Integrals Test

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Integrals Test - 46

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Question 1
1 / -0

$$\int ^{\pi /2}_{-\pi /2} cost . \sin(2t-\displaystyle \frac{\pi}{4})dt=$$

A
$$\displaystyle \frac{\sqrt{2}}{3}$$

B
$$-\displaystyle \frac{\sqrt{2}}{3}$$

C
$$\displaystyle \frac{\sqrt{3}}{1}$$

D
$$\displaystyle \frac{1}{\sqrt{3}}$$

Solution

$$I=\int { \left( cost \right) \left( sin\left( 2t-\cfrac { \pi }{ 4 } \right) \right) } dt=-\int { sin\left( \cfrac { \pi }{ 4 } -2t \right) \left( cost \right) dt } \\ =-\cfrac { 1 }{ 2 } \int { \left( sin\left( \cfrac { 1 }{ 4 } \left( \pi -12t \right) \right) +sin\left( \cfrac { 1 }{ 4 } \left( \pi -4t \right) \right) \right) dt } \\ =-\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -12t \right) \right) dt } -\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -4t \right) \right) dt } $$
Let $${ I }={ I }_{ 1 }+{ I }_{ 2 }$$, such that
$${ I }_{ 1 }=-\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -12t \right) \right) dt } $$
Substituting $$x=\pi -4t\Rightarrow dx=-4dt$$
$${ I }_{ 1 }=\cfrac { 1 }{ 8 } \int { sin\cfrac { x }{ 4 } dx } =-\cfrac { 1 }{ 2 } cos\cfrac { x }{ 4 } =-\cfrac { 1 }{ 2 } sin\left( 3t+\cfrac { \pi }{ 4 } \right) $$ ...(1)
And
$${ I }_{ 2 }=-\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -4t \right) \right) dt } $$
Substituting $$y=\pi -12t\Rightarrow dy=-12dt$$, we get
$${ I }_{ 2 }=\cfrac { 1 }{ 24 } \int { sin\cfrac { y }{ 4 } dy } =-\cfrac { 1 }{ 6 } cos\cfrac { y }{ 4 } \quad =-\cfrac { 1 }{ 6 } sin\left( t+\cfrac { \pi }{ 4 } \right) $$...(2)
Therefore from (1) and (2)
$$\int _{ -\cfrac { \pi }{ 2 } }^{ \cfrac { \pi }{ 2 } }{ \left( cott \right) sin\left( 2t-\cfrac { \pi }{ 4 } \right) dt } =\left[ -\cfrac { 1 }{ 2 } sin\left( 3t+\cfrac { \pi }{ 4 } \right) -\cfrac { 1 }{ 6 } sin\left( t+\cfrac { \pi }{ 4 } \right) \right] _{ -\cfrac { \pi }{ 2 } }^{ \cfrac { \pi }{ 2 } }=-\cfrac { \sqrt { 2 } }{ 3 } $$
Question 2
1 / -0

$$\displaystyle \int_{0}^{a}\frac{dx}{x+\sqrt{a^{2}-x^{2}}}=$$

A
$$\pi$$

B
$$\dfrac{\pi}{3}$$

C
$$-\pi$$

D
$$\dfrac{\pi}{4}$$

Solution

$$\displaystyle I=\int _{ 0 }^{ a }{ \cfrac { dx }{ x+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } } $$
Substituting $$x=a\sin t\Rightarrow dx=a\cos tdt$$
$$\displaystyle I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { \cos tdt }{ \sin t+\cos t } } $$ ...(1)
Using property $$\int _{ a }^{ b }{ f\left( x \right) dx } =\int _{ a }^{ b }{ f\left( a+b-x \right) dx }$$
$$\displaystyle I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { \sin tdt }{ \cos t+\sin t } } $$ ...(2)
Adding (1) and (2), we get
$$\displaystyle 2I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ dt } =\cfrac { \pi }{ 2 } \Rightarrow I=\cfrac { \pi }{ 4 } $$
Question 3
1 / -0

$$\displaystyle \int_{0}^{\pi/2}\sqrt{\cos x}\sin^{5}xdx=$$

A
$$\displaystyle \frac{34}{231}$$

B
$$\displaystyle \frac{64}{231}$$

C
$$\displaystyle \frac{30}{321}$$

D
$$\displaystyle \frac{128}{231}$$

Solution

$$\int_{0}^{\dfrac{\pi}{2}} \sqrt{cos x}sin^5 x   dx$$
$$\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x}(sin^4x)sin x   dx$$
$$-\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos   x}(1-cos^2 x)^2d cos x$$
$$-\left [ \int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x} dcos x+\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x} cos^4 x dx - 2\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x} cos^2 x   dx \right ]$$
$$-\left [ \dfrac{2}{3} cos^{\dfrac{3}{2}} x \int_{0}^{\dfrac{\pi}{2}}+\dfrac{2}{11} cos^{\dfrac{11}{2}}x\int_{0}^{\dfrac{\pi}{2}}-2(\dfrac{2}{7}) cos^{\dfrac{7}{2}}x \int_{0}^{\dfrac{\pi}{2}} \right ]$$
$$=-\left [\dfrac{2}{3} (-1)+\dfrac{2}{11}(-1)-2(\dfrac{2}{7})(-1)\right ]$$
$$=\dfrac{64}{231}$$
Question 4
1 / -0

$$\displaystyle \int_{0}^{\pi/2}\frac{1}{a+bcosx}dx=$$, where $$a>|b|$$

A
$$\displaystyle \frac{2}{\sqrt{a^{2}-b^{2}}}\tan^{{-1}}\sqrt{\frac{a+b}{a-b}}$$

B
$$\displaystyle \frac{2}{\sqrt{a^{2}-b^{2}}}cot^{-l} \sqrt{\frac{a-b}{a+b}}$$

C
$$\displaystyle \frac{2}{\sqrt{a^{2}-b^{2}}}\tan^{{-1}}\sqrt{\frac{a-b}{a+b}}$$

D
$$\displaystyle \frac{\pi}{\sqrt{a^{2}-b^{2}}}$$

Solution

Let $$\displaystyle I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { 1 }{ a+b\cos x } dx } $$
$$\tan\left( \cfrac { x }{ 2 } \right) =t\Rightarrow \cfrac { 1 }{ 2 } { \sec }^{ 2 }\left( \cfrac { x }{ 2 } \right) dx=dt$$
Gives $$\cos x=\cfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } } ,dx=\cfrac { 2dt }{ 1+{ t }^{ 2 } } $$
$$\displaystyle I=\int _{ 0 }^{ 1 }{ \cfrac { 1 }{ a+b\left(\cfrac { 1-{ t }^{ 2 } }{ 1+{ t }^{ 2 } }\right) } \cfrac { 2dt }{ 1+{ t }^{ 2 } } } =\cfrac { 2 }{ a+b } \int _{ 0 }^{ 1 }{ \cfrac { 1 }{ \cfrac { { t }^{ 2 }\left( a-b \right) }{ a+b } +1 } dt } $$
Substitute $$\cfrac { t\sqrt { a-b } }{ \sqrt { a+b } } =u\Rightarrow \cfrac { dt\sqrt { a-b } }{ \sqrt { a+b } } =du$$

$$\displaystyle I=\cfrac { 2 }{ \sqrt { a-b } \sqrt { a+b } } \int _{ 0 }^{ \cfrac { \sqrt { a-b } }{ \sqrt { a+b } } }{ \cfrac { 1 }{ 1+{ u }^{ 2 } } du }$$

$$ ={ \left[ \cfrac { 2{ \tan }^{ -1 }u }{ \sqrt { a-b } \sqrt { a+b } } \right] }_{ 0 }^{ \cfrac { \sqrt { a-b } }{ \sqrt { a+b } } }
=\cfrac { 2{ \tan }^{ -1 }\cfrac { \sqrt { a-b } }{ \sqrt { a+b } } }{ \sqrt { a-b } \sqrt { a+b } } -0\\ =\cfrac { 2 }{ \sqrt { { a }^{ 2 }-{ b }^{ 2 } } } { \tan }^{ -1 }\sqrt{\cfrac {a-b}{a+b}} $$
Question 5
1 / -0

$$\displaystyle \int_{1}^{2}\mathrm{x}^{2\mathrm{x}}[1+\log \mathrm{x}]\mathrm{d}\mathrm{x}=$$

A
$$\displaystyle \frac{9}{2}$$

B
$$\displaystyle \frac{11}{2}$$

C
$$\displaystyle \frac{13}{2}$$

D
$$\displaystyle \frac{15}{2}$$

Solution

Let $$ \displaystyle {x}^{x} = t $$
Diiferentiating both sides,
$$ \displaystyle {x}^{x}(1 + \log x) dx = dt $$
Substituting the above obtained values back into the expression,
$$ \displaystyle \int {x}^{2x}(1 + log x) dx = \int t dt $$
$$ \displaystyle = \frac{{t}^{2}}{2} + c $$
Putting the value of t back, we get the integral as
$$ \displaystyle \frac{{({x}^{x})}^{2}}{2} + c $$
Substituting the limits from 1 to 2, we get
$$= \displaystyle \left( \frac{{({2}^{2})}^{2}}{2} \right) - \left( \frac{{({1}^{1})}^{2}}{2}\right) $$
.
$$ \displaystyle = 8 - \frac{1}{2} $$
.
$$ \displaystyle = \frac{15}{2} $$
Question 6
1 / -0

lf $$\displaystyle \int_{0}^{\infty}e^{-\mathrm{x}^{2}}\mathrm{d}\mathrm{x}=\frac{\sqrt{\pi}}{2}$$, then $$\displaystyle \int_{0}^{\infty}e^{-ax^{2}}dx,\ \mathrm{a}>0$$ is

A
$$\displaystyle \frac{\sqrt{\pi}}{2}$$

B
$$\displaystyle \frac{\sqrt{\pi}}{2a}$$

C
$$2 \displaystyle \frac{\sqrt{\pi}}{a}$$

D
$$\displaystyle \frac{1}{2}\sqrt{\frac{\pi}{a}}$$

Solution

$$\int _{ 0 }^{ \infty }{ { e }^{ { -ax }^{ 2 } } } dx\\ =\int _{ 0 }^{ \infty }{ { e }^{ -(\sqrt { a } x)^{ 2 } } } dx$$
Let, $$\sqrt { a } .{ x }=1$$
$$\sqrt { a } .dx=dt\\ dx=\cfrac { dt }{ \sqrt { a } } \\ I=\int _{ 0 }^{ \infty }{ e^{ -t^{ 2 } } } .\cfrac { dt }{ \sqrt { a } } \\ =\cfrac { 1 }{ \sqrt { a } } \int _{ 0 }^{ \infty }{ e^{ -t^{ 2 } } } dt\\ =\cfrac { 1 }{ \sqrt { a } } \times \cfrac { \sqrt { \pi } }{ 2 } \\ =\cfrac { 1 }{ 2 } \sqrt { \cfrac { \pi }{ a } } $$
Hence the correct answer is $$\cfrac { 1 }{ 2 } \sqrt { \cfrac { \pi }{ a } } $$

Question 7

1 / -0

Observe the following Lists

List-I	List-II
A: $$\displaystyle \int_{-2}^{2}\frac{1}{4+x^{2}}dx$$	1) $$\displaystyle \frac{\pi}{3}$$
B: $$\displaystyle \int_{1}^{2}\frac{1}{x\sqrt{x^{2}-1}}dx$$	2) 0
C: $$\displaystyle \int_{0}^{\pi}\cos 3x.\cos 2xdx$$	3) $$\displaystyle \frac{\pi}{4}$$
	4) $$\displaystyle \frac{\pi}{2}$$

A-3, B-1, C-4

A-3, B-1, C-2

A-1, B-3, C-2

A-4, B-1, C-2

Solution

A) $$\displaystyle \int _{ -2 }^{ 2 }{ \frac { 1 }{ 4+{ x }^{ 2 } } } dx=\left[ \frac { 1 }{ 2 } \tan ^{ -1 }{ \frac { x }{ 2 } } \right] _{ -2 }^{ 2 }$$
$$\displaystyle =\left[ \frac { 1 }{ 2 } \tan ^{ -1 }{ 1 } -\frac { 1 }{ 2 } \tan ^{ -1 }{ \left( -1 \right) } \right] =\frac { \pi }{ 8 } +\frac { \pi }{ 8 } =\frac { \pi }{ 4 } $$

B) $$\displaystyle \int _{ 1 }^{ 2 }{ \frac { 1 }{ x\sqrt { { x }^{ 2 }-1 } } } dx=\left[ \sec ^{ -1 }{ x } \right] _{ 1 }^{ 2 }$$
$$\displaystyle =\left[ \sec ^{ -1 }{ 2 } -\sec ^{ -1 }{ 1 } \right] =\left[ \frac { \pi }{ 3 } -0 \right] =\frac { \pi }{ 3 } $$

C) $$\displaystyle \int _{ 0 }^{ \pi }{ \cos { 3x } \cos { 2x } } dx=\frac { 1 }{ 2 } \int _{ 0 }^{ \pi }{ \left( \cos { x } +\cos { 5x } \right) } dx$$
$$\displaystyle =\frac { 1 }{ 10 } \left[ 5\sin { x } +\sin { 5x } \right] _{ 0 }^{ \pi }=0$$

Question 8
1 / -0

If $$\displaystyle \int_{0}^{\pi/3}\frac{\cos x}{3+4\sin x}dx=k\log(\frac{3+2\sqrt{3}}{3})$$, then $$\mathrm{k}$$ is equal to

A
$$\dfrac{1}{2}$$

B
$$\dfrac{1}{3}$$

C
$$\dfrac{1}{4}$$

D
$$\dfrac{1}{8}$$

Solution

$$\displaystyle\int_{0}^{\pi/3}\dfrac{\cos{x}}{3+4\sin{x}}dx$$

Let $$z=3+4\sin{x}\implies dz=4\cos{x}dx\implies \cos{x}dx=\dfrac{dz}{4}$$

When $$ x=0,z=3$$ and when $$x=\dfrac{\pi}{3},z=3+2\sqrt{3}$$

Hence, integration becomes:-

$$\dfrac{1}{4}\displaystyle\int_{3}^{3+2\sqrt{3}}\dfrac{1}{z} dz$$

$$=\dfrac{1}{4}\left[\log{z}\right]_{3}^{3+2\sqrt{3}}$$

$$=\dfrac{1}{4}\log{\left(\dfrac{3+2\sqrt{3}}{3}\right)}$$

$$\implies k=\dfrac{1}{4}$$

Hence, answer is option-(C).
Question 9
1 / -0

The value of the integral
$$\displaystyle \int_{0}^{3\alpha}\text{cosec}(x -\alpha)\text{cosec}(x-2\alpha) dx$$ is

A
$$2\displaystyle \sec\alpha\log(\frac{1}{2}\text{cosec}\alpha)$$

B
$$2\displaystyle \sec\alpha\log(\frac{1}{2}\sec\alpha)$$

C
$$2 \text{cosec}\alpha\log(\sec\alpha)$$

D
$$2 \displaystyle \text{cosec}\alpha\log(\frac{1}{2}\sec\alpha)$$

Solution

$$\displaystyle \int_{0}^{3\alpha}\text{cosec}(x -\alpha)\text{cosec}(x-2\alpha) dx$$

$$=\displaystyle \int_{0}^{3\alpha}\frac{1}{\sin (x-\alpha)\sin (x-2\alpha)}dx$$

$$\displaystyle \frac{1}{\sin \alpha}\int_{0}^{3\alpha} \displaystyle \frac{\sin [(x-\alpha)-(x-2\alpha)]}{\sin (x-\alpha)\sin (x-2\alpha)}dx$$

$$\Rightarrow \displaystyle \frac{1}{\sin \alpha}\int_{0}^{3\alpha}[\cot(x-2\alpha)-\cot(x-\alpha)]dx$$

$$=\displaystyle \frac{1}{\sin \alpha}(\log [\sin (x-2\alpha)]|_{0}^{3\alpha}-\log (\sin (x-\alpha))|_{0}^{3\alpha})$$

$$=\displaystyle \frac{1}{\sin \alpha}[\log (\sin \alpha)-\log (\sin (-2\alpha))-\log (\sin (2\alpha))+\log (\sin (-\alpha))]$$

$$=2\displaystyle \frac{1}{\sin \alpha}\log \left | \displaystyle \frac{\sin \alpha}{2\sin \alpha \cos \alpha} \right |$$

$$=2 \text{cosec}\alpha\ \log\left(\dfrac{1}{2} \sec\alpha\right)$$
Question 10
1 / -0

If $$I_{n}=\displaystyle \int_{0}^{\frac{\pi}{4}}\tan^{n} dx,$$ then $$\dfrac{1}{I_{2}+I_{4}},\dfrac{1}{I_{3}+I_{5}},\dfrac{1}{I_{4}+I_{6}}\cdots$$ form

A
an A.P

B
a G.P

C
a H.P

D
an AGP

Solution

$$ I_n = \displaystyle \int_{0}^{\pi/4}\tan^{n}x\ dx$$

$$I_{n+2}=\displaystyle \int_{0}^{\pi/4}\tan^{n}x(\tan^{2}x)dx$$

$$=\displaystyle \int_{0}^{\pi/4}\tan^{n}x(\sec^{2}x-1)dx$$

$$=\displaystyle \int_{0}^{\pi/4}\tan^{n}x\ d(\tan x)-I_n$$

$$\Rightarrow I_{n+2} + I_n=\dfrac{1}{n+1}$$

$$I_{2}+I_{4}=\dfrac13$$

$$I_{3}+I_{5}=\dfrac14$$

$$I_{4}+I_{6}=\dfrac15$$

$$So,\dfrac{1}{I_{2}+I_{4}};\dfrac{1}{I_{3}+I_{5}};\dfrac{1}{I_{4}+I_{6}}$$ forms an AP

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Integrals Test - 46

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Integrals Test - 46

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SHARING IS CARING

Question 1

$$\displaystyle \frac{\sqrt{2}}{3}$$

$$-\displaystyle \frac{\sqrt{2}}{3}$$

$$\displaystyle \frac{\sqrt{3}}{1}$$

$$\displaystyle \frac{1}{\sqrt{3}}$$

Solution

Question 2

$$\pi$$

$$\dfrac{\pi}{3}$$

$$-\pi$$

$$\dfrac{\pi}{4}$$

Solution

Question 3

$$\displaystyle \frac{34}{231}$$

$$\displaystyle \frac{64}{231}$$

$$\displaystyle \frac{30}{321}$$

$$\displaystyle \frac{128}{231}$$

Solution

Question 4

$$\displaystyle \frac{2}{\sqrt{a^{2}-b^{2}}}\tan^{{-1}}\sqrt{\frac{a+b}{a-b}}$$

$$\displaystyle \frac{2}{\sqrt{a^{2}-b^{2}}}cot^{-l} \sqrt{\frac{a-b}{a+b}}$$

$$\displaystyle \frac{2}{\sqrt{a^{2}-b^{2}}}\tan^{{-1}}\sqrt{\frac{a-b}{a+b}}$$

$$\displaystyle \frac{\pi}{\sqrt{a^{2}-b^{2}}}$$

Solution

Question 5

$$\displaystyle \frac{9}{2}$$

$$\displaystyle \frac{11}{2}$$

$$\displaystyle \frac{13}{2}$$

$$\displaystyle \frac{15}{2}$$

Solution

Question 6

$$\displaystyle \frac{\sqrt{\pi}}{2}$$

$$\displaystyle \frac{\sqrt{\pi}}{2a}$$

$$2 \displaystyle \frac{\sqrt{\pi}}{a}$$

$$\displaystyle \frac{1}{2}\sqrt{\frac{\pi}{a}}$$

Solution

Question 7

A-3, B-1, C-4

A-3, B-1, C-2

A-1, B-3, C-2

A-4, B-1, C-2

Solution

Question 8

$$\dfrac{1}{2}$$

$$\dfrac{1}{3}$$

$$\dfrac{1}{4}$$

$$\dfrac{1}{8}$$

Solution

Question 9

$$2\displaystyle \sec\alpha\log(\frac{1}{2}\text{cosec}\alpha)$$

$$2\displaystyle \sec\alpha\log(\frac{1}{2}\sec\alpha)$$

$$2 \text{cosec}\alpha\log(\sec\alpha)$$

$$2 \displaystyle \text{cosec}\alpha\log(\frac{1}{2}\sec\alpha)$$

Solution

Question 10

an A.P

a G.P

a H.P

an AGP

Solution

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