Integrals Test

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Integrals Test - 18

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Question 1
1 / -0

$\displaystyle \int_{0}^{1}\tanh xdx=$

A
$\log(\mathrm{e}+1/\mathrm{e})$

B
$\log$ $(e-1/e)$

C
$\log(\mathrm{e}/2+1/2\mathrm{e})$

D
$\displaystyle \log(\frac{\mathrm{e}}{2}-\frac{1}{\mathrm{e}})$

Solution

$\int_{0}^{1} tan h x dx$
$\int tan h x dx=\int \dfrac{e^x-e^{-x}}{e^x+e^{-x}} dx =log (e^x+e^{-x})+c$
$\int_{0}^{1} tan h x dx = log (e^x+e^{-x})+c \int_{0}^{1}$
$=\left ( log (e+\dfrac{1}{e}) +c \right ) - \left ( log (2)+c \right )$
$=log (e+\dfrac{1}{e})-log 2$
$=log (\dfrac{e}{2}+\dfrac{1}{2e})$
$\int_{0}^{1}tan hx dx=log (\dfrac{e}{2}+\dfrac{1}{2e})$

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Question 2
1 / -0

Evaluate the integral
$\displaystyle \int_{1}^{e}\frac{(\ln x)^{3}}{x}dx$

A
$e^{4}/4$

B
$\dfrac{1}{4}$

C
$\displaystyle \frac{1}{4}({e}^{4}-1)$

D
$e^{4}-1$

Solution

$\displaystyle \int_{1}^{e}\frac{(ln x)^3}{x}dx$
$\displaystyle \int_{1}^{e}(ln x)^3d(ln x)=\frac{(ln x)^4}{4}+c$
$\displaystyle \int_{1}^{e}(ln x)^3d(ln x)=\left ( \frac{(1)^4}{4}+c \right ) - \left ( \frac{0}{4}+ c \right )$
$=\dfrac{1}{4}$
Thus $\displaystyle \int_{1}^{e}\frac{(ln x)^3}{x}dx=\frac{1}{4}$

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Question 3
1 / -0

$\displaystyle \int_{0}^{\pi}\frac{\tan x}{\sec x+\cos x}dx_{=}$

A
$\pi$

B
$\dfrac {-\pi}2$

C
$-\pi$

D
$2 \pi$

Solution

$\int_{0}^{\pi}\dfrac{\tan x}{\sec x+\cos x}dx$
$\int \dfrac{\tan x}{\sec x+\cos x}dx=\int \dfrac{d (\sec x)}{1+\sec ^2 x}$
$=\tan ^{-1}(\sec x)+c$
$\int_{0}^{\pi}\dfrac{\tan x}{\sec x+\cos x}dx=\left [ \tan (\sec x)+c \right ] |_{0}^{\pi}$
$=(\tan ^{-1}(\sec \pi)+c)- [ \tan ^{-1}(\sec 0) ]$
$=-\dfrac{\pi}{2}$

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Question 4
1 / -0

$\displaystyle \int_{0}^{1}\mathrm{e}^{\mathrm{x}}(\mathrm{e}^{\mathrm{x}}+1)^{3}\mathrm{d}\mathrm{x}=$

A
$\displaystyle \frac{e^{4}}{4}-4$

B
$\displaystyle \frac{(e+1)^{4}}{4}-4$

C
$\displaystyle \frac{(e+1)^{4}+16}{4}$

D
$\displaystyle \frac{(\mathrm{e}+1)^{4}}{4}+4$

Solution

Let D = $\int_{0}^{1}e^x(e^x+1)^3dx$
Consider $I = \int e^x(e^x+1)^3dx=\int (e^x+1)^3e^x.dx=\frac{(e^x+1)^4}{4}+c$
$I = \int e^x(e^x+1)^3dx=\frac{(e^x+1)^4}{4}+c$
$D = \int_0^1 e^x(e^x+1)^3dx$
$= \cfrac{(e^1+1)^4}{4} - 4$
Thus, $\int_{0}^{1}e^x(e^x+1)^3dx=\frac{(e+1)^4}{4}-4$

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Question 5
1 / -0

Evaluate: $\displaystyle \int_{\sqrt{8}}^{\sqrt{15}}x\sqrt{1+x^{2}}.dx$

A
$\dfrac{15}{8}$

B
$\dfrac{37}{3}$

C
$\dfrac{37}{6}$

D
$\displaystyle \frac{37}{9}$

Solution

$\displaystyle\int_{\sqrt{8}}^{\sqrt{15}} x\sqrt{1+x^2} dx$

Let $z=1+x^2$

$\implies dz=2xdx$
$\implies xdx=\dfrac{dz}{2}$

For, $x=\sqrt{8}, z=9$ and for $x=\sqrt{15}, z=16$

Hence, integration becomes-

$\displaystyle\int_{9}^{16} \sqrt{z} \dfrac{dz}{2}$

$$=\dfrac{1}{2} \left[\dfrac{z^\left(\tfrac{1}{2}+1\right)}{\dfrac{1}{2}+1}\right]_{9}^{16}$$

$$=\dfrac{1}{2}\times \dfrac{2}{3} \displaystyle\left[z^\left(\tfrac{3}{2}\right)\right]_{9}^{16}$$

$$= \dfrac{1}{3} \displaystyle\left[16^\left(\tfrac{3}{2}\right)-9^\left(\tfrac{3}{2}\right) \right]$$

$=\dfrac{1}{3} [64-27]$

$=\dfrac{37}{3}$

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Question 6
1 / -0

$\displaystyle \int_{0}^{1}\frac{xdx}{(x^{2}+1)^{2}}=$

A
$1/2$

B
$1/3$

C
$1/4$

D
$0$

Solution

$\int_{0}^{1}\dfrac{x dx}{(x^2+1)^2}$
$\int \dfrac{x dx}{(x^2+1)^2} =\dfrac{1}{2}\int \dfrac{dx^2}{(1+x^2)^2}=-\dfrac{1}{2}\left ( \dfrac{1}{1+x^2} \right )^{-1}$
$=-\dfrac{1}{2}(1+x^2)^{+c}$
$=\int_{0}^{1}\dfrac{x dx}{(x^2+1)^2}=\left [ -\dfrac{1}{2} (1+x^2)+c \right ] \int_{0}^{1}$
$=\left [ -\dfrac{1}{2} (2)+c \right ] - \left [ -\dfrac{1}{2}+c \right ]$
$=-\dfrac{1}{4}+\dfrac{1}{2} =\dfrac{1}{4}$
$\int_{0}^{1}\dfrac{x dx}{(x^2+1)^2}=\dfrac{1}{4}$

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Question 7
1 / -0

Evaluate: $\displaystyle \int_{0}^{\dfrac{\pi}{2}}e^{\sin^2 x}\sin 2xdx$

A
$e$

B
$e+1$

C
$e-1$

D
$2{e}+1$

Solution

$\displaystyle\int_{0}^{\dfrac{\pi}{2}} e^{\sin^2 x} \sin{2x} dx$

Let $z=\sin^2 x$
$\implies dz=2\sin{x} \cos{x}dx=\sin{2x}dx$

For $x=0, z=0$ and for $x=\dfrac{\pi}{2} ,z=1$ .

Hence, integration becomes-

$\displaystyle\int_{0}^{1} e^z dz$

$=\left[e^z\right]_{0}^{1}$

$=e-1$

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Question 8
1 / -0

$\displaystyle \int_{0}^{4}\sqrt{16-x^{2}}d_{X}=$

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

B
$\pi$

C
$16\pi$

D
$4\pi$

Solution

$\int_{0}^{4}\sqrt{16-x^2} dx$
$\int \sqrt{16-x^2} dx =\int \sqrt{4^2-x^2}dx$
$=\frac{x}{2} \sqrt{16-x^2}+ \frac{16}{2} sin^{-1} (\frac{x}{14})+c$
$=\int_{0}^{4}\sqrt{16-x^2}dx \left [ \frac{x}{2} \sqrt{16-x^2}+ \frac{16}{2}sin^{-1}\left ( \frac{x}{4} \right ) +c \right ]\int_{0}^{4}$
$=\left ( \frac{4}{2}\sqrt{16-16}+ 8 sin^{-1}\left ( \frac{4}{4}\right ) +c\right )- (0+0+c)$
$=8\left ( \dfrac{\pi}{2} \right ) =4\pi$
$\int_{0}^{4}\sqrt{16-x^2}dx=4\pi$

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Question 9
1 / -0

Find $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sec^{2}xdx}{(\sec x+\tan x)^{n}}$ , where $(\mathrm{n}>2)$

A
$\displaystyle \frac{1}{n^{2}-1}$

B
$\displaystyle \frac{n}{n^{2}-1}$

C
$\displaystyle \frac{n}{n^{2}+1}$

D
$\displaystyle \frac{2}{n^{2}-1}$

Solution

Let $\displaystyle I=\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \dfrac { \sec ^{ 2 }{ x } }{ { \left( \sec { x+\tan { x } } \right) }^{ n } } } dx$
Substitute $\sec { x } +\tan { x } =t\Rightarrow \left( \sec { x\tan { x } +\sec ^{ 2 }{ x } } \right) dx=dt$
$\displaystyle \Rightarrow \sec { x } dx=\dfrac { dt }{ t }$
$\displaystyle \therefore \sec { x } -\tan { x } =\dfrac { 1 }{ t } \Rightarrow \sec { x } =\dfrac { 1 }{ 2 } \left( t+\dfrac { 1 }{ t } \right)$
$\displaystyle \therefore I=\dfrac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \dfrac { \left( t-\dfrac { 1 }{ t } \right) \dfrac { dt }{ t } }{ { t }^{ n } } } =\dfrac { 1 }{ 2 } \int _{ 1 }^{ \infty }{ \left( \dfrac { 1 }{ { t }^{ n } } +\dfrac { 1 }{ { t }^{ +n-12 } } \right) } dt$

$\displaystyle =\dfrac { 1 }{ 2 } \left[ \dfrac { 1 }{ \left( 1-n \right) { t }^{ n-1 } } +\dfrac { 1 }{ \left( n+1 \right) { t }^{ n+1 } } \right] _{ 0 }^{ \infty }=\dfrac { n }{ { n }^{ 2 }-1 }$

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Question 10
1 / -0

$\displaystyle \int_{0}^{\pi/2}\frac{d_{X}}{4\cos^{2}x+9\sin^{2}x}=$

A
$\displaystyle \frac{\pi}{12}$

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

C
$\displaystyle \frac{\pi}{9}$

D
$\displaystyle \frac{\pi}{6}$

Solution

$\int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{4 cos^2 x + 9 sec^2 x}$
$\int \dfrac{dx}{4 cos^2 x + 9 sec^2 x}=\int \dfrac{sec^2 x dx} {4+9 tan^2 x}=\int \dfrac{d(tan x)}{4+9 tan^2 x}$
$=\dfrac{1}{4} tan^{-1}\left ( \dfrac{3}{2} tan x \right ) \dfrac{2}{3} +c$
$=\dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan x \right ) +c$
$\int_{0}^{\dfrac{\pi}{2}}\dfrac{dx}{4 cos^2 x + 9 sec^2 x}=\left [ \dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan x \right ) +c \right ]\int_{0}^{\dfrac{\pi}{2}}$
$=\left (\dfrac{1}{6} tan^{-1}\left ( \dfrac{3}{2} tan \dfrac{\pi}{2} \right ) +c \right ]- \left ( \dfrac{1}{6}tan^{-1}(0)+c \right )$
$=\dfrac{1}{6}\times \dfrac{\pi}{2}=\dfrac{\pi}{12}$

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Integrals Test - 18

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Integrals Test - 18

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Question 1

log⁡(e+1/e)\log(\mathrm{e}+1/\mathrm{e})log(e+1/e)

log⁡\loglog (e−1/e)(e-1/e)(e−1/e)

log⁡(e/2+1/2e)\log(\mathrm{e}/2+1/2\mathrm{e})log(e/2+1/2e)

log⁡(e2−1e)\displaystyle \log(\frac{\mathrm{e}}{2}-\frac{1}{\mathrm{e}})log(2e​−e1​)

Solution

Question 2

e4/4e^{4}/4e4/4

14\dfrac{1}{4}41​

14(e4−1)\displaystyle \frac{1}{4}({e}^{4}-1)41​(e4−1)

e4−1e^{4}-1e4−1

Solution

Question 3

π\piπ

−π2\dfrac {-\pi}22−π​

−π-\pi−π

2π2 \pi2π

Solution

Question 4

e44−4\displaystyle \frac{e^{4}}{4}-44e4​−4

(e+1)44−4\displaystyle \frac{(e+1)^{4}}{4}-44(e+1)4​−4

(e+1)4+164\displaystyle \frac{(e+1)^{4}+16}{4}4(e+1)4+16​

(e+1)44+4\displaystyle \frac{(\mathrm{e}+1)^{4}}{4}+44(e+1)4​+4

Solution

Question 5

158\dfrac{15}{8}815​

373\dfrac{37}{3}337​

376\dfrac{37}{6}637​

379\displaystyle \frac{37}{9}937​

Solution

Question 6

1/21/21/2

1/31/31/3

1/41/41/4

000

Solution

Question 7

eee

e+1e+1e+1

e−1e-1e−1

2e+12{e}+12e+1

Solution

Question 8

π4\displaystyle \frac{\pi}{4}4π​

π\piπ

16π 16\pi16π

4π 4\pi4π

Solution

Question 9

1n2−1\displaystyle \frac{1}{n^{2}-1}n2−11​

nn2−1\displaystyle \frac{n}{n^{2}-1}n2−1n​

nn2+1\displaystyle \frac{n}{n^{2}+1}n2+1n​

2n2−1\displaystyle \frac{2}{n^{2}-1}n2−12​

Solution

Question 10

π12\displaystyle \frac{\pi}{12}12π​

π4\displaystyle \frac{\pi}{4}4π​

π9\displaystyle \frac{\pi}{9}9π​

π6\displaystyle \frac{\pi}{6}6π​

Solution

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$\log(\mathrm{e}+1/\mathrm{e})$

$\log$ $(e-1/e)$

$\log(\mathrm{e}/2+1/2\mathrm{e})$

$\displaystyle \log(\frac{\mathrm{e}}{2}-\frac{1}{\mathrm{e}})$

$e^{4}/4$

$\dfrac{1}{4}$

$\displaystyle \frac{1}{4}({e}^{4}-1)$

$e^{4}-1$

$\pi$

$\dfrac {-\pi}2$

$-\pi$

$2 \pi$

$\displaystyle \frac{e^{4}}{4}-4$

$\displaystyle \frac{(e+1)^{4}}{4}-4$

$\displaystyle \frac{(e+1)^{4}+16}{4}$

$\displaystyle \frac{(\mathrm{e}+1)^{4}}{4}+4$

$\dfrac{15}{8}$

$\dfrac{37}{3}$

$\dfrac{37}{6}$

$\displaystyle \frac{37}{9}$

$1/2$

$1/3$

$1/4$

$0$

$e$

$e+1$

$e-1$

$2{e}+1$

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

$\pi$

$16\pi$

$4\pi$

$\displaystyle \frac{1}{n^{2}-1}$

$\displaystyle \frac{n}{n^{2}-1}$

$\displaystyle \frac{n}{n^{2}+1}$

$\displaystyle \frac{2}{n^{2}-1}$

$\displaystyle \frac{\pi}{12}$

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

$\displaystyle \frac{\pi}{9}$

$\displaystyle \frac{\pi}{6}$