left(\frac{2-\sin x}{2+\sin x}\right)$

f(-x)=-f(x)

Question 48

$\int_{-\pi / 4}^{\pi / 4} \sin ^{2} x dx$

Sol :

$=2 \int_{0}^{\pi / 4} \sin ^{2} x d x$

$=2 \int_{0}^{\pi/{4}} \frac{1-\cos 2x}{2} d x$

$=\left[x-\frac{\sin 2 x}{2}\right]_{0}^{\pi / 4}$

$=\frac{\pi}{4}-\frac{1}{2}-(0)$

$=\frac{\pi}{4}-\frac{1}{2}$

Question 49

$\int_{0}^{2\pi} \frac{1}{1+e^{\sin x}} d x$

Sol :

$=\int_{0}^{2 \pi} \frac{1}{1+e^{\sin x}} d x$..(1)

$I=\int_{0}^{2 \pi} \frac{1}{1+e^{\sin (2 \pi-x)}} dx$

$=\int_{0}^{2 \pi} \frac{1}{1+e^{-\sin x}} d x$

$=\int_{0}^{2 \pi} \frac{1}{1+\frac{1}{e^{\sin x}}} d x$

$=\int_{0}^{2 \pi} \frac{e^{\sin x}}{e^{\sin x}+1} d x$..(2)

From (1)+(2)

$2I=\int_{0}^{2 \pi} \frac{1+e^{\sin x}}{1+e^{\sin x}}dx$

$2 I= [x]_{0}^{2 \pi}$

2I=2π-0

2I=2π

I=π