Integrals Test

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Integrals Test - 32

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Weekly Quiz Competition

Question 1
1 / -0

What is $$\displaystyle \int \dfrac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} dx$$ equal to?

A
$$\sqrt{\dfrac{x^4 + x^2 + 1}{x}} + c$$

B
$$\sqrt{x^4 + 2 - \dfrac{1}{x^2}} + c$$

C
$$\sqrt{x^2 + \dfrac{1}{x^2} + 1} + c$$

D
$$\sqrt{\dfrac{x^4 - x^2 + 1}{x}} + c$$

Solution

$$I=\displaystyle \int \dfrac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} dx$$
$$=\displaystyle \int \dfrac{x^4 - 1}{x^3 \sqrt{x^2 + 1 + \dfrac{1}{x^2}}} dx$$
$$=\displaystyle \int \dfrac{x - \dfrac{1}{x^3}}{\sqrt{x^2 + 1 + \dfrac{1}{x^2}}} dx$$

Put $$x^2 + 1 + \dfrac{1}{x^2}=t$$
$$\therefore \left(2x-\dfrac{2}{x^3}\right)dx=dt$$

$$\therefore \left(x-\dfrac{1}{x^3}\right)dx=\dfrac{dt}{2}$$

$$\therefore I=\displaystyle \int \dfrac{dt}{2\sqrt t}$$

$$=\sqrt t +C$$

$$=\sqrt{x^2 + 1 + \dfrac{1}{x^2}}+C$$

Answer is option (C)
Question 2
1 / -0

The value of $$\displaystyle\int _{ 0 }^{ { x }/{ 4 } }{ \dfrac { \sec { x } }{ { \left( \sec { x } +\tan { x } \right) }^{ 2 } } dx } $$ is

A
$$1+\sqrt{2}$$

B
$$-11+\sqrt{2}$$

C
$$-\sqrt{2}$$

D
None of these

Solution

$$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \cfrac { \sec { x } }{ { \left( \sec { x } +\tan { x } \right) }^{ 2 } } } dx$$

$$=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \cfrac { { 1 }/{ \cos { x } } }{ { \left[ \cfrac { 1 }{ \cos { x } } +\cfrac { \sin { x } }{ \cos { x } } \right] }^{ 2 } } } dx$$

$$=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ \cfrac { { \cos ^{ 2 }{ x } }/{ \cos { x } } }{ { \left( 1+\sin { x } \right) }^{ 2 } } } dx$$
$$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { \left( 1+\sin { x } \right) }^{ -2 } } \cos { x } dx$$
Take $$1+\sin x=t \implies \cos x dx=dt$$
When $$x=0, t=1$$ and $$x=\dfrac{\pi}{4}, t=1+\dfrac{1}{\sqrt{2}}$$

$$\therefore \quad I=\int_{1}^{1+\frac{1}{\sqrt{2}}} t^{-2}\ dt$$
$$\therefore \quad I=\left|\dfrac{t^{-1}}{-1}\right|_{1}^{1+\frac{1}{\sqrt{2}}}$$

$$=-\left[ \cfrac { 1 }{ 1+\frac { 1 }{ \sqrt { 2 } } } -1 \right] $$
$$=-\cfrac { \sqrt { 2 } }{ \sqrt { 2 } +1 } +1$$$$=\cfrac { -\sqrt { 2 } +\sqrt { 2 } +1 }{ \sqrt { 2 } +1 } =\cfrac { 1 }{ \sqrt { 2 } +1 } $$
$$=\cfrac { 1 }{ \sqrt { 2 } +1 } \ast \cfrac { \sqrt { 2 } -1 }{ \sqrt { 2 } -1 } =\cfrac { \sqrt { 2 } -1 }{ 2-1 } $$
$$=\sqrt { 2 } -1$$
$$\therefore $$ None of these is correct option
Question 3
1 / -0

What is $$\displaystyle \int_0^1 {\frac{\tan^{-1}x}{1 + x^2} dx}$$ equal to ?

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

B
$$\dfrac{\pi}{8}$$

C
$$\dfrac{\pi^2}{8}$$

D
$$\dfrac{\pi^2}{32}$$

Solution

Let $$I=\displaystyle \int _{ 0 }^{ 1 } \dfrac { {\tan^{ -1 } }x }{ { 1+x^{ 2 } } } dx$$

Let, $$y={ {\tan^{ -1 } }x }$$ $$\Rightarrow dy=\dfrac { 1 }{ { 1+x^{ 2 } } } dx$$
Now for $$ x=1, y=\dfrac { \pi }{ 4 } $$ and for $$x=0, y=0$$
Putting the values in the integral we have,

$$I=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ y.dy }$$

$$= { \left[\dfrac { y^{ 2 } }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 4 } }$$

$$ =\dfrac { \pi ^{ 2 } }{ 32 } $$
Hence, D is correct.
Question 4
1 / -0

If $$I=\displaystyle\int _{ { 1 }/{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } \cdot \sin { \left( x-\dfrac { 1 }{ x } \right) } dx } $$, then $$I$$ is equal to

A
$$0$$

B
$$\pi$$

C
$$\pi - \dfrac{1}{\pi}$$

D
$$\pi + \dfrac{1}{\pi}$$

Solution

$$\displaystyle{I=\int _{ \frac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dx.....(i)}$$

Let $$t=\dfrac { 1 }{ x } $$

$$dt=-\dfrac { 1 }{ { x }^{ 2 } } dx\\ dx={ -x }^{ 2 }dt$$

Substituting in $$(i)$$

$$\displaystyle{I=\int _{ \pi }^{ \frac { 1 }{ \pi } }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } (-x^{ 2 })dt\\ \\ I=-\int _{ \pi }^{ \frac { 1 }{ \pi } }{ x } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dt\\ I=-\int _{ \pi }^{ \frac { 1 }{ \pi } }{ \dfrac { 1 }{ t } } .\sin { \left( \dfrac { 1 }{ t } -t \right) } dt\\ I=-\int _{ \pi }^{ \frac { 1 }{ \pi } }{ \dfrac { 1 }{ t } } .\sin { \left( -(t-\dfrac { 1 }{ t } ) \right) } dt}$$

using $$\sin { (-x)=-\sin { x } } $$

$$\displaystyle{I=\int _{ \pi }^{ \frac { 1 }{ \pi } }{ \dfrac { 1 }{ t } } .\sin { \left( t-\dfrac { 1 }{ t } \right) } dt}$$

Using $$\displaystyle{\int _{ a }^{ b }{ f\left( x \right) dx=-\int _{ b }^{ a }{ f\left( x \right) dx } } }$$ and changing the variable from $$t$$ to $$x$$

$$\displaystyle{I=-\int _{ \frac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dx.....(ii)}$$

Adding $$(i)$$ and $$(ii)$$

$$\displaystyle{2I=\int _{ \frac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dx-\int _{ \frac { 1 }{ \pi } }^{ \pi }{ \dfrac { 1 }{ x } } .\sin { \left( x-\dfrac { 1 }{ x } \right) } dx=0\\ \Rightarrow I=0}$$

So option $$(A)$$ is correct.
Question 5
1 / -0

$$\int { \left( \cfrac { 4{ e }^{ x }-25 }{ 2{ e }^{ x }-5 } \right) } dx=Ax+B\log { \left| 2{ e }^{ x }-5 \right| } +c$$, then

A
$$A=5,B=3$$

B
$$A=5,B=-3$$

C
$$A=-5,B=3$$

D
$$A=-5,B=-3$$

Solution

Given : $$\displaystyle \int { \left( \cfrac { 4{ e }^{ x }-25 }{ 2{ e }^{ x }-5 } \right) } dx=Ax+B\log { \left| 2{ e }^{ x }-5 \right| } +c$$

$$\implies \displaystyle \int { \left( \cfrac { 10 e^{x} -6{ e }^{ x }-25 }{ 2{ e }^{ x }-5 } \right) } dx=Ax+B\log { \left| 2{ e }^{ x }-5 \right| } +c$$

$$\implies \displaystyle \int {\left(5 - \cfrac { 6{e}^{x}}{ 2{ e }^{ x }-5 } \right) } dx=Ax+B\log { \left| 2{ e }^{ x }-5 \right| } +c$$

$$\implies \displaystyle \int 5. dx - 3\int {\left(\cfrac { 2{e}^{x}}{ 2{ e }^{ x }-5 } \right) } dx=Ax+B\log { \left| 2{ e }^{ x }-5 \right| } +c$$

$$\implies 5x -3 \log|2e^{x}-5|+c=Ax+B\log|2e^{x}-5|+c$$
Equating the like terms we get
$$A=5$$ and $$B=-3$$
Hence, option B is correct.
Question 6
1 / -0

If $$\displaystyle \int { \cfrac { f(x) }{ \log { (\sin { x } ) } } } dx=\log { \left[ \log { \sin { x } } \right] } +c$$, then $$f(x)=$$........

A
$$\cot { x } $$

B
$$\tan { x } $$

C
$$\sec { x } $$

D
$$co\sec { x } $$

Solution

Here, $$\displaystyle \int { \dfrac { f(x) }{ \log (\sin x) } dx } = \log (\log (\sin x))+c$$

R.H.S.=$$\log(\log (\sin x))$$

Differentiating the R.H.S

$$\Rightarrow$$ $$\dfrac { 1 }{ \log (\sin x) } \times \dfrac { 1 }{ \sin x } \times \cos x=\dfrac { \cot x }{ \log (\sin x) }$$

$$ \Rightarrow f(x)=\cot x$$
Question 7
1 / -0

If $$I_n = \int_0^{\pi/4} tan^n \theta d \theta$$, where n is a positive integer, then $$n(I_{n-1} + I_{n +1})$$ is equal to

A
1

B
n - 1

C
$$\dfrac{1}{n - 1}$$

D
None of these

Solution

$${ I }_{ n }=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { \left( { \tan { x } } \right) }^{ n }dx } $$
$$ { I }_{ n }^{ }=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { \left( { \tan { x } } \right) }^{ n-2 }\left( { \left( \sec { x } \right) }^{ 2 }-1 \right) dx } $$
$$ { I }_{ n }=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { \left( { \tan { x } } \right) }^{ n-2 } } .{ \left( \sec { x } \right) }^{ 2 }dx-\int _{ 0 }^{ \frac { \pi }{ 4 } }{ { \left( { \tan { x } } \right) }^{ n-2 } } dx$$
$$ Let\ \ J=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { \left( { \tan { x } } \right) }^{ n-2 } } .{ \left( \sec { x } \right) }^{ 2 }dx$$
$$ Let\ \ z=tanx$$
$$ \Rightarrow dz={ \left( \sec { x } \right) }^{ 2 }dx$$
$$ J=\displaystyle \int _{ 0 }^{ 1 }{ { z }^{ n-2 }dz } $$
$$ J=\frac { 1 }{ n-1 } $$
$$ { I }_{ n }=J-{ I }_{ n-2 }^{ }$$
$$ { I }_{ n }=\frac { 1 }{ n-1 } -{ I }_{ n-2 }^{ }$$
$$ \left( n-1 \right) { I }_{ n }+\left( n-1 \right) { I }_{ n-2 }^{ }=1$$
$$ n\rightarrow n+1$$
$$ \Rightarrow n{ I }_{ n+1 }^{ }+n{ I }_{ n-1 }^{ }=1$$
Question 8
1 / -0

The solution of the equation $$\dfrac { dy }{ dx } =\dfrac { x\left( 2\log { x } +1 \right) }{ \sin { y } +y\cos { y } } $$ is

A
$$y\sin { y } ={ x }^{ 2 }\log { x } +\dfrac { { x }^{ 2 } }{ y } +c$$

B
$$y\cos { y } ={ x }^{ 2 }\left( \log { x } +1 \right) +c$$

C
$$y\cos { y } ={ x }^{ 2 }\log { x } +\dfrac { { x }^{ 2 } }{ 2 } +c$$

D
$$y\sin { y } ={ x }^{ 2 }\log { x } +c$$

Solution

Given equation is $$\dfrac {dy}{dx}=\dfrac {x(2\log x+1)}{\sin y+y \cos y}$$
Therefore, $$\left( y\cos { y } +\sin { y } \right) dy=\left( 2x\log { x } +x \right) dx$$
$$\Rightarrow y\sin { y } -\displaystyle\int { \sin { y } dy } +\displaystyle\int { \sin { y } dx } ={ x }^{ 2 }\log { x } -\displaystyle\int { { x }^{ 2 }\cdot \dfrac { 1 }{ x } dx } +\displaystyle\int { xdx } +c$$
$$\Rightarrow y\sin { y } ={ x }^{ 2 }\log { x } +c$$
Question 9
1 / -0

The value of $$\displaystyle \int_{1/n}^{(an - 1)/n} \dfrac {\sqrt {x}}{\sqrt {a - x} + \sqrt {x}} dx$$ is equal to

A
$$\dfrac {a}{2}$$

B
$$\dfrac {n\cdot a + 2}{2n}$$

C
$$\dfrac {n\cdot a - 2}{2n}$$

D
$$\dfrac {n\cdot a}{2}$$

Solution

Let, $$I =\displaystyle \int_{1/2}^{(an - 1)/n} \dfrac {\sqrt {x}}{\sqrt {a - x} + \sqrt {x}}dx$$
$$=\displaystyle \int_{\dfrac {1}{n}}^{a - \dfrac {1}{n}} \dfrac {\sqrt {x}}{\sqrt {a - x} + \sqrt {x}} dx$$ ....(i)
$$= \displaystyle \int_{\dfrac {1}{n}}^{a - \dfrac {1}{n}} \dfrac {\left (\sqrt {\dfrac {1}{n} + a - \dfrac {1}{n} - x}\right )}{\sqrt {a - \left (\dfrac {1}{n} + a - \dfrac {1}{n} - x\right )} + \sqrt {\dfrac {1}{n} + a - \dfrac {1}{n} - x}}dx$$
$$\Rightarrow I =\displaystyle \int_{\dfrac {1}{n}}^{a - \dfrac {1}{n}} \dfrac {\sqrt {a - x}}{\sqrt {x} + \sqrt {a - x}} dx$$ .... (ii)
On adding Eqs. (i) and (ii), we get
$$2I =\displaystyle \int_{\dfrac {1}{n}}^{a - \dfrac {1}{n}} 1dx = [x]_{\dfrac {1}{n}}^{a - \dfrac {1}{n}}$$
$$= a - \dfrac {1}{n} - \dfrac {1}{n} = \dfrac {na - 2}{n}\Rightarrow I = \dfrac {na - 2}{2n}$$.
Question 10
1 / -0

The value of $$\displaystyle\int _{ 0 }^{ \pi }{ \dfrac { dx }{ 5+3\cos { x } } } $$ is

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

B
$$\dfrac{ \pi }{ 8 }$$

C
$$\dfrac{ \pi }{ 2 }$$

D
Zero

Solution

Consider, $$\dfrac{1}{5+3\cos x}$$
$$=\dfrac{1}{5+\dfrac{3-3\tan^{2}(\frac{x}{2})}{1+\tan^{2}(\frac{x}{2})}}$$ ...... $$\left[\because \cos 2x=\dfrac{1-\tan^{2}x}{1+\tan^{2}x}\right]$$

$$=\dfrac{1+\tan^{2}\frac{x}{2}}{5+5\tan^{2}\frac{x}{2}+3-3\tan^{2}\frac{x}{2}}$$

$$=\dfrac{\sec^{2}\frac{x}{2}}{8+2\tan^{2}\frac{x}{2}}$$

Now, $$\displaystyle \int_{0}^{\pi}\dfrac{1}{5+3\cos x}$$

$$=\displaystyle \int_{0}^{\pi} \dfrac{\sec^{2}\frac{x}{2}}{8+2\tan^{2}\frac{x}{2}}dx$$
Take $$\tan\dfrac{x}{2}=t$$$$\implies \dfrac{1}{2}.\sec^{2}\dfrac{x}{2}dx=dt$$

Then, we get
$$\displaystyle \int \dfrac{dt}{4+t^{2}}dx$$

$$=\dfrac{1}{2}\left[\tan^{-1}\left(\dfrac{t}{2}\right)\right]$$
$$=\dfrac{1}{2}\left[\tan^{-1}\left(\dfrac{\tan(\frac{x}{2})}{2}\right)\right]_{0} ^{\pi}$$

$$=\dfrac{1}{2}\left(\dfrac{\pi}{2}\right)$$

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

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Integrals Test - 32

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Integrals Test - 32

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Question 1

$$\sqrt{\dfrac{x^4 + x^2 + 1}{x}} + c$$

$$\sqrt{x^4 + 2 - \dfrac{1}{x^2}} + c$$

$$\sqrt{x^2 + \dfrac{1}{x^2} + 1} + c$$

$$\sqrt{\dfrac{x^4 - x^2 + 1}{x}} + c$$

Solution

Question 2

$$1+\sqrt{2}$$

$$-11+\sqrt{2}$$

$$-\sqrt{2}$$

None of these

Solution

Question 3

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

$$\dfrac{\pi}{8}$$

$$\dfrac{\pi^2}{8}$$

$$\dfrac{\pi^2}{32}$$

Solution

Question 4

$$0$$

$$\pi$$

$$\pi - \dfrac{1}{\pi}$$

$$\pi + \dfrac{1}{\pi}$$

Solution

Question 5

$$A=5,B=3$$

$$A=5,B=-3$$

$$A=-5,B=3$$

$$A=-5,B=-3$$

Solution

Question 6

$$\cot { x } $$

$$\tan { x } $$

$$\sec { x } $$

$$co\sec { x } $$

Solution

Question 7

1

n - 1

$$\dfrac{1}{n - 1}$$

None of these

Solution

Question 8

$$y\sin { y } ={ x }^{ 2 }\log { x } +\dfrac { { x }^{ 2 } }{ y } +c$$

$$y\cos { y } ={ x }^{ 2 }\left( \log { x } +1 \right) +c$$

$$y\cos { y } ={ x }^{ 2 }\log { x } +\dfrac { { x }^{ 2 } }{ 2 } +c$$

$$y\sin { y } ={ x }^{ 2 }\log { x } +c$$

Solution

Question 9

$$\dfrac {a}{2}$$

$$\dfrac {n\cdot a + 2}{2n}$$

$$\dfrac {n\cdot a - 2}{2n}$$

$$\dfrac {n\cdot a}{2}$$

Solution

Question 10

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

$$\dfrac{ \pi }{ 8 }$$

$$\dfrac{ \pi }{ 2 }$$

Zero

Solution

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