MCQ Questions for Class 12 Commerce Maths Integrals Quiz 6

The value of the integral

$\displaystyle \int_0^{1} {{1 + 2 x}}dx$ is

0%
$2$
0%
$3$
0%
$4$
0%
$\frac{ - 1}{4}$

The value of the integral

$\displaystyle \int\limits_0^1 {\dfrac{{{x^3}}}{{1 + {x^8}}}\,\,dx}$ is

0%
$\frac{\pi }{{16}}$
0%
$\frac{\pi }{{4}}$
0%
$\frac{\pi }{{8}}$
0%
none of these

Explanation

$\displaystyle \int _{ 0 }^{ 1 }{ \frac { { x }^{ 3 } }{ 1+{ x }^{ 8 } } dx }$

$=\displaystyle \int _{ 0 }^{ 1 }{ \frac { { x }^{ 3 } }{ 1+({ x }^{ 4 })^{ 2 } } dx }$

$=\displaystyle \dfrac { 1 }{ 4 } \int _{ 0 }^{ 1 }{ \frac { 1 }{ 1+(t)^{ 2 } } dt\quad (substituting\quad t={ x }^{ 4 }\Rightarrow dt=4x^3dx) }$

$=\dfrac { 1 }{ 4 } { \left[ { tan }^{ -1 }(t) \right] }_{ 0 }^{ 1 }$

$=\dfrac 14\left[\dfrac{\pi}{4}-0\right]$

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

Solve:

$\int\limits_0^{\pi /6} {\dfrac{{\cos 2x}}{{{{\left( {\cos x - \sin x} \right)}^2}}}dx}$

0%
$- \log \left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)$
0%
$- \log \left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)$
0%
$\log \left( {\dfrac{{\sqrt 3 + 1}}{2}} \right)$
0%
None of these

Explanation

$\int _{ 0 }^{ \pi /6 }{ \dfrac { \cos 2x }{ \left( \cos x-\sin x \right) ^{ 2 } } dx } \\ \int _{ 0 }^{ \pi /6 }{ \dfrac { \cos ^{ 2 } x-\sin ^{ 2 } x }{ \left( \cos x-\sin x \right) ^{ 2 } } dx } \\ \int _{ 0 }^{ \pi /6 }{ \dfrac { (\cos x-\sin x)(\cos x+\sin x) }{ \left( \cos x-\sin x \right) ^{ 2 } } dx } \\ \int _{ 0 }^{ \pi/6 }{ \dfrac { (\cos x+\sin x) }{ (\cos x-\sin x) } dx } \\ put\quad t=\cos x-\sin x\\ Hence, dt=-(\cos x+\sin x)dx\\ -\displaystyle\int _{ 1 }^{ \frac { \sqrt { 3 } -1 }{ 2 }}{\dfrac {dt }{t}} \\ -\bigg[\log { t } \bigg]_{ 1 }^{ \dfrac { \sqrt { 3 } -1 }{ 2 } }\\ -\left[ \log { \left( \dfrac { \sqrt { 3 } -1 }{ 2 } \right) -\log { \left( 1 \right) } } \right] \\ - \log { \left( \dfrac { \sqrt { 3 } -1 }{ 2 } \right) } \quad \quad \quad \quad \because \log { \left( 1 \right) =0 } \\ \\$

Solve

$\displaystyle\int\limits_0^{\pi /2} {{{\sin }^4}x{{\cos }^3}xdx}$

0%
$\dfrac {6}{35}$
0%
$\dfrac {2}{21}$
0%
$\dfrac {2}{15}$
0%
$\dfrac {2}{35}$

Explanation

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

$\int_{0}^{\pi/2}\sin^{4} x(1-\sin^{2} x)\cos x dx$

$\because \cos^{2} x=1-\sin^{2} x$

$\int_{0}^{\pi/2}(\sin^{4} x-\sin^{6} x)\cos x dx$

substitute

$\sin x=t$ hence,

$dt=\cos x\ dx$

and at

$x=0; \ t=\sin (0)=0 ;$ at

$x=\pi/2; \ t=\sin(\pi/2)=1$

$\int_{0}^{1}(t^{4} -t^{6} ) dt$

$\bigg[\dfrac{t^{5}}{5}-\dfrac{t^{7}}{7}\bigg]_{0}^{1}$

$\bigg[\dfrac{1}{5}-\dfrac{1}{7} \bigg]=\dfrac{2}{35}$

Evaluate :

$\displaystyle\int {\dfrac{9cosx-sinx}{4sinx+5cosx}dx}$

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$x+\ln(4\sin x+\cos x)+c$
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$x+\ln(\sin x-5\cos x)+c$
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$x+\ln(4\sin x+5\cos x)+c$
0%
None of these

Explanation

$\to \displaystyle \int \dfrac{9\cos x - \sin x}{4\sin x + 5 \cos x} dx$

It is in the form

$\dfrac{f(x)}{g(x)}$ , so we will write

$f(x)$ as

$A(g(x)) + B(g'(x))$ .

$\to f(x) = 9\cos x - \sin x$

$\to g(x) = 4\sin x + 5 \cos x$

$g'(x) = 4 \cos x - 5 \sin x$

So,

$f(x) = g(x) + g'(x)$

$\to \displaystyle \int 1+ \dfrac{g'(x)}{g(x)} dx \Rightarrow x + ln (g(x))+C$

Answer

$= x + ln (4\sin x + 5\cos x) + C$

1048864_1177850_ans_edd56aa53b5f407cbaafdd20171be0dc.png

$\int { \cfrac { \sin { x } }{ \sin { \left( x-\alpha \right) } } } dx=Ax+B\log { \sin { \left( x-\alpha \right) } } +c$ , then the value of

$(A,B)$ is-

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$\left( \sin { \alpha } ,\cos { \alpha } \right)$
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$\left( \cos { \alpha } ,\sin { \alpha } \right)$
0%
$\left( -\sin { \alpha } ,\cos { \alpha } \right)$
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$\left(- \cos { \alpha } ,\sin { \alpha } \right)$

Explanation

$\displaystyle \int \frac{sin(x-\alpha +\alpha )}{sin(x-\alpha )}dx$

$\displaystyle \int \frac{sin(x-\alpha )cos\alpha +cos(x-\alpha )sin\alpha }{sin(x-\alpha )}$

$\displaystyle \int (cos\alpha +sin\alpha \,cot(x-\alpha ))dx$

$xcos\alpha +sin\alpha ln sin(x-\alpha )+c$

$A = cos\alpha$

$B = sin\alpha$

1084285_1173392_ans_6ec6706d0f664d52868d64482df94d43.jpg

Evaluate:

$\displaystyle\int _ { 0 } ^ { \pi / 2 } \dfrac { \sin x \cos x } { \cos ^ { 2 } x + 3 \cos x + 2 }dx$

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$\ln\left(\dfrac {5}{3}\right)$
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$\ln\left(\dfrac {4}{3}\right)$
0%
$\ln\left(\dfrac {1}{3}\right)$
0%
None of these

$\displaystyle \int_{0}^{2}\left( \sqrt {\dfrac{4-x}{x} }- \sqrt {\dfrac{x}{4-x}} \right) dx$ is equal to:

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$0$
0%
$8$
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$4$
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$16$

$f(x)=\int { \left( \dfrac { x^2+\sin^2x }{ 1+x^2 } \right) } \sec^2 x dx$ and f(0)=0,then f (1) equals:

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$1-\dfrac { \pi }{ 4 }$
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$-\dfrac { \pi }{ 4 }$
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$\tan 1-\dfrac { \pi }{ 4 }$
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$\tan 1+1$

Explanation

$f(x)=\int { \left( \dfrac { x^2+sin^2x }{ 1+x^2 } \right) } sec^2x dx$

$= \int { \left( \dfrac { x^2+1-(1-sin^2x) }{ 1+x^2 }sec^2x \right) } dx$

$=\int (sec^2x-\dfrac {cos^2x sec^2x} {1+x^2})dx$

$=\int sec^2x- \int \dfrac {1} {1+x^2} dx$

$f(x)= tanx - tan^{-1}x+ \alpha$

$f(0)=0\Rightarrow \alpha=0$

$f(1)=tan 1 -\dfrac { \pi }{ 4 }$

$\int _ { - 4 } ^ { - 5 } e ^ { ( x + 5 ) ^ { 2 } } d x + 3 \int _ { 1 / 3 } ^ { 2 / 3 } e ^ { 9 ( x - 2 / 3 ) ^ { 2 } } d x$ is equal to-

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$e ^ { 5 }$
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$e ^ { 4 }$
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3 $e ^ { 2 }$
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0

The integrating factor of the differential equation

$\dfrac{dy}{dx}\left(x\log _e\:x\right)+y=2\log _e\:x$ is given by

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$x$
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$e^x$
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$\log _e\:x$
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$\log _e\left(\log _e\:x\right)$

Explanation

$\dfrac{dy}{dx}\left(x\log_{e}{x}\right)+y=2\log_{e}{x}$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{1}{\left(x\log_{e}{x}\right)}y=\dfrac{2\log_{e}{x}}{\left(x\log_{e}{x}\right)}$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{1}{\left(x\log_{e}{x}\right)}y=\dfrac 2x$ which is of the form

$\dfrac{dy}{dx}+py=q$

Integrating factor

$={e}^{\int{p}dx}$ where

$p=\dfrac{1}{\left(x\log_{e}{x}\right)}$

Integrating factor

$={e}^{\int{\dfrac{1}{\left(x\log_{e}{x}\right)}}dx}$

Take

$t=\log_{e}{x}\Rightarrow dt=\dfrac{1}{x}dx$

$\int{\dfrac{1}{\left(x\log_{e}{x}\right)}}dx=\int{\dfrac{dt}{t}}=\ln{\left|t\right|}\ where\ t=\log_{e}{x}$

Integrating factor

$={e}^{\int{p}dx}={e}^{\ln t}=t=\log_{e}{x}$

$\displaystyle\int {{{{2^x}} \over {\sqrt {1 - {4^x}} }}dx = K{{\sin }^{ - 1}}} ({2^x}) + C,$ then K is equal to

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$\ell n2$
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$\displaystyle{1 \over 2}\ell n2$
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$\displaystyle{1 \over 2}$
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$\displaystyle{1 \over {\ell n2}}$

Explanation

$I=\displaystyle\int \dfrac{2^x}{\sqrt{1-4^x}}dx$

Let

$2^x=z$

$\Rightarrow 2^x\log 2dx=dz$

$\Rightarrow I=\displaystyle\int \dfrac{\dfrac{1}{\log2}}{\sqrt{1-z^2}}dz$

$=\dfrac{1}{\log 2}\cdot \sin^{-1}z+c$

$=\dfrac{1}{\log 2}\cdot \sin^{-1}(e^x)+c$

$=k\sin^{-1}(2^x)+c$

then

$k=\dfrac{1}{\log 2}$ .

$\int \{1 + 2 \tan x (\tan x + \sec x)\}^{\dfrac{1}{2}} dx =$

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$log (\sec x + \tan x) + c$
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$log (\sec x + \tan x)^{\dfrac{1}{2}} + c$
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$log \,\sec x (\sec x + \tan x) + c$
0%
None of these

Explanation

$\displaystyle\int \sqrt{1+2\tan x(\tan x+\sec x)}dx$

$=\displaystyle\int \sqrt{1+2\tan^2x+2\tan x\sec x}dx$

$=\displaystyle\int \sqrt{\sec^2x-\tan^2x+2\tan^2x+2\tan x\sec x}dx$

$=\displaystyle\int \sqrt{\sec^2x+\tan^2x+2\tan x\sec x}dx$

$=\displaystyle\int \sqrt{(\sec x+\tan x)^2}dx$

$=\displaystyle\int (\sec x+\tan x)dx$

$=\displaystyle\int \sec xdx+\displaystyle\int \tan xdx$

$=log(\sec x+\tan x)+log(\sec x)+c$

$=log(\sec x(\sec x+\tan x)+c$

$\therefore \displaystyle\int \sqrt{1+2\tan x(\tan x+\sec x)}dx=log(\sec x(\sec x+\tan x))+c$ .

The value of the defined integral

$\displaystyle \int^{\pi/2}_{0}(\sin x+\cos x)\sqrt {\dfrac {e^{x}}{\sin x}}dx$ equals

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$2\sqrt {e^{\pi/2}}$
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$\sqrt {e^{\pi/2}}$
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$2\sqrt {e^{\pi/2}}.\cos 1$
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$\dfrac {1}{2}e^{\pi/4}$

$\int \log _ { 10 } x d x =$

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$x \log _ { 10 } x + c$
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$x \left( \log _ { 10 } x + \log _ { 10 } e \right) + c$
0%
$\log _ { 10 } x + c$
0%
$x \left( \log _ { 10 } x - \log _ { 10 } e \right) + c$

Explanation

$\int log_{10}xdx\\=\int(\dfrac{log_ex}{log_e10})dx\\=(\dfrac{1}{log_e10})\int logx\cdot1\cdot dx\\(\dfrac{1}{log_e10})[(logx)\cdot x-\int(\dfrac{1}{x}\cdot x dx)]\\=(\dfrac{d1}{log_e10})(xlogx-x)+c\\=x((\dfrac{log_ex}{log_e10})-(\dfrac{1}{log_e10}))+c\\=x(log_{10}x-log_{10}e)+C$

$\int { f\left( x \right)dx=f\left( x \right) } ,$ then

$\int { \left\{ f\left( x \right) \right\}^2 }$ dx is equal to :

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$\frac { 1 }{ 2 } \left\{ f\left( x \right) \right\}^2$
0%
${ \left\{ f\left( x \right) \right\}^3 }$
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$\dfrac { { \left\{ f\left( x \right) \right\} ^{ 3 } } }{ 3 }$
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${ \left\{ f\left( x \right) \right\}^2 }$

Evaluate:

$\displaystyle \int _ { 0 } ^ { \pi / 4 } \sec ^ { 7 } {\theta} \sin ^ { 3 } {\theta} {d \theta} =$

0%
$\dfrac { 1 } { 12 }$
0%
$\dfrac { 3 } { 12 }$
0%
$\dfrac { 5 } { 12 }$
0%
$\dfrac { 7 } { 12 }$

Explanation

$\displaystyle \int_{0}^{\pi /4} sec^{7}\theta \sin^{3}\theta d\theta$

$=\displaystyle \int_{0}^{\pi /4} \frac{\sec^{4}\theta }{\cos^{3}\theta }sin^{3}\theta d\theta$

$=\displaystyle\int_{0}^{\pi /4} \tan^{3}\theta \sec^{4}\theta d\theta$

$= \displaystyle\int_{0}^{\pi /4} \tan^{3}\theta \sec^{2}\theta .\sec^{2}\theta d\theta$

$=\displaystyle\int_{0}^{\pi /4} tan^{3}\theta (1+tan^{2}\theta )\sec^{2}\theta d\theta$

$u = tan\theta \Rightarrow du = \sec^{2}\theta d\theta$

$=\displaystyle\int_{0}^{1} u^{3}(1+u^{2})du$

$=\displaystyle\int_{0}^{1} (u^{5}+u^{3})du$

$= \dfrac{u^{6}}{6}+\dfrac{u^{4}}{4}|_{0}^{1}$

$= \dfrac{1}{6}+\dfrac{1}{4} = \dfrac{5}{12}$

$\displaystyle \int \frac { 1 - x ^ { 2 } } { \left( 1 + x ^ { 2 } \right) \sqrt { 1 + x ^ { 4 } } } d x$ is equal to

0%
$\sqrt { 2 } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$
0%
$\frac { 1 } { \sqrt { 2 } } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$
0%
$\frac { 1 } { 2 } \sin ^ { - 1 } \left\{ \frac { \sqrt { 2 } x } { x ^ { 2 } + 1 } \right\} + c$
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$\frac { 1 } { \sqrt { 2 } } \sin ^ { - 1 } \left\{ \frac { x ^ { 2 } + 1 } { \sqrt { 2 } x } \right\} + c$

Integral of

$f ( x ) = \sqrt { 1 + x ^ { 2 } }$ with respect to

$x ^ { 2 }$ is

0%
$\frac { 2 } { 3 } \frac { \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } } { x } + k$
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$\frac { 2 } { 3 } \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } + k$
0%
$\frac { 2 } { 3 } x \left( 1 + x ^ { 2 } \right) ^ { 3 / 2 } + k$
0%
None of these

Explanation

$f(x)=\sqrt{1+x^2}$

$\int \sqrt{1+x^2} dx^2$

$=\dfrac{(1+x^2)^{\frac{3}{2}}}{\frac{3}{2}}$

$=\dfrac{2}{3}(1+x^2)^{\frac{3}{2}}+k$

$\int { { e }^{ x^{ 3 } }+{ x }^{ 2-1 }(3{ x }^{ 4 }+{ 2x }^{ 3 }+{ 2x }^{ 2 }\quad x=h(x)+c }$ then the value of

$h(1)h(-1)$ .

0%
1
0%
-1
0%
2
0%
-2

$\int _{ 0 }^{ 1 }{ \dfrac { dx }{ \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+2 \right) } }$ =

0%
$\dfrac { \pi }{ 4 } +\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } }$
0%
$\dfrac { \pi }{ 2 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } }$
0%
$\dfrac { \pi }{ 4 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } }$
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$\dfrac { \pi }{ 3 } -\dfrac { 1 }{ \sqrt { 2 } } { tan }^{ -1 }\dfrac { 1 }{ \sqrt { 2 } }$

The value of the integral

$\displaystyle \int_{-\pi/2}^{\pi/2} \left(x^{2}+\log \dfrac{\pi-x}{\pi+x}\right) \cos x dx$ is

0%
$0$
0%
$\dfrac{\pi^{2}}{2}-4$
0%
$\dfrac{\pi^{2}}{2}+4$
0%
$\dfrac{\pi^{2}}{2}$

Explanation

$I=\displaystyle \int_{-\pi/2}^{\pi/2}\left[x^2+log\left( \dfrac{\pi-x}{\pi+x}\right)\right]cosxdx$

As,

$\displaystyle \int_{-a}^{a}f(x)dx=0,$ when

$f(-x)=-f(x)$

$\therefore I=\displaystyle \int_{-\pi/2}^{\pi/2}x^2cosxdx+0=2\displaystyle \int_{0}^{\pi/2}(x^2cosx)dx$

$=2\left\{ (x^2sinx)_{0}^{\pi/2}-\displaystyle \int_{0}^{\pi/2}(x^3cosx) \right\}$

$2\left[ \dfrac{\pi^2}{4}-2\left\{(-xcosx)_{0}^{\pi/2}-\displaystyle \int_{0}^{\pi/2}1.(-cosx)dx \right\}\right]$

$=2\left[ \dfrac{\pi^2}{4}-2(sinx)_{0}^{\pi/2}\right]=2\left[\dfrac{\pi^2}{4}-2\right]=\left( \dfrac{\pi^2}{2}-4\right)$

The value of

$\int _{ -1 }^{ 1 }{ \dfrac { { cot }^{ -1 }x }{ \pi } } dx$

0%
1
0%
2
0%
3
0%
0

$\displaystyle \int _{ 0 }^{ { \pi }^{ 2 } }{ \dfrac { \sin { \sqrt { x } } }{ \sqrt { x } } } dx$ is equal to

0%
$2$
0%
$1$
0%
$1/2$
0%
$4$

The integral

$\displaystyle{\int}_{\pi/12}^{\pi/4}\dfrac{8\cos 2x}{\left(\tan x+\cot x\right)^{3}}dx$ equals:

0%
$\dfrac{15}{128}$
0%
$\dfrac{13}{156}$
0%
$\dfrac{15}{64}$
0%
$\dfrac{13}{32}$

Solve:

$\int {\dfrac{{dx}}{{\left( {x - 3} \right)\sqrt {x + 1} }}}$

0%
$\cos { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$
0%
$\sin { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$
0%
$-\sin { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$
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$-\cos { { h }^{ -1 } } \left( \dfrac { 1-x }{ \sqrt { 3 } \left( 1+x \right) } \right) +c$

The integral

$\int _{ 2a/4 }^{ a/2 }{ (2\quad cosecx{ ) }^{ 17 } }$ dx is equal to:

0%
$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ 2({ e }^{ u }+{ e }^{ -u }{ ) }^{ 16 } } du$
0%
$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ ({ e }^{ u }+{ e }^{ -u }{ ) }^{ 17 } } du$
0%
$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ ({ e }^{ u }-{ e }^{ -u }{ ) }^{ 17 } } du$
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$\int _{ 0 }^{ log(1+\sqrt { 2 } ) }{ 2({ e }^{ u }-{ e }^{ -u }{ ) }^{ 16 } } du$

$\begin{matrix} lim \\ n\rightarrow \infty \end{matrix}\int _{ 0 }^{ 1 }{ \frac { { nx }^{ { n- }1 } }{ { 1+x }^{ 2 } } dx= }$

0%
0
0%
1
0%
2
0%
$\frac { 1 }{ 2 }$

The value of the integral

$\int _{ -\pi /2 }^{ \pi /2 }{ \left[ { x }^{ 2 }+log\frac { \pi -x }{ \pi +x } \right] }$ cos x dx is

0%
0
0%
$\frac { { \pi }^{ 2 } }{ 2 } -4$
0%
$\frac { { \pi }^{ 2 } }{ 2 } +4$
0%
$\frac { { \pi }^{ 2 } }{ 2 }$

$\int _ { 0 } ^ { 1 } \frac { d x } { \sqrt { x + 1 } + \sqrt { x } } d x =$

0%
$\frac { 4 } { 3 } ( \sqrt { 2 } + 1 )$
0%
$\frac { 4 } { 3 } ( \sqrt { 2 } - 1 )$
0%
$\frac { 3 } { 4 } ( \sqrt { 2 } - 1 )$
0%
$\frac { 3 } { 4 } ( \sqrt { 2 } - 2 )$

The value of the definite integral

$\overset { { a }_{ 1 } }{ \underset { { a }_{ 2 } }{ \int { \frac { d\theta }{ 1+tan\theta } } } } =\frac { 501\pi }{ K }$ where

$\ a _{ 2 }=\quad \frac { 1003\pi }{ 2008 }$ and

${ \ a }_{ 1 }=\frac { \pi }{ 2008 }$ The value of K equalls

0%
2007
0%
2006
0%
2009
0%
2008

Explanation

$\begin{array}{l} \int _{ { a_{ 2 } } }^{ { a_{ 1 } } }{ \frac { { d\theta } }{ { 1+\tan \theta } } } \\ I=\int _{ \frac { \pi }{ { 2008 } } }^{ \frac { { 1003\pi } }{ { 2008 } } }{ \frac { { d\theta } }{ { 1+\tan \theta } } } \\ I=\int _{ \frac { \pi }{ { 2008 } } }^{ \frac { { 1003\pi } }{ { 2008 } } }{ \frac { { \tan \theta } }{ { 1+\tan \theta } } }d\theta \\ 2I=\int _{ \frac { \pi }{ { 2008 } } }^{ \frac { { 1003\pi } }{ { 2008 } } }{ d\theta } \\ 2I=\frac { { 1002\pi } }{ { 2008 } } \\ I=\frac { { 501 } }{ { 2008 } } \pi \\ k=2008 \end{array}$

Option

$D$ is correct answer.

For

$x\in R,\ f(x)=|\log 2-\sin x|$ and

$g(x)=f(f(x))$ , then

0%
$g'(0)=\cos (\log 2)$
0%
$g'(0)=-\cos (\log 2)$
0%
$g$ is differentible at $x=0$ and $g'(0)=-\sin (\log 2)$
0%
$g$ is not differentiable at $x=0$

Explanation

In the neighourwood of

$x=0$

$f(x)=\log 2-\sin x$

$g(x)=f(x)=\log 2-\sin f(x)$

$=\log 2-\sin (\log 2-\sin x)$

$g'(x)=\cos (\log2-\sin x)(-\cos x)$

$g'(0)=\cos (log 2)$

option

$A$ is correct answer.

The integral

$\displaystyle \int { \left( 1+2{ x }^{ 2 }+\frac { 1 }{ x } \right) } { e }^{ { x }^{ 2-\frac { 1 }{ x } } }dx$ is equal to

0%
$(2x-1).e^{x^{2-\dfrac {1}{x}}}+c$
0%
$(2x+1).e^{x^{2-\dfrac {1}{x}}}+c$
0%
$xe^{x^{2-\dfrac {1}{x}}}+c$
0%
$-xe^{x^{2-\dfrac {1}{x}}}+c$

Explanation

$\begin{array}{l} Let \\ I=\int { { e^{ { x^{ 2 } }-\frac { 1 }{ x } } }\left( { 1+2{ x^{ 2 } }+\frac { 1 }{ x } } \right) dx } \\ =\int { { e^{ { x^{ 2 } }-\frac { 1 }{ x } } }\left( { x\left( { 2x+\frac { 1 }{ { { x^{ 2 } } } } } \right) +1 } \right) dx } \\ =\int { { e^{ g } }\left( { fg'+f' } \right) dx } \\ ={ e^{ g } }f+C \\ ={ e^{ x-\frac { 1 }{ { { x^{ 2 } } } } } }\times x+C \\ Hence,\, option\, C\, is\, the\, correct\, answer. \end{array}$

$\displaystyle \int \dfrac {1}{\sqrt {\sin^{3}x\sin(x+a)}}dx$ is equal to

0%
$2\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \tan x}+c$
0%
$-2\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \cot x}+c$
0%
$\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \tan x}+c$
0%
$-\cos ec\alpha \sqrt {\cos \alpha +\sin \alpha \tan x}+c$

Explanation

$\begin{array}{l} \int { \frac { { dx } }{ { \sqrt { { { \sin }^{ 3 } }x\sin \left( { x+a } \right) } } } } \\ =\int { \frac { { dx } }{ { \sqrt { { { \sin }^{ 3 } }x\sin x\left( { \cos a+\cot x\sin a } \right) } } } } \\ =\int { \frac { { \cos e{ c^{ 2 } }xdx } }{ { \sqrt { \cos a+\sin a\cot x } } } } \\ \cos a+\sin a\cot =t \\ \frac { { dt } }{ { dx } } =-\sin a\cos e{ c^{ 2 } }x \\ -\frac { 1 }{ { \sin a } } dt=\cos e{ c^{ 2 } }xdx \\ =-\frac { 1 }{ { \sin a } } \int { \frac { { dt } }{ { \sqrt { t } } } } \\ =-\frac { 1 }{ { \sin a } } \int { { t^{ \frac { { -1 } }{ 2 } } }dt } \\ =\frac { { -2 } }{ { \sin a } } { t^{ \frac { 1 }{ 2 } } }+C \\ =-2\cos eca\sqrt { \cos a+\sin a\cot x } +C \end{array}$

$\int {\dfrac{1}{{9{x^2} - 25}}dx = \_\_\_\_\_\_ + c.}$

0%
$\dfrac{1}{{30}}\log \left| {\dfrac{{3x + 5}}{{3x - 5}}} \right|$
0%
$\log \left| {x + \sqrt {3x - 5} } \right|$
0%
$\dfrac{1}{{30}}\log \left| {\dfrac{{3x - 5}}{{3x + 5}}} \right|$
0%
$\log \left| {x - \sqrt {3x - 5} } \right|$

Explanation

$I=\int { \dfrac { 1 }{ {9{ x^{ 2 } }-25 } } dx }$

$I=\dfrac { 1 }{ 9 } \int { \dfrac { 1 }{ { \left( { { x^{ 2 } }-\dfrac { { 25 } }{ 9 } } \right) } } dx }$

$I=\dfrac { 1 }{ 9 } \int { \dfrac { 1 }{ { { { x^{ 2 } }-\left(\dfrac { 5 }{ 3 }\right)^2 } } } dx }$

We know that

$\int \dfrac { dx }{ x^ 2-a^2}=\dfrac{1}{2a}\log\left(\dfrac{x-a}{x+a}\right)+C$

Therefore,

$I =\dfrac { 1 }{ 9 } .\dfrac { 1 }{ { 2\times \dfrac { 5 }{ 3 } } } \log \left| { \dfrac { { x-\dfrac { 5 }{ 3 } } }{ { x+\dfrac { 5 }{ 3 } } } } \right| +C$

$I=\dfrac { 3 }{ { 90 } } \log \left| { \dfrac { { \dfrac { { 3x-5 } }{ 3 } } }{ { \dfrac { { 3x+5 } }{ 3 } } } } \right| +C$

$I =\dfrac { 1 }{ { 30 } } \log \left| { \dfrac { { 3x-5 } }{ { 3x+5 } } } \right| +C$

Hence, this is the answer.

$\int {{e^{3{{\log }_e}x}}.{{\left( {{x^4} + 1} \right)}^{ - 1}}dx = \_\_\_\_\_\_\_\_\_ + C.}$

0%
$\log \left( {{x^4} + 1} \right)$
0%
$\frac{1}{4}\log \left( {{x^4} + 1} \right)$
0%
$-\log \left( {{x^4} + 1} \right)$
0%
$\frac{{ - 3}}{{{{\left( {{x^4} + 1} \right)}^3}}}$

Explanation

$I=\int { { e^{ 3\log_ex } }.{ { \left( { { x^{ 4 } }+1 } \right) }^{ -1 } }dx }$

$I=\int { { e^{ \log_ex^3 } }.{ { \left( { { x^{ 4 } }+1 } \right) }^{ -1 } }dx }$

$I=\int \dfrac{ { x^3 }}{{ { \left( { { x^{ 4 } }+1 } \right) } }}dx$

Let

$t=x^4+1$

$dt=4x^3dx$

Therefore,

$I=\dfrac{1}{4}\int \dfrac{1}{t}dt$

$I=\dfrac{1}{4}\log(t)+C$

Put the value of

$t$ , we get

$I=\dfrac{1}{4}\log(x^4+1)+C$

Hence, this is the answer.

$\int {\sqrt {1 - \cos x} \,dx = \_\_\_\_\_\_\_ + C;\,2\pi < x < 3\pi }$

0%
$- 2\sqrt 2 \cos \dfrac{x}{2}$
0%
$- \sqrt 2 \cos \dfrac{x}{2}$
0%
$2\sqrt 2 \cos \dfrac{x}{2}$
0%
$\dfrac{{ - 1}}{2}\sqrt 2 \cos \dfrac{x}{2}$

Explanation

We have,

$I= \int { \sqrt { 1-\cos x } dx }$

$I= \int { \sqrt { \left( { { { \sin }^{ 2 } }\dfrac { x }{ 2 } +{ { \cos }^{ 2 } }\dfrac { x }{ 2 } -{ { \cos }^{ 2 } }\dfrac { x }{ 2 } +{ { \sin }^{ 2 } }\dfrac { x }{ 2 } } \right) } dx }$

$I= \int { \sqrt { 2{ { \sin }^{ 2 } }\dfrac { x }{ 2 } } dx }$

$I= \sqrt { 2 } \int { \sin \dfrac { x }{ 2 } dx }$

$I= \sqrt { 2 } \dfrac{1}{\dfrac{1}{2}}(-\cos \dfrac{x}{2})+C$

$I= -2\sqrt { 2 }\cos \dfrac{x}{2}+C$

Hence, this is the answer.

$\int _{ 0 }^{ 400\pi }{ \sqrt { 1-\cos { 2x } } }$

0%
$200\sqrt 2$
0%
$400\sqrt 2$
0%
$800\sqrt 2$
0%
$none$

Explanation

$I=\int _{ 0 }^{ 400\pi }{ \sqrt { 1-\cos x } dx } \\ =400\int _{ 0 }^{ \pi }{ \sqrt { 1-\cos 2x } dx } \\ =800\int _{ 0 }^{ \pi }{ \sqrt { 2 } \sqrt { { { \sin }^{ 2 } }x } dx } \\ =800\sqrt { 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sin xdx } \\ =800\sqrt { 2 } \left[ { \cos x } \right] _{ \frac { \pi }{ 2 } }^{ 0 } \\ =800\sqrt { 2 }$

$\\ Hence,\, the\, option\, C\, is\, the\, correct\, answer.$

$g\left( x \right) =\int { { x }^{ x }\log _{ e }{ (ex)dx } }$ then

$g\left( \pi \right)$ equals

0%
$\pi \log _{ e }{ \pi }$
0%
${ \pi }^{ \pi }\log _{ e }{ (e\pi } )$
0%
${ \pi }^{ \pi }\log _{ e }{ (\pi } )$
0%
${\pi}^\pi$

Explanation

$\begin{array}{l} g\left( x \right) =\int _{ }^{ }{ { x^{ x } } } \left( { 1+\log { e^{ x } } } \right) dx \\ =\int _{ }^{ }{ d\left( { { x^{ x } } } \right) } \\ g\left( x \right) ={ x^{ x } } \\ g\left( \pi \right) ={ \pi ^{ \pi } } \end{array}$

$f(a-x)=-f(x)$ , then

$\displaystyle \int_{0}^{a}f(x)dx=0$ .

0%
True
0%
False

Explanation

$\displaystyle\int_{-a}^{a}{f\left(x\right)dx}=\displaystyle\int_{-a}^{0}{f\left(x\right)dx}+\displaystyle\int_{0}^{a}{f\left(x\right)dx}$

$\displaystyle\int_{a}^{a}{f\left(x\right)dx}=I+\displaystyle\int_{0}^{a}{f\left(x\right)dx}$

Now

$I=\displaystyle\int_{-a}^{0}{f\left(x\right)dx}$

Put

$x=-t$

$\Rightarrow dx=-dt$

When

$x=-a,\,t=a$ and when

$x=0,\,t=0$

$I=\displaystyle\int_{a}^{0}{f\left(-t\right)\left(-dt\right)}$

$=-\displaystyle\int_{a}^{0}{f\left(-t\right)dt}$

$=\displaystyle\int_{0}^{a}{f\left(-t\right)dt}...........$ since

$\displaystyle\int_{a}^{b}{f\left(x\right)dx}=-\displaystyle\int_{b}^{a}{f\left(x\right)dx}$

$=\displaystyle\int_{0}^{a}{f\left(-x\right)dt}..........$ since

$\displaystyle\int_{a}^{b}{f\left(x\right)dx}=-\displaystyle\int_{b}^{a}{f\left(t\right)dt}$

Thus,

$\displaystyle\int_{-a}^{a}{f\left(x\right)dx}=\displaystyle\int_{0}^{a}{f\left(-x\right)dx}+\displaystyle\int_{0}^{a}{f\left(x\right)dx}$

$=\displaystyle\int_{0}^{a}{\left[f\left(-x\right)+f\left(x\right)\right]dx}$

$=2\displaystyle\int_{0}^{a}{f\left(x\right)dx}$

$f\left(x\right)$ is an odd function, then

$f\left(-x\right)=-f\left(x\right)$

$\Rightarrow \displaystyle\int_{-a}^{a}{f\left(x\right)dx}=\displaystyle\int_{0}^{a}{\left[-f\left(x\right)+f\left(x\right)\right]dx}=0$

Hence proved.

Thus, the given statement is true.

$\displaystyle \int_{0}^{1}\sin^{-1}x dx=\dfrac {\pi}{2}-1$

0%
True
0%
False

$\displaystyle\int{{e}^{{\tan}^{-1}{x}}\left(\dfrac{1+x+{x}^{2}}{1+{x}^{2}}\right)dx}$ is equal to

0%
${ -e }^{ { tan }^{ -1 }x }+c$
0%
${ e }^{ { tan }^{ -1 }x }+c$
0%
${ -xe }^{ { tan }^{ -1 }x }+c$
0%
${ xe }^{ { tan }^{ -1 }x }+c$

Explanation

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

Let

$x=\tan{t}\rightarrow dx={\sec}^{2}{t}dt$

$=\displaystyle\int{{e}^{{\tan}^{-1}{\left(\tan{t}\right)}}\left(\dfrac{1+\tan{t}+{\tan}^{2}{t}}{1+{\tan}^{2}{t}}\right){\sec}^{2}{t}dt}$

$=\displaystyle\int{{e}^{t}\left(\dfrac{\tan{t}+{\sec}^{2}{t}}{{\sec}^{2}{t}}\right){\sec}^{2}{t}dt}$

$=\displaystyle\int{{e}^{t}\left(\tan{t}+{\sec}^{2}{t}\right)dt}$

$=\displaystyle\int{{e}^{t}\tan{t}dt}+\displaystyle\int{{e}^{t}{\sec}^{2}{t}dt}$

$\tan{t}\displaystyle\int{{e}^{t}dt}-\displaystyle\int{\dfrac{d}{dt}\left(\tan{t}\right)\left(\displaystyle\int{{e}^{t}dt}\right)dt}+\displaystyle\int{{e}^{t}\left({\sec}^{2}{t}\right)dt}+c$

$={e}^{t}\tan{t}-\displaystyle\int{{e}^{t}\left({\sec}^{2}{t}\right)dt}+\displaystyle\int{{e}^{t}\left({\sec}^{2}{t}\right)dt}+c$

$={e}^{t}\tan{t}+c$

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

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

Hence

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

Integrate:

$\displaystyle \int \dfrac{x}{\sqrt{x+4}}dx$

0%
$\dfrac 23(x+4)^{\tfrac 32}-8\sqrt{x+4}$
0%
$\dfrac 23(x+4)^{\tfrac 32}+8\sqrt{x+4}$
0%
$\dfrac 23(x+4)^{\tfrac 32}+4\sqrt{x+4}$
0%
None of these

Explanation

Given,

$I=\displaystyle \int \dfrac{x}{\sqrt{x+4}}dx$

let

$u=\sqrt{x+4}$

$\Rightarrow du=\dfrac{1}{\sqrt{x+4}}$

$\Rightarrow \displaystyle \int \:2\left(u^2-4\right)du$

$\displaystyle =2\left(\int \:u^2du-\int \:4du\right)$

$=2\left(\dfrac{u^3}{3}-4u\right)$

$=2\left(\dfrac{\left(\sqrt{x+4}\right)^3}{3}-4\sqrt{x+4}\right)$

$=2\left(\dfrac{1}{3}\left(x+4\right)^{\frac{3}{2}}-4\sqrt{x+4}\right)$

$=\dfrac{2}{3}\left(x+4\right)^{\frac{3}{2}}-8\sqrt{x+4}$

$\int _{ 0 }^{ 1 }{ \frac { x }{ { \left( { x }^{ 2 }+1 \right) }^{ \frac { 3 }{ 2 } } } dx } =........$

0%
$\dfrac{1}{3}$
0%
$\dfrac{2}{3}$
0%
$\dfrac{3}{2}$
0%
$1-\dfrac{1}{\sqrt{2}}$

Explanation

$\displaystyle\int _{ 0 }^{ 1 }{ \dfrac { x }{ { \left( { x }^{ 2 }+1 \right) }^{ \frac { 3 }{ 2 } } } dx } =\dfrac{1}{2}\int ^{1}_0 \dfrac{d(x^2+1)}{(x^2+1)^{3/2}}=\dfrac{1}{2}\bigg[\dfrac{(x^2+1)^{-3/2+1}}{-3/2+1}\bigg]^{1}_{0}=\bigg[\dfrac{1}{2}\times \dfrac{(x^2+1)^{-1/2}}{-1/2}\bigg]^{1}_{0}=-[\dfrac{1}{\sqrt{2}}-1\bigg]=1-\dfrac{1}{\sqrt{2}}$

$\int _{ -1 }^{ 1/2 }{ \dfrac { { e }^{ x }\left( 2-{ x }^{ 2 } \right) dx }{ \left( 1-x \right) \sqrt { 1-{ x }^{ 2 } } } }$ is equal to

0%
$\dfrac { \sqrt { e } }{ 2 } \left( \sqrt { 3 } +1 \right)$
0%
$\dfrac { \sqrt { 3e } }{ 2 }$
0%
$\sqrt { 3e }$
0%
$\sqrt { \dfrac { e }{ 3 } }$

$\int _{ log2 }^{ x }{ \dfrac { dx }{ \sqrt { { e }^{ x }-1 } } } =\dfrac { \pi }{ 6 } ,$ then x is equal to _________.

0%
4
0%
in 8
0%
in 4
0%
None of these

$\displaystyle\int^{\dfrac{\pi}{2}}_0\dfrac{\cot x}{\cot x+cosec x}dx=m(\pi +n)$ , then

$mn$ is equal to?

0%
$-1$
0%
$1$
0%
$\dfrac{1}{2}$
0%
$-\dfrac{1}{2}$

Explanation

$\displaystyle\int^{\pi /2}_0\dfrac{\cot xdx}{\cot x+cosec x}$

$\displaystyle\int^{\pi /2}_0\dfrac{\cos x}{\cos x+1}=\displaystyle\int \dfrac{2\cos^2\dfrac{x}{2}-1}{2\cos^2\dfrac{x}{2}}$

$\displaystyle\int^{\pi/2}_0\left(1-\dfrac{1}{2}\sec^2\dfrac{x}{2}\right)dx$

$\left[x-\tan \dfrac{x}{2}\right]^{\dfrac{\pi}{2}}_0$

$\Rightarrow$

$\dfrac{1}{2}[\pi -2]$

$\Rightarrow$

$m=\dfrac{1}{2}, n=-2$

$mn=-1$ .

What is

$\displaystyle \int \dfrac{dx}{2x^2 - 2x + 1}$ equal to ?

0%
$\dfrac{\tan^{-1} (2x - 1)}{2} + c$
0%
$2 \tan^{-1} (2x - 1) += c$
0%
$\dfrac{\tan^{-1} (2x + 1)}{2} + c$
0%
$\tan^{-1} (2x - 1) + c$

Explanation

Given,

$\int \dfrac{1}{2x^2-2x+1}dx$

complete the square

$=\int \dfrac{1}{2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}}dx$

apply

$u=x-\frac{1}{2}$

$\int \dfrac{2}{4u^2+1}du$

apply

$u=\frac{1}{2}v$

$=2\cdot \int \dfrac{1}{2\left(v^2+1\right)}dv$

$=2\cdot \dfrac{1}{2}\tan ^{-1} \left(v\right)$

$=\tan ^{-1} \left(2\left(x-\dfrac{1}{2}\right)\right)$

$=\tan ^{-1} \left(2x-1\right)$

$=\tan ^{-1}\left(2x-1\right)+C$

The value of

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

0%
$2\pi$
0%
$\pi^2$
0%
$2\pi^2$
0%
$4\pi$

Explanation

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

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

Select and write the most appropriate answer from the given alternatives for question :
If

$\displaystyle \int^k_0 4x^3dx=16$ , then the value of

$k$ is _____.

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

Given:

$\displaystyle \int^K_0 4x^3dx=16$

$4\left(\dfrac{x^4}{4}\right)^K_0=16$

$(x^4)^k_0=16$

$k^4=16$

$k^4=2^4$

$\boxed{k=2}$

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 6 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

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