Integrals Test

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Integrals Test - 56

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Question 1
1 / -0

The value of the integral $$\displaystyle \int_{0}^{\pi/4}\dfrac {\sin x + \cos x}{3 + \sin 2x}dx$$ is equal to

A
$$\log_{e}2$$

B
$$\log_{e}3$$

C
$$\dfrac {1}{4}\log_{e} 2$$

D
$$\dfrac {1}{4}\log_{e} 3$$

Solution

Let
$$I = \displaystyle \int_{0}^{\pi/4}\dfrac {\sin x + \cos x}{3 + \sin 2x}dx$$

$$=\displaystyle \int_{0}^{\pi/4} \dfrac {\sin x + \cos x}{3 + 2\sin x \cos x} dx$$

$$=\displaystyle \int_{0}^{\pi/4} - \dfrac {\sin x + \cos x}{(\sin x - \cos x)^{2} - 4} dx$$ ...... (i)

Put $$\sin x - \cos x = t$$
$$\Rightarrow (\cos x + \sin x) dx = dt$$
when $$x = 0\Rightarrow t = -1$$
and
$$x = \dfrac {\pi}{4}\Rightarrow t = 0$$

$$\therefore$$ Eq. (i) becomes,

$$I = -\displaystyle \int_{-1}^{0} \dfrac {dt}{t^{2} - 4} = -\dfrac {1}{4} \left [\log \left |\dfrac {t - 2}{t + 2}\right | \right ]_{-1}^{0}$$

$$= -\dfrac {1}{4}(\log 1 - \log 3)$$

$$=-\dfrac {1}{4} (0 - \log 3) = \dfrac {1}{4} \log 3$$
Question 2
1 / -0

The value of $$\displaystyle \int_{0}^{\pi /2} \dfrac {\sin 2t}{\sin^{4}t + \cos^{4}t} dt $$

A
$$\pi$$

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

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

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

Solution

Let $$I = \displaystyle \int_{0}^{\pi /2} \dfrac {\sin 2t}{\sin^{4}t + \cos^{4}t} dt = \int_{0}^{\pi /2} \dfrac {2\sin t . \cos t}{\sin^{4}t + \cos^{4}t} dt$$

Divide Nr. and Dr. by $$\cos^{4}t =\displaystyle \int_{0}^{\pi /2} \dfrac {2\tan t. \sec^{2}t}{(\tan^{2}t)^{2} + 1} dt$$

put $$\tan^{2}t = x \Rightarrow 2\tan t\sec^2tdt=dx$$

$$\therefore I=\displaystyle \int_{0}^{\infty} \dfrac {1}{1 + x^{2}} dx = \tan^{-1} x \mid_{0}^{\infty} = \dfrac {\pi}{2}$$
Question 3
1 / -0

$$\displaystyle\int \dfrac{cos^{n-1}}{sin^{n+1}}dx, n\neq 0$$ is ____

A
$$\dfrac{cot^nx}{n}$$

B
$$\cfrac{-cot^{n-1}x}{n-1}$$

C
$$\cfrac{-cot^nx}{n}$$

D
$$\cfrac{cot^{n-1}x}{n-1}$$

Solution

$$\int \dfrac{cos^{n-1}x}{sin^{n+1}x}dx$$

$$=\int (\dfrac{cosx}{sinx})^{n}.\dfrac{1}{cosx.sinx}.dx$$

$$=\int cot^{n}x.\dfrac{sinx}{cosx}.\dfrac{1}{sin^{2}x}dx$$

$$=\int cot^{n}x.tanx.cosec^2x dx$$

$$=\int cot^{n-1}x.cosec^{2} x dx$$

Let $$cot(x)=t$$, then $$cosec^{2}xdx=-dt$$
Hence
$$I=-\int t^{n-1}.dt$$
$$=\dfrac{-t^{n}}{n}$$
$$=-\dfrac{cot^{n}x}{n}$$
Question 4
1 / -0

$$\int { { ({ x }^{ 2 }+5) }^{ 3 } } dx$$

A
$$\cfrac { { x }^{ 7 } }{ 7 } +3{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$$

B
$$\cfrac { { x }^{ 6 } }{ 7 } +3{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$$

C
$$\cfrac { { x }^{ 7 } }{ 7 } +3{ x }^{ 5 }-25{ x }^{ 3 }+125x+C$$

D
$$\cfrac { { x }^{ 7 } }{ 7 } +4{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$$

Solution

$$\int (x^2 + 5)^3 dx$$
$$= \int (x^6 + 3(x^2)^2 \times 5 + 3x^2 \times 5^2 + 5^3) dx$$
$$= \int (x^6 + 15x^4 + 75x^2 + 125) dx$$
$$= \cfrac{x^7}{7} + 15 \times \cfrac{x^5}{5} + 75 \times \cfrac{x^3}{3} + 125x + c$$
$$= \cfrac{x^7}{7} + 3x^5 + 25x^3 + 125x +c$$
Question 5
1 / -0

$$\displaystyle \int {\dfrac{dx}{\sin^2x \cos^2x}}$$

A
$$\tan x-\cot x+c$$

B
$$\tan x-x+1$$

C
$$\tan x-x$$

D
$$\tan x+x$$

Solution

$$\displaystyle \int \dfrac{4dx}{4\sin ^{2}x\cos ^{2}x}$$

$$=\displaystyle \int \dfrac{4dx}{\sin ^{2}2x}$$

$$=4\displaystyle \int \text{cosec}^{2}2x dx$$

$$=-2\cot 2x+C$$

$$=\dfrac{-2(\cos 2x)}{\sin 2x}+C$$

$$=\dfrac{2(\sin ^{2}x-\cos ^{2}x)}{2\sin x.\cos x}+C$$

$$=\dfrac{\sin ^{2}x}{\sin x.\cos x}-\dfrac{\cos ^{2}x}{\sin x.\cos x}+C$$

$$=\tan x-\cot x+C$$
Question 6
1 / -0

If $$\displaystyle \int_{\log 2}^{x} \dfrac {dy}{\sqrt {e^{y} - 1}} = \dfrac {\pi}{6}$$, then $$x$$ is equal to

A
$$\log_{e}4$$

B
$$\log_{e} 2$$

C
$$4$$

D
$$2$$

Solution

$$\displaystyle \int_{\log 2}^{x} \dfrac {dy}{\sqrt {e^{y} - 1}} = \dfrac {\pi}{6}$$

Put $$t = e^{y} - 1$$
$$\dfrac {dt}{dy} = e^{y}\Rightarrow dy = \dfrac {dt}{t + 1}$$

$$2\displaystyle \int_{1}^{e^{x} - 1} \dfrac {dt}{(t + 1)2\sqrt {t}} = \dfrac {\pi}{6}$$

$$\Rightarrow 2\displaystyle\int_{1}^{e^{x} - 1} \dfrac {d(\sqrt {t})}{(\sqrt {t})^{2} + 1} = \dfrac {\pi}{6}$$

$$\tan^{-1} (\sqrt {t})|_{1}^{e^{x} - 1} = \dfrac {\pi}{12}$$

$$\tan^{-1} (\sqrt {e^{x} - 1}) - \tan^{-1} 1 = \dfrac {\pi}{12}$$

$$\tan^{-1} (\sqrt {e^{x} - 1}) = \dfrac {\pi}{12} + \dfrac {\pi}{4} = \dfrac {\pi}{3}$$

$$\sqrt {e^{x} - 1} = \sqrt {3}$$

$$e^{x} - 1 = 3$$

$$e^{x} = 4\Rightarrow x = \log_{e}4$$
Question 7
1 / -0

Evaluate: $$\displaystyle \int \dfrac{(x-1)e^x}{(x+1)^3}dx=$$

A
$$\dfrac{e^x}{x+1}$$

B
$$\dfrac{e^x}{(x+1)^2}$$

C
$$\dfrac{e^x}{(x+1)^3}$$

D
$$\dfrac{x\cdot e^x}{(x+1)}$$

Solution

Consider, $$\displaystyle I=\int \dfrac{(x-1)e^{x}}{(x+1)^{3}}dx$$

$$\displaystyle I =\int \dfrac{xe^{x}-e^{x}-2e^x+2e^x}{(x+1)^{3}}dx$$

$$\displaystyle I =\int \dfrac{(x+1)e^{x}-2e^{x}}{(x+1)^{3}}dx$$

$$\displaystyle I=\int \dfrac{e^{x}}{(x+1)^{2}}dx-\int\dfrac{2e^{x}}{(x+1)^{3}}dx$$

Applying ILATE rule to the first integral, gives us

$$\displaystyle I =\dfrac{e^{x}}{(x+1)^{2}}-\int \left(\dfrac{-2e^{x}}{(x+1)^{3}}\right)dx-\int\dfrac{2e^{x}}{(x+1)^{3}}dx$$

$$\displaystyle I =\dfrac{e^{x}}{(x+1)^{2}}+\int \left(\dfrac{2e^{x}}{(x+1)^{3}}\right)dx-\int\dfrac{2e^{x}}{(x+1)^{3}}dx$$

$$\displaystyle I =\dfrac{e^{x}}{(x+1)^{2}}+C$$
Question 8
1 / -0

Let $$f$$ be a positive function. Let
$${ I }_{ 1 }=\int _{ 1-k }^{ k }{ xf\left\{ x(1-x) \right\} } dx$$
$${ I }_{ 2 }=\int _{ 1-k }^{ k }{ f\left\{ x(1-x) \right\} } dx$$
where $$2k-1>0$$. Then $$\cfrac { { I }_{ 1 } }{ { I }_{ 2 } } $$

A
$$2$$

B
$$k$$

C
$$\cfrac { 1 }{ 2 } $$

D
$$1$$

Solution

$${ I }_{ 1 }=\int _{ 1-k }^{ k }{ xf\left\{ x(1-x) \right\} } dx$$
$$=\int _{ 1-k }^{ k }{ (1-x)f\left\{ (1-x)(1-(1-x)) \right\} } dx$$
$$=\int _{ 1-k }^{ k }{ f\left\{ x(1-x) \right\} } dx-\int _{ 1-k }^{ k }{ xf\left\{ x(1-x) \right\} } dx$$
$$={ I }_{ 2 }-{ I }_{ 1 }$$
$$\therefore 2{ I }_{ 1 }={ I }_{ 2 }\quad \Rightarrow \cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =\cfrac { 1 }{ 2 } $$
Question 9
1 / -0

The value of $$\displaystyle\int _{ 0 }^{ { 1 }/{ \sqrt { 2 } } }{ \dfrac { \sin ^{ -1 }{ x } }{ { \left( 1-{ x }^{ 2 } \right) }^{ { 3 }/{ 2 } } } dx } $$ is

A
$$\dfrac { \pi }{ 2 } -\log { 2 } $$

B
$$\dfrac { \pi }{ 4 } -\dfrac { 1 }{ 2 } \log { 2 } $$

C
$$\dfrac { \pi }{ 2 } +\dfrac { 1 }{ 2 } \log { 2 } $$

D
$$\pi -\dfrac { 1 }{ 2 } \log { 2 } $$

Solution

Let $$l=\displaystyle\int _{ 0 }^{ { 1 }/{ \sqrt { 2 } } }{ \dfrac { \sin ^{ -1 }{ x } }{ { \left( 1-{ x }^{ 2 } \right) }^{ { 3 }/{ 2 } } } dx }$$

When $$x=0$$, then $$\sin { t } =0=\sin { 0 } \Rightarrow t=0$$

When $$x=\dfrac { 1 }{ \sqrt { 2 } }$$, then $$\sin { t } =\dfrac { 1 }{ \sqrt { 2 } } =\sin { \dfrac { \pi }{ 4 } }$$

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

Put $$ x=\sin { t } dx=\cos { t } dt$$

$$\therefore l=\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \dfrac { t }{ \cos ^{ 3 }{ t } } \cdot \cos { t } dt } =\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ t\sec ^{ 2 }{ t } dt }$$

$$ ={ \left[ t\cdot \tan { t } \right] }_{ 0 }^{ { \pi }/{ 4 } }-\displaystyle\int _{ 0 }^{ { \pi }/{ 4 } }{ \tan { t } dt }$$

$$ =\dfrac { \pi }{ 4 } -{ \left[ \log { \sec { t } } \right] }_{ 0 }^{ { \pi }/{ 4 } }$$

$$=\dfrac { \pi }{ 4 } -\log { \sqrt { 2 } } -0=\dfrac { \pi }{ 4 } -\log { { 2 }^{ { 1 }/{ 2 } } } =\dfrac { \pi }{ 4 } -\dfrac { 1 }{ 2 } \log { 2 } $$
Question 10
1 / -0

Let $$f(x)$$ be a positive function. Let
$$I_{1} = \int_{1 - k}^{k} xf\left \{x(1 - x)\right \} dx$$,
$$I_{2} = \int_{1 - k}^{k} f\left \{x(1 - x) \right \} dx$$,
where $$2k - 1 > 0$$, then $$\dfrac {I_{1}}{I_{2}}$$ is

A
$$2$$

B
$$k$$

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

D
$$1$$

Solution

$$I_{1} = \displaystyle \int_{1 - k}^{k} xf \left \{x (1 - x)\right \}dx$$
$$= \displaystyle \int_{1 - k}^{k} (1 - x)f\left \{(1 - x)(1 -(1- x))\right \}dx$$
$$(\because \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx)$$
$$=\displaystyle \int_{1 - k}^{k} (1 - x)f\left \{(1 - x)x\right \}dx$$
$$= \displaystyle \int_{1 - k}^{k} f\left \{x(1 - x)\right \} dx$$
$$- \displaystyle \int_{1 - k}^{k} xf \left \{x(1 - x)\right \}dx$$
Therefore, $$ I_{1} = I_{2} - I_{1}\Rightarrow 2I_{1} = I_{2}$$
$$\Rightarrow \dfrac {I_{1}}{I_{2}} = \dfrac {1}{2}$$

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Re-Attempt Weekly Quiz Competition

Integrals Test - 56

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Integrals Test - 56

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SHARING IS CARING

Question 1

$$\log_{e}2$$

$$\log_{e}3$$

$$\dfrac {1}{4}\log_{e} 2$$

$$\dfrac {1}{4}\log_{e} 3$$

Solution

Question 2

$$\pi$$

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

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

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

Solution

Question 3

$$\dfrac{cot^nx}{n}$$

$$\cfrac{-cot^{n-1}x}{n-1}$$

$$\cfrac{-cot^nx}{n}$$

$$\cfrac{cot^{n-1}x}{n-1}$$

Solution

Question 4

$$\cfrac { { x }^{ 7 } }{ 7 } +3{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$$

$$\cfrac { { x }^{ 6 } }{ 7 } +3{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$$

$$\cfrac { { x }^{ 7 } }{ 7 } +3{ x }^{ 5 }-25{ x }^{ 3 }+125x+C$$

$$\cfrac { { x }^{ 7 } }{ 7 } +4{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$$

Solution

Question 5

$$\tan x-\cot x+c$$

$$\tan x-x+1$$

$$\tan x-x$$

$$\tan x+x$$

Solution

Question 6

$$\log_{e}4$$

$$\log_{e} 2$$

$$4$$

$$2$$

Solution

Question 7

$$\dfrac{e^x}{x+1}$$

$$\dfrac{e^x}{(x+1)^2}$$

$$\dfrac{e^x}{(x+1)^3}$$

$$\dfrac{x\cdot e^x}{(x+1)}$$

Solution

Question 8

$$2$$

$$k$$

$$\cfrac { 1 }{ 2 } $$

$$1$$

Solution

Question 9

$$\dfrac { \pi }{ 2 } -\log { 2 } $$

$$\dfrac { \pi }{ 4 } -\dfrac { 1 }{ 2 } \log { 2 } $$

$$\dfrac { \pi }{ 2 } +\dfrac { 1 }{ 2 } \log { 2 } $$

$$\pi -\dfrac { 1 }{ 2 } \log { 2 } $$

Solution

Question 10

$$2$$

$$k$$

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

$$1$$

Solution

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