Integrals Test

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Integrals Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

Evaluate the integral
$$\displaystyle \int_{0}^{\pi/4}\frac{\sin x+\cos x}{3+\sin 2x} dx$$

A
$$-\displaystyle \frac{1}{4} \log 3$$

B
$$\displaystyle \frac{1}{4} \log 3$$

C
$$-\displaystyle \frac{1}{3} \log 4$$

D
$$\displaystyle \frac{1}{2}\log 3$$

Solution

$$\displaystyle \int_{0}^{\pi/2}\dfrac{sinx+cosx}{3+sin2x}dx.$$
$$\displaystyle \int_{0}^{\pi/2}\dfrac{sinx+cosx}{4-(sinx-cosx)^{2}}$$
$$dx= \displaystyle \int_{0}^{\pi/2}\dfrac{d(sinx-cosx)}{(2)^{2}-(sinx-cosx)^{2}}$$
$$=\dfrac{1}{4}\log\left | \dfrac{2+(sinx-cosx)}{2-(sinx-cosx)} \right |_{0}^{\pi}$$
$$=\dfrac{1}{4}\left ( \log\left | \dfrac{2+0}{2-0} \right | -\log\left|\dfrac{2-1}{2+1}\right | \right )$$
$$=\dfrac{1}{4} (-\log \dfrac{1}{3})$$
$$= \dfrac{1}{4} \log(3)$$
Question 2
1 / -0

Evaluate: $$\displaystyle \int_{0}^{1}\frac{x^{3}}{1+x^{8}} dx$$

A
$$\displaystyle \frac{\pi}{4}$$

B
$$\displaystyle \frac{\pi}{8}$$

C
$$\displaystyle \frac{\pi}{16}$$

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

Solution

Let, $$z=x^4\implies dz=4x^3dx\implies x^3dx=\dfrac{dz}{4}$$

When $$ x=0,z=0$$ and when $$x=1,z=1$$

Hence,integration becomes:-

$$\displaystyle\dfrac{1}{4}\int_{0}^{1} \dfrac{dz}{1+z^2}$$

$$ =\dfrac{1}{4}\displaystyle\left[\tan^{-1}{z}\right]_{0}^{1}$$

$$=\dfrac{1}{4}\displaystyle\left[\dfrac{\pi}{4}-0\right]$$

$$=\dfrac{\pi}{16}$$

Hence, answer is option-(C).
Question 3
1 / -0

The value of $$x$$ such that $$\displaystyle \int_{\sqrt{2}}^{\mathrm{x}}\frac{1}{x\sqrt{x^{2}-1}}dx=\frac{\pi}{12}$$ is

A
$$\mathrm{x}=3$$

B
$$\mathrm{x}=4$$

C
$$\mathrm{x}=1$$

D
$$\mathrm{x}=2$$

Solution

Let $$ x=\sec{\theta} \implies dx=\sec{\theta}\tan{\theta}d\theta$$

When $$ x=\sqrt{2}, \theta=\dfrac{\pi}{4}$$ and when $$x=x, \theta=\sec^{-1} {x}$$

Now, $$\displaystyle\int_{\pi/4}^{\sec^{-1} {x}} \dfrac{1}{\sec{\theta} \sqrt{\sec^2{\theta} -1}}\sec{\theta}\tan{\theta}d\theta=\dfrac{\pi}{12}$$

$$\implies\displaystyle\int_{\pi/4}^{\sec^{-1} {x}} \dfrac{1}{\sec{\theta} \tan{\theta}}\sec{\theta}\tan{\theta}d\theta=\dfrac{\pi}{12}$$

$$\implies\displaystyle\int_{\pi/4}^{\sec^{-1} {x}} d\theta=\dfrac{\pi}{12}$$

$$\implies\sec^{-1} {x} - \dfrac{\pi}{4}=\dfrac{\pi}{4}$$

$$\implies\sec^{-1} {x}=\dfrac{\pi}{4}+\dfrac{\pi}{12}=\dfrac{\pi}{3}$$

$$\implies x=\sec{\dfrac{\pi}{3}}=2$$

$$\implies x=2$$

Hence,answer is option-(D).
Question 4
1 / -0

Evaluate: $$\displaystyle \int_{0}^{\pi/4}\sec^{6}x \ dx$$

A
$$\dfrac {8}{15}$$

B
$$\dfrac {28}{15}$$

C
$$\dfrac {35}{8}$$

D
$$\dfrac {44}{15}$$

Solution

$$\displaystyle\int_{0}^{\pi/4} \sec^6{x}dx$$

$$=\displaystyle\int_{0}^{\pi/4}\sec^4{x}.sec^2{x}dx$$

$$\displaystyle\int_{0}^{\pi/4} \left(\tan^2{x}+1\right)^2 \sec^2{x}dx$$

Let, $$z=\tan{x}\implies dz=\sec^2{x}dx$$

And when $$ x=0, z=0$$ and when $$ x=\dfrac{\pi}{4}, z=1$$

Now, integration becomes:-

$$\displaystyle\int_{0}^{1} \left(z^2+1\right)^2dz$$

$$=\displaystyle\int_{0}^{1} \left(z^4+2z^2+1\right)dz$$

$$=\dfrac{1}{5}\left[z^5\right]_{0}^{1}+\dfrac{2}{3}\left[z^3\right]_{0}^{1}+\left[z\right]_{0}^{1}$$

$$=\dfrac{1}{5}+\dfrac{2}{3}+1$$

$$=\dfrac{28}{15}$$

Hence, answer is option-(B).
Question 5
1 / -0

lf $$f(x)=\left\{\begin{array}{l}e^{\cos x}\sin x, for |x|\leq 2\\2 ; otherwise\end{array}\right.$$, then $$\displaystyle \int_{-2}^{3}f(x)dx$$ is

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

$$I=\displaystyle\int_{-2}^{3}f(x)dx$$

$$I=\displaystyle\int_{-2}^{2}e^{\cos{x}}\sin{x}dx+\int_{2}^{3}2dx$$

consider, $$f(x)=e^{\cos{x}}\sin{x}$$

so, $$f(-x)=e^{\cos{(-x)}}\sin{(-x)}$$

$$\Rightarrow$$ $$f(-x)=-e^{\cos{x}}\sin{x}=-f(x)$$

$$\Rightarrow$$ $$e^{\cos{x}}\sin{x}$$ is an odd function

hence, $$\displaystyle\int_{-2}^{2}e^{\cos{x}}\sin{x}dx=0$$

Thus, $$I=\displaystyle\int_{2}^{3} 2dx$$

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

$$I=2$$

Hence, answer is option-(C).
Question 6
1 / -0

The function $$F(x)=\displaystyle \int_{0}^{x}\log(t+\sqrt{1+t^{2}})dt$$ is

A
a periodic function

B
an even function

C
an odd function

D
even or odd

Solution

$$f(x)=\int_{0}^{x}log(t+\sqrt{1+t^{2}})dt$$
$$f(-x)=\int_{0}^{-x} log(t+\sqrt{1+t^{2}})dt$$
let t=-y
=$$-\int_{0}^{x}log(-y+\sqrt{1+}y^{2})dy$$
$$=-\int_{0}^{x}log(\sqrt{1+y^{2}}-y)dy$$
$$+\int_{0}^{4}log(\sqrt{1+y^{2}}+y)dy$$
$$f(-x)=\int_{0}^{x}log\sqrt{1+t^{2}+t}dt$$
So, f(x)=f(-x) even function
Question 7
1 / -0

The value of $$\displaystyle \int_{0}^{1}\frac{2^{2x+1}-5^{2x-1}}{10^{x}} dx$$ is

A
$$\displaystyle \frac{3}{5}\left[ \displaystyle \frac { 2 }{ \log _{ e } (\displaystyle \frac { 2 }{ 5 } ) } +\displaystyle \frac { 1 }{ 2\log _{ e } (\displaystyle \frac { 5 }{ 2 } ) } \right] $$

B
$$-\displaystyle \frac{3}{5}\left[ \displaystyle \frac { 2 }{ \log _{ e } (\displaystyle \frac { 2 }{ 5 } ) } +\displaystyle \frac { 1 }{ 2\log _{ e } (\displaystyle \frac { 5 }{ 2 } ) } \right] $$

C
$$\displaystyle \frac{3}{5}\left[\displaystyle \frac { 2 }{ \log _{ e } (\displaystyle \frac { 2 }{ 5 } ) } -\displaystyle \frac { 1 }{ 2\log _{ e } (\displaystyle \frac { 5 }{ 2 } ) } \right] $$

D
$$\displaystyle \frac{1}{2}[\log 2-\log 3]$$

Solution

$$\int_{0}^{1}\left ( \dfrac{2^{2x+1}+5^{2x-1}}{10x} \right )dx$$
$$2\int_{0}^{1}\left ( \dfrac{4}{10} \right )^{xdx}-1/5\int_{0}^{1}\dfrac{25}{10}^{x}dx$$
$$=2\dfrac{(\dfrac{4}{10})^{x}}{log\dfrac{(4)}{10}}\int_{0}^{1}-\dfrac{\dfrac{1}{5}(\dfrac{25}{10})}{log(\dfrac{25}{10})}$$
$$=\dfrac{2\left [ (\dfrac{4}{10})-1 \right ]}{log(2/5)}-\dfrac{1/5\left [ (\dfrac{25}{10}-1) \right ]}{log(5/2)}$$
$$\dfrac{2(-3/5)}{l0g(2/5)}-\dfrac{1/5(3/2)}{log(5/2)}$$
$$=-3/5\left [ \dfrac{2}{log(2/5)}+\dfrac{1}{2log}(5/2) \right ].$$
Question 8
1 / -0

lf $$\displaystyle \int_{0}^{1}\frac{\tan^{-1}x}{x}dx=k\int_{0}^{\pi/2}\frac{x}{\sin x}dx$$, then the value of $$k$$ is

A
$$1$$

B
$$\displaystyle \frac{1}{4}$$

C
$$4$$

D
$${2}$$

Solution

Let $$I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { x }{ sinx } dx } $$

Substitute $$tan\left( \cfrac { x }{ 2 } \right) =t\Rightarrow \cfrac { 1 }{ 2 } { sec }^{ 2 }\left( \cfrac { x }{ 2 } \right) dx=dt$$

gives $$sinx=\cfrac { 2t }{ 1+{ t }^{ 2 } } ,dx=\cfrac { 2dt }{ 1+{ t }^{ 2 } } $$

$$I=\int _{ 0 }^{ 1 }{ \cfrac { 2{ tan }^{ -1 }t }{ \cfrac { 2t }{ 1+{ t }^{ 2 } } } \cfrac { 2dt }{ 1+{ t }^{ 2 } } }
=2\int _{ 0 }^{ 1 }{ \cfrac { { tan }^{ -1 }t }{ t } dt } =2\int _{ 0 }^{ 1 }{ \cfrac { { tan }^{ -1 }x }{ x } dx } \\ \Rightarrow k=2$$
Question 9
1 / -0

The value of the integral $$\displaystyle \int_{0}^{\infty}\frac{1}{1+x^{4}} dx$$ is

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

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

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

D
$$\pi/4$$

Solution

$$I=\int _{ 0 }^{ \infty }{ \cfrac { 1 }{ x^{ 4 }+1 } dx } =\int _{ 0 }^{ \infty }{ \left( \cfrac { \sqrt { 2 } x-2 }{ 4\left( -x^{ 2 }+\sqrt { 2 } x-1 \right) } +\cfrac { \sqrt { 2 } x+2 }{ 4\left( x^{ 2 }+\sqrt { 2 } x+1 \right) } \right) dx } \\ =\cfrac { 1 }{ 4 } \int { \cfrac { \sqrt { 2 } x+2 }{ x^{ 2 }+\sqrt { 2 } x+1 } dx } +\cfrac { 1 }{ 4 } \int { \cfrac { \sqrt { 2 } x-2 }{ -x^{ 2 }+\sqrt { 2 } x-1 } dx } \\ =\cfrac { 1 }{ 4\sqrt { 2 } } \int { \cfrac { \sqrt { 2 } +2x }{ x^{ 2 }+\sqrt { 2 } x+1 } dx } +\cfrac { 1 }{ 4 } \int { \cfrac { 1 }{ x^{ 2 }+\sqrt { 2 } x+1 } dx } +\cfrac { 1 }{ 4 } \int { \cfrac { \sqrt { 2 } x-2 }{ -x^{ 2 }+\sqrt { 2 } x-1 } dx } $$
Let $${ I }_{ 1 }=\cfrac { 1 }{ 4\sqrt { 2 } } \int { \cfrac { \sqrt { 2 } +2x }{ x^{ 2 }+\sqrt { 2 } x+1 } dx } $$
Substituting $$t=x^{ 2 }+\sqrt { 2 } x+1\Rightarrow dt=\left( 2x+\sqrt { 2 } \right) dx$$, we get
$${ I }_{ 1 }=\cfrac { 1 }{ 4\sqrt { 2 } } \int { \cfrac { 1 }{ t } dt } =\cfrac { \log { t } }{ 4\sqrt { 2 } } =\cfrac { \log { \left( x^{ 2 }+\sqrt { 2 } x+1 \right) } }{ 4\sqrt { 2 } } $$
Now let $${ I }_{ 2 }=\cfrac { 1 }{ 4 } \int { \cfrac { 1 }{ x^{ 2 }+\sqrt { 2 } x+1 } dx } $$
Substituting $$t=x+\cfrac { 1 }{ \sqrt { 2 } } \Rightarrow dt=dx$$, we get
$${ I }_{ 2 }=\cfrac { 1 }{ 4 } \int { \cfrac { 1 }{ t^{ 2 }+\cfrac { 1 }{ 2 } } dt } =\cfrac { 1 }{ 4 } \int { \cfrac { 2 }{ 2t^{ 2 }+1 } } =tan^{ -1 }\left( \cfrac { \sqrt { 2 } x+1 }{ 2\sqrt { 2 } } \right) $$
And let $${ I }_{ 3 }=\cfrac { 1 }{ 4 } \int { \cfrac { \sqrt { 2 } x-2 }{ -x^{ 2 }+\sqrt { 2 } x-1 } dx } $$
$$=\cfrac { 1 }{ 4 } \int { \left( \cfrac { \sqrt { 2 } x-2 }{ \sqrt { 2 } \left( -x^{ 2 }+\sqrt { 2 } x-1 \right) } -\cfrac { 1 }{ -x^{ 2 }+\sqrt { 2 } x-1 } \right) dx } \\ =-\cfrac { log\left( -x^{ 2 }+\sqrt { 2 } x-1 \right) }{ 4\sqrt { 2 } } -\cfrac { tan\left( 1-\sqrt { 2 } x \right) }{ 2\sqrt { 2 } } \\ $$
Therefore resubstituting $${ { I }=I }_{ 1 }+{ I }_{ 2 }+{ I }_{ 3 }$$
$$\int _{ 0 }^{ \infty }{ \cfrac { 1 }{ x^{ 4 }+1 } dx } =\left[ \cfrac { \log { \left( x^{ 2 }+\sqrt { 2 } x+1 \right) } }{ 4\sqrt { 2 } } +tan^{ -1 }\left( \cfrac { \sqrt { 2 } x+1 }{ 2\sqrt { 2 } } \right) -\cfrac { log\left( -x^{ 2 }+\sqrt { 2 } x-1 \right) }{ 4\sqrt { 2 } } -\cfrac { tan\left( 1-\sqrt { 2 } x \right) }{ 2\sqrt { 2 } } \right] _{ 0 }^{ \infty }\\ =\cfrac { \pi }{ 2\sqrt { 2 } } $$
Question 10
1 / -0

$$\displaystyle \int e^{tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx=$$

A
$$\displaystyle x^{2}e^{\tan^{-1}x}+c$$

B
$$\displaystyle x e^{\tan^{-1}x}+c$$

C
$$\displaystyle e^{\tan^{-1}x}+c$$

D
$$\displaystyle \frac{1}{2}e^{\tan^{-1}x}+c$$

Solution

Let $$I=\displaystyle \int e^{\tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx$$

$$=\displaystyle \int e^{\tan^{-1}x}\left[1+\frac {x}{1+x^2}\right]dx$$

$$=\displaystyle \int \left[e^{\tan^{-1}x}+\frac {xe^{\tan^{-1}x}}{1+x^2}\right]dx$$
For $$f(x)=e^{\tan^{-1} x}$$, above integral $$I$$ reduces in the form of $$\displaystyle \int [f(x)+x f^{'}(x)]dx$$
But we know that, $$\dfrac{d}{dx} (xf(x)+c) = \displaystyle [f(x)+x f^{'}(x)]$$
$$\therefore \displaystyle \int [f(x)+x f^{'}(x)]dx=xf(x)+c$$

$$\therefore I = \displaystyle xe^{\tan^{-1}x}+c$$
$$\therefore I=\displaystyle \int e^{\tan^{-1}x}\left[\frac{1+x+x^{2}}{1+x^{2}}\right]dx=xe^{\tan ^{-1} x}+c$$
Hence, option 'B' is correct.

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Integrals Test - 25

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Integrals Test - 25

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Question 1

$$-\displaystyle \frac{1}{4} \log 3$$

$$\displaystyle \frac{1}{4} \log 3$$

$$-\displaystyle \frac{1}{3} \log 4$$

$$\displaystyle \frac{1}{2}\log 3$$

Solution

Question 2

$$\displaystyle \frac{\pi}{4}$$

$$\displaystyle \frac{\pi}{8}$$

$$\displaystyle \frac{\pi}{16}$$

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

Solution

Question 3

$$\mathrm{x}=3$$

$$\mathrm{x}=4$$

$$\mathrm{x}=1$$

$$\mathrm{x}=2$$

Solution

Question 4

$$\dfrac {8}{15}$$

$$\dfrac {28}{15}$$

$$\dfrac {35}{8}$$

$$\dfrac {44}{15}$$

Solution

Question 5

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 6

a periodic function

an even function

an odd function

even or odd

Solution

Question 7

$$\displaystyle \frac{3}{5}\left[ \displaystyle \frac { 2 }{ \log _{ e } (\displaystyle \frac { 2 }{ 5 } ) } +\displaystyle \frac { 1 }{ 2\log _{ e } (\displaystyle \frac { 5 }{ 2 } ) } \right] $$

$$-\displaystyle \frac{3}{5}\left[ \displaystyle \frac { 2 }{ \log _{ e } (\displaystyle \frac { 2 }{ 5 } ) } +\displaystyle \frac { 1 }{ 2\log _{ e } (\displaystyle \frac { 5 }{ 2 } ) } \right] $$

$$\displaystyle \frac{3}{5}\left[\displaystyle \frac { 2 }{ \log _{ e } (\displaystyle \frac { 2 }{ 5 } ) } -\displaystyle \frac { 1 }{ 2\log _{ e } (\displaystyle \frac { 5 }{ 2 } ) } \right] $$

$$\displaystyle \frac{1}{2}[\log 2-\log 3]$$

Solution

Question 8

$$1$$

$$\displaystyle \frac{1}{4}$$

$$4$$

$${2}$$

Solution

Question 9

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

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

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

$$\pi/4$$

Solution

Question 10

$$\displaystyle x^{2}e^{\tan^{-1}x}+c$$

$$\displaystyle x e^{\tan^{-1}x}+c$$

$$\displaystyle e^{\tan^{-1}x}+c$$

$$\displaystyle \frac{1}{2}e^{\tan^{-1}x}+c$$

Solution

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