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KC Sinha Solution Class 12 Chapter 21 (निश्चित समकलनों के गुणधर्म) Property of Definite Integral Exercise 21.1

 Exercise 21.1

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

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