Integrals Test

Result

Integrals Test - 36

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

$$\int _{ 0 }^{ \pi /2 }{ \cfrac { \cos { 2x } }{ { \left( \sin { x } +\cos { x } \right) }^{ 2 } } } dx=......$$

A
$$\cfrac { \pi }{ 4 } $$

B
$$\cfrac { \pi }{ 2 } $$

C
$$0$$

D
$$-\cfrac { \pi }{ 4 } $$

Solution

$$ I=\int _{ 0 }^{ \pi /2 }{ \cfrac { \cos { 2x } }{ { \left( \sin { x } +\cos { x } \right) }^{ 2 } } } dx$$
$$=\int _{ 0 }^{ \pi /2 }{ \cfrac { \cos { 2x } }{ 1+2\sin { x } \cos { x } } } dx (\therefore cos^{2}{x} + sin^{2}{x} = 1) $$
$$= \int _{ 0 }^{ \pi /2 }{ \cfrac { \cos { 2x } }{ 1+\sin { 2x } } } dx$$
$$\cfrac { 1 }{ 2 } \int _{ 0 }^{ \pi /2 }{ \cfrac { \cfrac { d }{ dx } \left( 1+\sin { 2x } \right) }{ 1+\sin { 2x } } } dx (\therefore substitute\ cos{2x} = 1+sin{2x}) $$
$$=\cfrac { 1 }{ 2 } { \left[ \log { \left| 1+\sin { 2x } \right| } \right] }_{ 0 }^{ \pi /2 }=\cfrac { 1 }{ 2 } \left[ \log { 1 } -\log { 1 } \right] \quad $$
$$\therefore I=0$$
Question 2
1 / -0

$$\displaystyle \int _0^4\dfrac{2x+3}{x^2+3x+2}dx$$

A
$$\log 3$$

B
$$\log 5$$

C
$$\log 15$$

D
$$\log 2$$

Solution

Let $$x^2+3x+2=t\implies 2x+3=dt$$

$$x\to 0 \>to \>4\\t\to 2\to 30\\\displaystyle \int _2^{30}\dfrac 1tdt\\\left.\log t\right]_2^{30}\\\log 30-\log 2\\\\\log \dfrac {30}2\\\\log 15$$
Question 3
1 / -0

$$ \int { P\left( x \right) { e }^{ kx }dx=Q\left( x \right) { e }^{ 4x }+C }$$, where $$P(x)$$ is polynomial of degree $$n$$ and $$Q(x)$$ is polynomial of degree $$7$$. Then the value of $$ n+7+k+\lim _{ x\rightarrow \infty }{ \dfrac { P\left( x \right) }{ Q\left( x \right) } }$$ is:

A
$$18$$

B
$$19$$

C
$$20$$

D
$$22$$
Question 4
1 / -0

$$\int { \cos { \left( \log { x } \right) } } dx=............\quad +\quad c$$

A
$$\dfrac { x }{ 2 } \left[ \cos { \left( \log { x } \right) } +\sin { \left( \log { x } \right) } \right] $$

B
$$\dfrac { x }{ 4 } \left[ \cos { \left( \log { x } \right) } +\sin { \left( \log { x } \right) } \right] $$

C
$$\dfrac { x }{ 2 } \left[ \cos { \left( \log { x } \right) } -\sin { \left( \log { x } \right) } \right] $$

D
$$\dfrac { x }{ 2 } \left[ \sin { \left( \log { x } \right) } +\cos { \left( \log { x } \right) } \right] $$

Solution

Consider the given integral.

$$I=\int{\cos \left( \log x \right)dx}$$

Let $$t=\log x$$

$$ \dfrac{dt}{dx}=\dfrac{1}{x} $$

$$ xdt=dx $$

Therefore,

$$ I=\int{{{e}^{t}}\cos tdt} $$

$$ I=\cos t{{e}^{t}}-\int{\left( -\sin t \right){{e}^{t}}}dt $$

$$ I={{e}^{t}}\cos t+\int{\left( \sin t \right){{e}^{t}}}dt $$

$$ I={{e}^{t}}\cos t+\sin t{{e}^{t}}-\int{\cos t{{e}^{t}}}dt $$

$$ I={{e}^{t}}\cos t+\sin t{{e}^{t}}-I $$

$$ 2I={{e}^{t}}\left( \cos t+\sin t \right)+C $$

$$ I=\dfrac{{{e}^{t}}}{2}\left( \cos t+\sin t \right)+C $$

On putting the value of $$'t'$$, we get

$$ I=\dfrac{{{e}^{\log x}}}{2}\left( \cos \left( \log x \right)+\sin \left( \log x \right) \right)+C $$

$$ I=\dfrac{x}{2}\left( \cos \left( \log x \right)+\sin \left( \log x \right) \right)+C $$

Hence, this is the answer.
Question 5
1 / -0

$$\displaystyle\int^{\pi}_0\sqrt{2}(1+\cos x)^{7/2}dx=?$$

A
$$\dfrac{32}{35}$$

B
$$\dfrac{64}{35}$$

C
$$\dfrac{256}{35}$$

D
$$\dfrac{512}{35}$$
Question 6
1 / -0

The value of the integral $$\displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +1}$$, where $$0 < \alpha < \dfrac{\pi}{2}$$, is equal to

A
$$\sin \alpha$$

B
$$\alpha \sin \alpha$$

C
$$\dfrac{\alpha}{2\sin \alpha}$$

D
$$\dfrac{\alpha}{2}\sin \alpha$$

Solution

$$\displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +1}$$

$$\implies \displaystyle\int^1_0\dfrac{dx}{x^2+2x\cos \alpha +\cos^2 \alpha+\sin^2 \alpha}$$

$$\implies \displaystyle \int_0^1 \cfrac{dx}{(x+ \cos \alpha)^2 + \sin^2 \alpha}$$

$$\implies \displaystyle \cfrac{1}{\sin \alpha} \Bigg[\tan^{-1} \cfrac{x+\cos x}{\sin x}\Bigg]_0^1 $$ $$\left[\because \int \cfrac{1}{x^2+a^2}=\dfrac{1}{a}tan^{-1}(\dfrac{x}{a}) \right]$$

$$\implies \displaystyle \cfrac{1}{\sin \alpha} \Bigg[\tan^{-1} \cfrac{1+\cos \alpha}{\sin \alpha} - \tan^{-1} \cfrac{\cos \alpha}{\sin \alpha}\Bigg] $$

$$\implies \cfrac{1}{\sin \alpha} \Bigg[ \tan^{-1} \cfrac{\cfrac{1+\cos \alpha}{\sin \alpha} - \cfrac{\cos \alpha}{\sin \alpha}}{1+\cfrac{\cos \alpha + \cos^2 \alpha}{\sin^2 \alpha}}\Bigg]$$

$$\implies \cfrac{1}{\sin \alpha} \tan^{-1} \cfrac{\sin \alpha}{1+\cos \alpha}$$

$$\implies \cfrac{1}{\sin \alpha} \tan^{-1} \tan \cfrac{\alpha}{2}$$

$$\implies \cfrac{\alpha}{2 \sin \alpha}$$
Question 7
1 / -0

The value of $$\int \dfrac {1}{\sqrt {\sin^{3}x \cos^{5}x}} dx$$ is

A
$$\dfrac {-2}{\sqrt {\tan x}} + \dfrac {2}{3} (\tan x)^{3/2} + C$$

B
$$\dfrac {2}{\sqrt {\tan x}} - \dfrac {2}{3} (\tan x)^{3/2} + C$$

C
$$\dfrac {-2}{\sqrt {\tan x}} + \dfrac {2}{3} (\tan x)^{1/2} + C$$

D
None of these

Solution

Let $$I=\int { \cfrac { 1 }{ \sqrt { \sin ^{ 3 }{ x } \cos ^{ 5 }{ x } } } dx } $$
$$I=\int { \cfrac { 1 }{ \sqrt { \cfrac { \sin ^{ 3 }{ x } }{ \cos ^{ 3 }{ x } } \cos ^{ 8 }{ x } } } dx } =\cfrac { 1 }{ \cos ^{ 4 }{ x } \sqrt { \tan ^{ 3 }{ x } } } dx$$
$$I=\int { \cfrac { \sec ^{ 4 }{ x } }{ \sqrt { \tan ^{ 3 }{ x } } } dx } =\int { \cfrac { \sec ^{ 3 }{ x } .\sec ^{ 2 }{ x } }{ \sqrt { \tan ^{ 3 }{ x } } } dx } $$
$$I=\int { \cfrac { \left( 1+\tan ^{ 2 }{ x } \right) }{ \sqrt { \tan ^{ 3 }{ x } } } \sec ^{ 2 }{ x } dx } $$
Let $$\tan { x } =t$$
$$\tan { x } =t$$
$$\tan { x } =t$$
$$\sec ^{ 2 }{ x } dx=dt$$
So, $$I=\int { \cfrac { 1+{ t }^{ 2 } }{ \sqrt { { t }^{ 3 } } } dt } =\int { \cfrac { 1+{ t }^{ 2 } }{ { t }^{ 3/2 } } dt } $$
$$I=\int { { t }^{ -3/2 }dt } +\int { { t }^{ 1/2 }dt } $$
$$I=\cfrac { { t }^{ -1/2 } }{ -1/2 } +\cfrac { { t }^{ 3/2 } }{ 3/2 } +c$$
So, $$I=\cfrac { -2 }{ \sqrt { \tan { x } } } +\cfrac { 2 }{ 3 } { \left( \tan { x } \right) }^{ 3/2 }+c$$
where C is arbitary constant
Question 8
1 / -0

$$\displaystyle \int \dfrac{x \,\ell nx}{(x^2 - 1)^{3/2}}dx$$ equals

A
$$arc \,\sec x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$$

B
$$\sec^{-1} x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$$

C
$$\cos^{-1} x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$$

D
$$\sec x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$$

Solution

$$\displaystyle \displaystyle \int \dfrac {x ln x}{(x^2-1)^{3/2}}dx$$
$$lnx\displaystyle \int { \dfrac { x }{ { \left( { x }^{ 2 }-1 \right) }^{ 3/2 } } dx } -\displaystyle \int { \left[ \left( \dfrac { d }{ dx } lnx \right) \displaystyle \int { \dfrac { x }{ { \left( { x }^{ 2 }-1 \right) }^{ 3/2 } } } \right] dx } $$
$$=lnx\dfrac { 1 }{ 2 } \dfrac { \left( -2 \right) }{ \sqrt { { x }^{ 2 }-1 } } -\displaystyle \int { \dfrac { 1 }{ x } \left( \dfrac { -1 }{ \sqrt { { x }^{ 2 }-1 } } \right) dx } $$
$$=lnx\dfrac { 1 }{ 2 } \dfrac { \left( -2 \right) }{ \sqrt { { x }^{ 2 }-1 } } -\displaystyle \int { \dfrac { 1 }{ x } \left( \dfrac { -1 }{ \sqrt { { x }^{ 2 }-1 } } \right) dx } $$
$$=-\dfrac { lnx }{ \sqrt { { x }^{ 2 }-1 } } +\sec ^{ -1 }{ x } +C$$
$$\therefore \displaystyle \int { \dfrac { xlnx }{ { \left( { x }^{ 2 }-1 \right) }^{ 3/2 } } dx } =arc\sec { x-\dfrac { lnx }{ \sqrt { { x }^{ 2 }-1 } } +C } $$
Question 9
1 / -0

$$\displaystyle \int_{1/e}^{e}{|\ln x|dx}$$ equals

A
$$e^{-1}-1$$

B
$$2\left (1-\dfrac {1}{e}\right)$$

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

D
$$e-1$$

Solution

We know,
By parts integrals
$$\displaystyle \int u.v \ dx=u\int v\ dx$$ $$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$$

$$ \displaystyle \int lnx \cdot 1dx = lnx\int dx-\int \frac{1}{x}\int dx\,dx $$

$$ \displaystyle = xlnx-x+c $$

$$ \displaystyle \int _{1/e}^{e}\ |lnx|dx = \int_{1/e}^{1}|lnx|dx+\int _{1}^{e} |lnx|dx $$

$$ \displaystyle =\int _{1/e}^{-1}lnxdx+\int_{1}^{e} lnxdx$$

$$\displaystyle = -[x(lnx-1)]_{1/e}^{1}+[x|lnx-1|]_{1}^{e} $$

$$ \displaystyle = -[-1-\frac{1}{e}(-2)]+[0-1(-1)] $$

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

$$ \therefore $$ Option B is correct
Question 10
1 / -0

$$\int _{ 0 }^{ \pi /4 }{ x.\sec ^{ 2 }{ x } dx=? }$$

A
$$\frac {\pi}{4}+\log {\sqrt {2}}$$

B
$$\frac {\pi}{4}-\log {\sqrt {2}}$$

C
$$1+\log {\sqrt {2}}$$

D
$$1-\frac{1}{2}\log {2}$$

Solution

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Integrals Test - 36

Result

Integrals Test - 36

Score

-

Rank

-

SHARING IS CARING

Question 1

$$\cfrac { \pi }{ 4 } $$

$$\cfrac { \pi }{ 2 } $$

$$0$$

$$-\cfrac { \pi }{ 4 } $$

Solution

Question 2

$$\log 3$$

$$\log 5$$

$$\log 15$$

$$\log 2$$

Solution

Question 3

$$18$$

$$19$$

$$20$$

$$22$$

Question 4

$$\dfrac { x }{ 2 } \left[ \cos { \left( \log { x } \right) } +\sin { \left( \log { x } \right) } \right] $$

$$\dfrac { x }{ 4 } \left[ \cos { \left( \log { x } \right) } +\sin { \left( \log { x } \right) } \right] $$

$$\dfrac { x }{ 2 } \left[ \cos { \left( \log { x } \right) } -\sin { \left( \log { x } \right) } \right] $$

$$\dfrac { x }{ 2 } \left[ \sin { \left( \log { x } \right) } +\cos { \left( \log { x } \right) } \right] $$

Solution

Question 5

$$\dfrac{32}{35}$$

$$\dfrac{64}{35}$$

$$\dfrac{256}{35}$$

$$\dfrac{512}{35}$$

Question 6

$$\sin \alpha$$

$$\alpha \sin \alpha$$

$$\dfrac{\alpha}{2\sin \alpha}$$

$$\dfrac{\alpha}{2}\sin \alpha$$

Solution

Question 7

$$\dfrac {-2}{\sqrt {\tan x}} + \dfrac {2}{3} (\tan x)^{3/2} + C$$

$$\dfrac {2}{\sqrt {\tan x}} - \dfrac {2}{3} (\tan x)^{3/2} + C$$

$$\dfrac {-2}{\sqrt {\tan x}} + \dfrac {2}{3} (\tan x)^{1/2} + C$$

None of these

Solution

Question 8

$$arc \,\sec x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$$

$$\sec^{-1} x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$$

$$\cos^{-1} x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$$

$$\sec x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$$

Solution

Question 9

$$e^{-1}-1$$

$$2\left (1-\dfrac {1}{e}\right)$$

$$1-\dfrac {1}{e}$$

$$e-1$$

Solution

Question 10

$$\frac {\pi}{4}+\log {\sqrt {2}}$$

$$\frac {\pi}{4}-\log {\sqrt {2}}$$

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

$$1-\frac{1}{2}\log {2}$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.