Integrals Test

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Integrals Test - 13

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Question 1
1 / -0

Let $$F(x)=f(x)+f\left ( \dfrac{1}{x} \right )$$, where $$f(x)=\int_{l}^{x}\dfrac{logt}{l+t}dt$$. Then $$F(e)$$ equals

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

B
0

C
1

D
2

Solution

$$\displaystyle \because f\left( x \right) =\int _{ 1 }^{ x }{ \frac { x\log { t } }{ 1+t } dt } $$ and $$\displaystyle F\left( e \right) =f\left( e \right) +f\left( \frac { 1 }{ e } \right) $$
$$\displaystyle \Rightarrow F\left( e \right) =\int _{ 1 }^{ e }{ \frac { \log { t } }{ 1+t } dt } +\int _{ 1 }^{ 1/e }{ \frac { x\log { t } }{ 1+t } dt } $$
$$\displaystyle =\int _{ 1 }^{ e }{ \frac { \log { t } }{ 1+t } dt } +\int _{ 1 }^{ e }{ \frac { \log { t } }{ t\left( 1+t \right) } dt } =\int _{ 1 }^{ e }{ \frac { \log { t } }{ t } dt } $$
$$\displaystyle ={ \left[ \frac { { \left( \log { t } \right) }^{ 2 } }{ 2 } \right] }_{ 1 }^{ e }=\frac { 1 }{ 2 } \left[ { \left( \log { e } \right) }^{ 2 }-{ \left( \log { 1 } \right) }^{ 2 } \right] =\frac { 1 }{ 2 } $$
Question 2
1 / -0

For x > 0, let f(x) = $$\displaystyle \int_{1}^{x}{ \frac{log t}{1 + t}} dt $$. Then f(x) + f$$\displaystyle \left( \frac{1}{x} \right)$$ is equal to:

A
$$\displaystyle \frac{1}{4} log x^2$$

B
log x

C
$$\displaystyle \frac{1}{2} (log x)^2$$

D
$$\displaystyle \frac{1}{4} (log x)^2$$

Solution

$$f(x)=\int _{ 1 }^{ x }{ \cfrac { \log { t } }{ 1+{ t } } } $$ $$--(1)$$
$$f\left( \cfrac { 1 }{ x } \right) =\int _{ 1 }^{ 1/x }{ \cfrac { \log { t } }{ 1+{ t } } } $$
Put t=1/u
$$f\left( \cfrac { 1 }{ x } \right) =\displaystyle\int _{ 1 }^{ x }{ \cfrac { -u.\log { u } }{ (1+{ u) } } \cfrac { -du }{ { u }^{ 2 } } } $$
$$f\left( \cfrac { 1 }{ x } \right) =\displaystyle \int _{ 1 }^{ x }{ \cfrac { \log { u } }{ u(1+{ u) } } du } $$ $$--(2)$$
Add (1) and (2)
$$=\displaystyle \int _{ 1 }^{ x }{ \cfrac { \log { u } }{ u } du }$$
$$z=logu\\ dz=\cfrac { du }{ u } $$
$$\int _{ 1 }^{ x }{ zdz } $$
$$={ \cfrac { 1 }{ 2 } (\log { x } })^{ 2 }$$
Question 3
1 / -0

The value of $$\displaystyle\int^{2\pi}_{0}\dfrac{x\sin^8x}{\sin^8x+\cos^8x}dx$$ is equal to?

A
$$2\pi$$

B
$$\pi^2$$

C
$$2\pi^2$$

D
$$4\pi$$

Solution

Let $$I = \displaystyle \int_0^{2 \pi} \dfrac{x \sin^8 x}{\sin^8 x + \cos^8 x}$$ ___(1)

$$I = \displaystyle \int_0^{2 \pi} \dfrac{(2\pi - x) \sin^8 (2\pi - x)}{\sin^8 (2 \pi - x) + \cos^8 (2 \pi - x)}$$

$$I = \displaystyle \int_0^{2 \pi} \dfrac{(2\pi - x) \sin^8 x}{\sin^8 x + \cos^8 x}$$ ___(2)

Add (1) + (2) we get,

$$2I = \displaystyle \int_0^{2 \pi} \dfrac{2 \pi \sin^8 x}{\sin^8 x + \cos^8 x}$$

$$2I = 2 \times 2 \pi \displaystyle \int_0^{\pi} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x}$$

$$I = 2 \pi \displaystyle \int_0^{\pi} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$$

$$I = 4 \pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 x}{\sin^8 x + \cos^8 x} dx$$ __(3)

$$I = 4\pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 \left(\dfrac{\pi}{2} - x \right)}{\sin^8 \left(\dfrac{\pi}{2} - x \right) + \cos^8 \left(\dfrac{\pi}{2} - x \right)}$$

$$I = 4 \pi \displaystyle \int_0^{\pi/2} \dfrac{\cos^8 (x)}{\sin^8 x + \cos^8 x} $$ __(4)

add (3) + (4) we get

$$I = 2 \pi \displaystyle \int_0^{\pi/2} \dfrac{\sin^8 x + \cos^8 x}{\sin^8 x + \cos^8 x} dx = 2\pi \displaystyle \int_0^{\pi/2} 1 dx = 2 \pi \times \dfrac{\pi}{2} = \pi^2$$
Question 4
1 / -0

The integral $$\displaystyle\int^{{\pi}/{4}}_{{\pi}/{12}}\frac{8\cos 2x}{(\tan x+\cot x)^3}dx$$ equals?

A
$$\displaystyle\frac{15}{128}$$

B
$$\displaystyle\frac{13}{32}$$

C
$$\displaystyle\frac{13}{256}$$

D
$$\displaystyle\frac{15}{64}$$

Solution

$$I=\displaystyle\int^{{\pi}/{4}}_{{\pi}/{12}} {\displaystyle\cfrac{8\cos 2x \sin^3x\cos^3 x}{{(\sin^2x+\cos^2x)}^3}} dx$$

$$I=\displaystyle\int^{{\pi}/{4}}_{{\pi}/{12}}{\displaystyle\cfrac{2.8\cos 2x \sin^3 2x}{2}}\frac{dx}{8}$$

Put $$\sin 2x =t \Rightarrow 2 \cos2x dx=dt$$

When $$x=\displaystyle\frac{\pi}{12}, t=\frac{1}{2}$$

$$x=\displaystyle\frac{\pi}{4}, t=1$$

$$I=\displaystyle\int^1_{\frac{1}{2}}\dfrac{t^3}2\ {dt}$$

$$=\displaystyle\frac{1}{2}\left(\displaystyle\frac{t^4}{4}\right)^1_{\frac{1}{2}}$$

$$=\displaystyle\frac{1}{8}\left(1-\displaystyle\frac{1}{16}\right)$$

$$=\displaystyle\frac{15}{128}$$.
Question 5
1 / -0

The value of $$\displaystyle \int_{0}^{1}\frac{8\log(1+\mathrm{x})}{1+\mathrm{x}^{2}}$$ dx is

A
$$\pi \log 2$$

B
$$\displaystyle \frac{\pi}{8}\log 2$$

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

D
$$\log 2$$

Solution

$$\displaystyle \mathrm{I}=\int_{0}^{1}\frac{8\log(1+\mathrm{x})}{1+\mathrm{x}^{2}} dx$$

Let $$\mathrm{x}=\tan\theta=$$ dx $$=\sec^{2}\theta \mathrm{d}\theta$$

$$\displaystyle \mathrm{I}=\int_{0}^{\pi/4}8\log(1+\tan\theta)\mathrm{d} \theta$$

$$\displaystyle \mathrm{I}=8\int_{0}^{\pi/4}\log(1+\tan(\frac{\pi}{4}-\theta))\mathrm{d} \theta$$

$$=8\displaystyle \int_{0}^{\pi/4}\log(\frac{2}{1+\tan\theta})\mathrm{d}\theta$$

$$=8\displaystyle \int_{0}^{\pi/4}(\log 2-\log(1+\tan\theta))\mathrm{d} \theta$$

$$\displaystyle \mathrm{I}=4\int_{0}^{\pi/4}\log 2\mathrm{d}\theta=\pi log2 $$
Hence, option 'A' is correct.
Question 6
1 / -0

If $$\displaystyle\int \dfrac{\cos x \, dx}{\sin^3x(1+\sin^6x)^{2/3}} = f(x)(1 + \sin^6x)^{1/\alpha} + c$$
Where $$c$$ is a constant of integration, then $$\lambda f\left(\dfrac{\pi}{3}\right)$$ is equal to:

A
$$\dfrac{9}{8}$$

B
$$-2$$

C
$$2$$

D
$$-\dfrac{9}{8}$$

Solution

$$\displaystyle\int \dfrac{\cos x \, dx}{\sin^3x(1+\sin^6x)^{2/3}} = f(x)(1 + \sin^6x)^{1/\alpha} + c.....given$$

Let $$\sin x = t$$

$$\cot x dx= dt$$

$$\displaystyle I= \int \dfrac{dt}{t^3(1+t^6)^{2/3}}$$

$$\displaystyle I=\int \dfrac{dt}{t^3.t^4\left(1 + \dfrac{1}{t^6}\right)^{2/3}}$$

$$\displaystyle I=\int \dfrac{dt}{t^7\left(1 + \dfrac{1}{t^6}\right)^{2/3}}$$

Put $$ 1+\dfrac{1}{t^6}=r^3$$

$$\Rightarrow \dfrac{dt}{t^7} = \dfrac{-1}{2}r^2dr$$

$$= \dfrac{-1}{2}\displaystyle\int \dfrac{r^2dr}{r^2}$$

$$=\dfrac{-1}{2}r +c$$

$$= \dfrac{-1}{2}\left(\dfrac{\sin^6x + 1}{\sin^6x}\right)^{1/3}+c .......$$As $$r=\left (1+\dfrac{1}{t^6}\right )^{1/3}$$ and $$t=\sin x$$

$$=\dfrac{-1}{2\sin^2x} (1 + \sin^6x)^{1/3}+c$$

$$f(x) = -\dfrac{1}{2} cosec^{2}x$$ and $$\lambda = 3$$

Now for $$x=\dfrac{\pi}3$$

$$cosec \dfrac{\pi}3=\dfrac{2}{\sqrt3}$$

$$\lambda f \left(\dfrac{\pi}{3}\right) = \dfrac{-1}2\times \left (\dfrac{2}{\sqrt3}\right )^2\times 3$$

$$\boxed{\lambda f \left(\dfrac{\pi}{3}\right) = -2}.....Answer$$

Hence option $$'B'$$ is the answer.
Question 7
1 / -0

The following integral $$\displaystyle \int_{\pi/4}^{\pi/2} (2 cosec x)^{17}dx$$ is equal to

A
$$\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du$$

B
$$\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u + e^{-u})^{17} du$$

C
$$\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u - e^{-u})^{17} du$$

D
$$\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u - e^{-u})^{16} du$$

Solution

$$\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (2 cosec x)^{17}dx$$

Let
$$e^u+e^{-u}= 2 cosec x$$
$$ x = \displaystyle \frac{\pi}{4} \Rightarrow u = ln (1+\sqrt{2})$$

$$x = \dfrac{\pi}{2} \Rightarrow u = 0$$
$$\Rightarrow cosec x + cot x = e^u$$ and $$cosec x - cot x = e^{-u} $$
$$\Rightarrow x = \displaystyle \frac{e^u - e^{-u}}{2} (e^u - e^{-u}) dx = -2 cosec x cot x dx$$

$$\Rightarrow - \displaystyle \int(e^u + e^{-u})^{17} \frac{(e^u - e^{-u})}{2 cosec x cot x}du$$

$$=-2 \int_{ln(1+\sqrt{2})}^{0} (e^u + e^{-u})^{16}du$$

$$= \int_{0}^{ln (1+\sqrt{2})} 2(e^u + e^{-u})^{16}du$$
Question 8
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Directions For Questions

Given that for each $$\displaystyle a \in (0, 1), \lim_{h \rightarrow 0^+} \int_h^{1-h} t^{-a} (1 -t)^{a-1}dt$$ exists. Let this limit be $$g(a)$$
In addition, it is given that the function $$g(a)$$ is differentiable on $$(0, 1)$$
Then answer the following question.

...view full instructions

The value of $$g\displaystyle \left ( \frac{1}{2} \right )$$ is?

A
$$\pi$$

B
$$2 \pi$$

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

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

Solution

$$g \displaystyle \left ( \frac{1}{2} \right ) = \lim_{h \rightarrow 0^+} \int_h^{1 - h} t^{-1/2} (1-t)^{-1/2} dt$$
$$=

\int_0^1 \displaystyle \frac{dt}{\sqrt{t - t^2}} = \int_0^1

\frac{dt}{\sqrt{\frac{1}{4} - \left ( t-\frac{1}{2} \right )^2}} =

sin^{-1} \left ( \frac{t - \frac{1}{2}}{\frac{1}{2}} \right ) |_0^1$$
$$= sin^{-1} 1 -sin^{-1} (-1) = \pi$$
Question 9
1 / -0

The value of the integral $$\displaystyle \int_0^1 \dfrac{x^{\alpha}-1}{\log x}dx$$ is

A
$$\log (\alpha+1)$$

B
$$2\log (\alpha+1)$$

C
$$3\log \alpha$$

D
none of these

Solution

$$I(\alpha)=\displaystyle\int_0^1\dfrac{x^{\alpha}-1}{\text{log }x}dx$$

Differentiating with respect to $$\alpha,$$ we get$$,$$
$$I'(\alpha)=\displaystyle\int_0^1x^\alpha dx=\dfrac{1}{\alpha +1}$$
We need to solve this simple differential equation $$:I'(\alpha)=\dfrac{1}{\alpha +1}$$
On solving, we get,
$$I(\alpha)=\text{log}(\alpha +1)+C$$

Now, from the initial equation, we can find that $$I(0)=0$$
$$I(0)=\text{log}(1)+C=0$$ $$\Rightarrow C=0$$

So, $$I(\alpha)=\text{log}(\alpha +1)$$
Question 10
1 / -0

$$\int _{ 0 }^{ \pi }{ \cfrac { xdx }{ 4\cos ^{ 2 }{ x } +9\sin ^{ 2 }{ x } } } =$$

A
$$\cfrac { { \pi }^{ 2 } }{ 12 } $$

B
$$\cfrac { { \pi }^{ 2 } }{ 4 } $$

C
$$\cfrac { { \pi }^{ 2 } }{ 6 } $$

D
$$\cfrac { { \pi }^{ 2 } }{ 3 } $$

Solution

To find the Integral of

$$ \int_{0}^{\Pi }\dfrac{xdx}{4cos^{2}x + 9sin^{2}x} $$

Let ,
$$I = \int_{0}^{\Pi }\dfrac{xdx}{4cos^{2}x + 9sin^{2}x} $$ --------> Equation 1

As we know that ,
$$ \int_{a}^{b } f(x)dx = \int_{a}^{b } f(a+b -x)dx $$

$$ I = \int_{0}^{\Pi }\dfrac{(\Pi -x)dx}{4cos^{2}(\Pi -x) + 9sin^{2}(\Pi -x)} $$

$$ I = \int_{0}^{\Pi }\dfrac{(\Pi -x)dx}{4cos^{2}x + 9sin^{2}x} $$  ------> Equation 2

On Adding Equation 1 and 2 we get
$$ 2I = \int_{0}^{\Pi }\dfrac{(\Pi -x + x)dx}{4cos^{2}x + 9sin^{2}x} $$

$$ I = \dfrac{\Pi }{2}\int_{0}^{\Pi }\dfrac{dx}{4cos^{2}x + 9sin^{2}x} $$

$$ I = \dfrac{\Pi }{18}\int_{0}^{\Pi }\dfrac{dx}{\dfrac{4}{9}cos^{2}x + sin^{2}x} $$

$$ I = \dfrac{\Pi }{18}\int_{0}^{\Pi }\dfrac{sec^{2}xdx}{tan^{2}x + \dfrac{4}{9}} $$

$$ I = \dfrac{2\Pi }{18}\int_{0}^{\Pi/2 }\dfrac{sec^{2}xdx}{(tanx)^{2} + \left ( \dfrac{2}{3} \right )^{2}} $$

$$ I = \dfrac{\Pi }{9}\int_{0}^{\Pi/2 }\dfrac{sec^{2}xdx}{(tanx)^{2} + \left ( \dfrac{2}{3} \right )^{2}} $$

Let $$ tanx = t $$
So that $$sec^{2}x dx =dt$$

When $$ x = 0 , t = 0 $$

When   $$ x = \dfrac{\Pi }{2} $$ , $$t = ∞$$

Above Integral Becomes ,

$$ I = \dfrac{\Pi }{9}\int_{0}^{\infty }\dfrac{dt}{t^{2} + \left ( \dfrac{2}{3} \right )^{2}} $$

$$ I =\dfrac{\Pi }{6} \left [ tan^{-1}\left \{ \dfrac{t}{\dfrac{2}{3}} \right \}\right ]_{t= 0}^{\infty } $$

$$ I =\dfrac{\Pi }{6} \left [ tan^{-1}\left ( \dfrac{3t}{2} \right )\right ]_{t = 0}^{\infty } $$

$$ I =\dfrac{\Pi }{6}\times \left [ tan^{-1}\infty - tan^{-1}(0)\right ]$$

$$ I =\dfrac{\Pi }{6}\times \left [ \dfrac{\Pi }{2} - 0 \right ]$$

$$ I = \dfrac{\Pi }{6}\times \dfrac{\Pi }{2} $$

$$ I = \dfrac{\Pi^{2} }{12} $$

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Re-Attempt Weekly Quiz Competition

Integrals Test - 13

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Integrals Test - 13

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Question 1

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

0

1

2

Solution

Question 2

$$\displaystyle \frac{1}{4} log x^2$$

log x

$$\displaystyle \frac{1}{2} (log x)^2$$

$$\displaystyle \frac{1}{4} (log x)^2$$

Solution

Question 3

$$2\pi$$

$$\pi^2$$

$$2\pi^2$$

$$4\pi$$

Solution

Question 4

$$\displaystyle\frac{15}{128}$$

$$\displaystyle\frac{13}{32}$$

$$\displaystyle\frac{13}{256}$$

$$\displaystyle\frac{15}{64}$$

Solution

Question 5

$$\pi \log 2$$

$$\displaystyle \frac{\pi}{8}\log 2$$

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

$$\log 2$$

Solution

Question 6

$$\dfrac{9}{8}$$

$$-2$$

$$2$$

$$-\dfrac{9}{8}$$

Solution

Question 7

$$\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u + e^{-u})^{16} du$$

$$\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u + e^{-u})^{17} du$$

$$\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u - e^{-u})^{17} du$$

$$\displaystyle \int_{0}^{log(1+\sqrt{2})} 2(e^u - e^{-u})^{16} du$$

Solution

Question 8

$$\pi$$

$$2 \pi$$

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

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

Solution

Question 9

$$\log (\alpha+1)$$

$$2\log (\alpha+1)$$

$$3\log \alpha$$

none of these

Solution

Question 10

$$\cfrac { { \pi }^{ 2 } }{ 12 } $$

$$\cfrac { { \pi }^{ 2 } }{ 4 } $$

$$\cfrac { { \pi }^{ 2 } }{ 6 } $$

$$\cfrac { { \pi }^{ 2 } }{ 3 } $$

Solution

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