MCQ Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 6

$\int _{ 0 }^{ \pi }{ x\ln { \left( \sin { x } \right) } } dx=$

0%
$\dfrac { \pi }{ 2 } \ln { 2 }$
0%
$\dfrac { -\pi^{2} }{ 2 } \ln { 2 }$
0%
$\dfrac { -\pi }{ 2 } \ln { 2 }$
0%
$-2p\ \ln { 2 }$

Explanation

$I = \int_0^\pi {x\ln \left( {\sin x} \right)\ln }$

$\Rightarrow I = \int_0^\pi {\left( {\pi - x} \right)\ln } \left[ {\sin \left( {\pi - x} \right)} \right]dx$

$\Rightarrow I = \int_0^\pi {\pi \ln \sin xdx - \int_0^\pi {x\ln \sin xdx} }$

$\Rightarrow I = \pi \int_0^\pi {\ln \sin xdx - I}$

$\Rightarrow 2I = \pi \int_0^\pi {\ln \sin xdx}$

$\Rightarrow 2I = 2\pi \int_0^\pi {\ln \sin xdx}$

$\Rightarrow I = \pi \int_0^\pi {\ln \sin xdx} - eq(1)$

$\Rightarrow I = \pi \int_0^{\pi /2} {\ln \sin \left( {\dfrac{\pi }{2} - x} \right)} dx$

$\Rightarrow I = \pi \int_0^{\pi /2} {\ln \cos x} dx - eq\left( 2 \right)$

$\Rightarrow 2I = \pi \int_0^{\pi /2} {\ln \dfrac{{2\sin a\cos x}}{2}} dx$

$\Rightarrow 2I = \pi \int_0^{\pi /2} {\ln {{\sin }^2}xdx - \pi \int_0^{\pi /2} {\ln 2dx} }$

$\Rightarrow 2I = \pi {\int_0^{\pi /2} {\ln {{\sin }^2}xdx - \pi \ln 2\left[ x \right]} _0}^{\pi /2}$

$\Rightarrow 2I = \pi \int_0^{\pi /2} {\ln {\mathop{\rm sintdt}\nolimits} = \pi \ln 2\left[ {\dfrac{\pi }{2}} \right]}$

$\Rightarrow 2I = \dfrac{\pi }{2}\int_0^\pi {\ln \sin xdx - \dfrac{{{\pi ^2}}}{2}{{\ln }^2}}$

$\Rightarrow 2I = \dfrac{\pi }{2} \times 2\int_0^{\pi /2} {\ln {\mathop{\rm sintdt}\nolimits} - \dfrac{{{\pi ^2}}}{2}{{\ln }^2}}$

$\Rightarrow 2I = \pi \int_0^{\pi /2} {\ln {\mathop{\rm sintdt}\nolimits} - \frac{{{\pi ^2}}}{2}{{\ln }^2}}$

$\Rightarrow 2I = \pi \times \frac{I}{\pi } - \dfrac{{{\pi ^2}}}{2}{\ln ^2}$

$\Rightarrow 2I - I = \dfrac{{ - {\pi ^2}}}{2}{\ln ^2}$

$\Rightarrow I = \dfrac{{ - {\pi ^2}}}{2}\ln 2$

Evaluate

$\displaystyle\int^{\pi}_0\dfrac{dx}{(1+\sin x)}$

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

Explanation

Let

$I=\displaystyle\int^{\pi}_0\dfrac{dx}{(1+\sin x)}$

$=\displaystyle\int^{\pi}_0\dfrac{(1-\sin x)}{(1-\sin^2x)}dx$

$=\displaystyle\int^{\pi}_0\dfrac{(1-\sin x)}{\cos^2x}dx$ .

$=\displaystyle\int^{\pi}_0[\sec^2x-\sec x\tan x]dx$

$=[\tan x-\sec x]^{\pi}_0$

$=(0+1)-(0-1)$

$=2$ .

Evaluate

$\displaystyle\int^2_0\dfrac{dx}{\sqrt{4-x^2}}$

0%
$1$
0%
$\sin^{-1}\dfrac{1}{2}$
0%
$\dfrac{\pi}{4}$
0%
None of these

$\displaystyle \int_{0}^{1}\sin \left(2\tan^{-1}\sqrt{\dfrac{1+x}{1-x}}\right)dx=$

0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{2}$
0%
$\pi$

Explanation

$\int_0^1 {\sin \left( {2{{\tan }^{ - 1}}\sqrt {\cfrac{{1 + x}}{{1 - x}}} } \right)dx}$

$x = \cos \theta$

$\Rightarrow dx = - \sin \theta d\theta$

$= - \int_{\cfrac{\pi }{2}}^0 {\sin \left( {2{{\tan }^{ - 1}}\sqrt {\cfrac{{1 + \cos \theta }}{{1 - \cos \theta }}} } \right)\sin \theta d\theta }$

$= \int_0^{\cfrac{\pi }{2}} {\sin \left( {2{{\tan }^{ - 1}}\sqrt {\cfrac{{2{{\cos }^2}\cfrac{\theta }{2}}}{{2{{\sin }^2}\cfrac{\theta }{2}}}} } \right)\sin \theta d\theta }$ (Since

$1 + \cos 2\theta = 2{\cos ^2}\theta ,1 - \cos 2\theta = 2{\sin ^2}\theta$ )

$= \int_0^{\cfrac{\pi }{2}} {\sin \left( {2{{\tan }^{ - 1}}\left( {\cot \cfrac{\theta }{2}} \right)} \right)} \sin \theta d\theta$

$= \int_0^{\cfrac{\pi }{2}} {\sin \left( {2{{\tan }^{ - 1}}\left( {\tan \left( {\cfrac{\pi }{2} - \cfrac{\theta }{2}} \right)} \right)} \right)\sin \theta d\theta }$ (Since

$\tan \left( {\cfrac{\pi }{2} - \theta } \right) = \cot \theta$ )

$= \int_0^{\cfrac{\pi }{2}} {\sin \left( {2 \times \left( {\cfrac{\pi }{2} - \cfrac{\theta }{2}} \right)} \right)} \sin \theta d\theta$

$= \int_0^{\cfrac{\pi }{2}} {\sin \left( {\pi - \theta } \right)\sin \theta d\theta }$

$= \int_0^{\cfrac{\pi }{2}} {{{\sin }^2}\theta d\theta }$

$= \cfrac{1}{2}\int_0^{\cfrac{\pi }{2}} {2{{\sin }^2}\theta d\theta }$

$= \cfrac{1}{2}\int_0^{\cfrac{\pi }{2}} {\left( {1 - \cos 2\theta } \right)d\theta }$

$= \cfrac{1}{2}\left( {\theta - \cfrac{{\sin 2\theta }}{2}} \right)_0^{\cfrac{\pi }{2}}$

$= \cfrac{1}{2}\left( {\cfrac{\pi }{2}} \right)$

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

$\int_{0}^{2\pi}{ln(1+\cos{x})}dx=$

0%
$\pi\ln{2}$
0%
$-\pi\ln{2}$
0%
$-2\pi\ln{2}$
0%
$2\pi\ln{2}$

Explanation

$I = \int_0^{2\pi } {\ln \left( {1 + \cos x} \right)dx}$

$\Rightarrow I = 2\int_0^\pi {\ln \left( {1 + \cos x} \right)dx}$ ---(1) (

${\because \int_0^{2a} {f\left( x \right)dx = 2\int_0^a {f\left( x \right)dx,} } }$ if

${f\left( {2a - x} \right) = f\left( x \right)}$ )

$\Rightarrow I = 2\int_0^\pi {\ln \left( {1 + \cos \left( {\pi - x} \right)} \right)dx}$

$\Rightarrow I = 2\int_0^\pi {\ln \left( {1 - \cos x} \right)dx}$ ---(2)

Adding (1) and (2) we get

$\Rightarrow 2I = 2\int_{}^{} {\ln \left[ {\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)} \right]dx}$

$\Rightarrow I = \int_0^\pi {\ln \left( {1 - {{\cos }^2}x} \right)dx}$

$\Rightarrow I = \int_0^\pi {\ln {{\sin }^2}xdx}$

$\Rightarrow I = 2\int_0^\pi {\ln \sin xdx}$ ---(A)

$\Rightarrow I = 4\int_0^{\frac{\pi }{2}} {\ln \sin xdx}$ ---(3) (

${\because \int_0^{2a} {f\left( x \right)dx = 2\int_0^a {f\left( x \right)dx,} } }$

${f\left( {2a - x} \right) = f\left( x \right)}$ )

$\Rightarrow I = 4\int_0^{\frac{\pi }{2}} {\ln \sin \left( {\frac{\pi }{2} - x} \right)dx}$

$\Rightarrow I = 4\int_0^{\frac{\pi }{2}} {ln(cosx)dx}$ ---(4)

Adding (3) and (4)

$\Rightarrow 2I = 4\int_0^{\frac{\pi }{2}} {\ln \left( {\frac{{2\sin x\cos x}}{2}} \right)dx}$

$\Rightarrow I = 2\int_0^{\frac{\pi }{2}} {\ln \sin 2xdx - 2\int_0^{\frac{\pi }{2}} {\ln 2dx} }$

$\Rightarrow I = 2\int_0^{\frac{\pi }{2}} {\ln \sin 2xdx - 2\ln 2\left[ x \right]_0^{\frac{\pi }{2}}}$

putting

$2x=t$

$\Rightarrow 2dx = dt$

$\Rightarrow I = \int_0^\pi {\ln \sin tdt} - 2\ln 2\left[ {\frac{\pi }{2}} \right]$

$\Rightarrow I = \int_0^\pi {\ln \sin x dx} - \pi \ln 2$ (since

${\int_a^b {f\left( t \right)dt = \int_a^b {f\left( x \right)dx} } }$ )

$\Rightarrow I = \frac{I}{2} - \pi \ln 2$ (from (A))

$\Rightarrow \frac{I}{2} = - \pi \ln 2$

$\Rightarrow I = - 2\pi \ln 2$

Evaluate

$\displaystyle \int_\cfrac{\pi}{4}^\cfrac{\pi}{2}(\sqrt{\tan x}+\sqrt{\cot x})dx=$

0%
$\dfrac{\pi}{2\sqrt{2}}$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{\sqrt{2}}$
0%
$\dfrac{\pi}{3}$

Explanation

$\int_{\cfrac{\pi }{4}}^{\cfrac{\pi }{2}} {\left( {\sqrt {\tan x} + \sqrt {\cot x} } \right)dx}$

$= \int_{\cfrac{\pi }{4}}^{\cfrac{\pi }{2}} {\left( {\sqrt {\tan x} + \cfrac{1}{{\sqrt {\tan x} }}} \right)dx}$

$= \int_{\cfrac{\pi }{4}}^{\cfrac{\pi }{2}} {\left( {\cfrac{{1 + \tan x}}{{\sqrt {\tan x} }}} \right)dx}$

putting

$\tan x = {t^2}$

$\Rightarrow {\sec ^2}xdx = 2tdt$

$\Rightarrow \left( {1 + {{\tan }^2}x} \right)dx = 2tdt$

$\Rightarrow \left( {1 + {t^4}} \right)dx = 2tdt$

$\Rightarrow dx = \cfrac{{2tdt}}{{1 + {t^4}}}$

$= \int_1^\infty {\cfrac{{\left( {1 + {t^2}} \right)}}{{t\left( {1 + {t^4}} \right)}} \times 2tdt}$

$= 2\int_1^\infty {\cfrac{{1 + {t^2}}}{{1 + {t^4}}}} dt$

$= 2\int_1^\infty {\cfrac{{\left( {1 + \cfrac{1}{{{t^2}}}} \right)}}{{\left( {{t^2} + \cfrac{1}{{{t^2}}}} \right)}}dt}$

$= 2\int_1^\infty {\cfrac{{\left( {1 + \cfrac{1}{{{t^2}}}} \right)}}{{\left( {{t^2} + \cfrac{1}{{{t^2}}} - 2 + 2} \right)}}dt}$

$= 2\int_1^\infty {\cfrac{{\left( {1 + \cfrac{1}{{{t^2}}}} \right)}}{{{{\left( {t - \cfrac{1}{t}} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}}}dt}$

putting

$t - \cfrac{1}{t} = y$

$\ \Rightarrow \left( {1 + \cfrac{1}{{{t^2}}}} \right)dt = dy$

$= 2\int_0^\infty {\cfrac{{dy}}{{{y^2} + {{\left( {\sqrt 2 } \right)}^2}}}}$

$= 2 \times \cfrac{1}{{\sqrt 2 }}\left[ {{{\tan }^{ - 1}}\left( {\cfrac{y}{{\sqrt 2 }}} \right)} \right]_0^\infty$

$= \sqrt 2 \left[ {{{\tan }^{ - 1}}\left( \infty \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right]$

$= \sqrt 2 \left( {\cfrac{\pi }{2}} \right)$

$= \cfrac{\pi }{{\sqrt 2 }}$

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

0%
$\log 3$
0%
$\log 5$
0%
$\log 15$
0%
$\log 2$

Explanation

Let

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

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

Let

$I = \int\limits_1^3 {\sqrt {{x^4} + {x^2}} dx,}$ then I=

0%
$1 > 6\sqrt {10}$
0%
$1 < 2\sqrt 2$
0%
$2\sqrt 2 < 1 < 6\sqrt {10}$
0%
none of above

Explanation

$\sum { = \int\limits_1^3 {\sqrt {{x^4} + {x^2}} dx} }$

$= \int\limits_1^3 {x\sqrt {{x^2} + 1} dx}$

$= \frac{1}{2}\int\limits_{\sqrt 2 }^{\sqrt {10} } {{7^2}dx}$

$= \frac{1}{2}{\left( {\frac{{{7^3}}}{3}} \right)_{\sqrt 2 }}^{\sqrt {10} }$

$= \frac{1}{6}\left( {10\sqrt {10} - 2\sqrt 2 } \right)$

$= \frac{1}{3}\left( {5\sqrt {10} - \sqrt 2 } \right)$

$= 4.8$

$\therefore 2\sqrt 2 < I < 6\sqrt {10}$

$\int_\limits {0}^\cfrac{\pi}{4}\dfrac{1}{1+2\sin^2x}dx$

0%
$\dfrac{\pi}{3\sqrt3}$
0%
$-\dfrac{\pi}{3\sqrt3}$
0%
$-\dfrac{\pi}{2\sqrt3}$
0%
$-\dfrac{\pi}{\sqrt3}$

$\displaystyle\int^{\pi/4}_0\sec^4xdx$ .

0%
$\dfrac{3}{4}$
0%
$\dfrac{-3}{4}$
0%
$\dfrac{4}{3}$
0%
$\dfrac{5}{4}$

Explanation

We are given

$I=\displaystyle \int_{0}^{\pi/4}{\sec^4x dx}$

$I=\displaystyle \int_{0}^{\pi/4}{\sec^{2}.\sec^2x dx}$ (

$\therefore$ we know that

$\sec^2\theta=1+\tan^2\theta$ )

$I=\displaystyle \int_{0}^{\pi/4}{\sec^2x.(1+\tan^2x) dx}$

Let, take

$\tan x=t\Rightarrow \sec^2xdx=dt$

( differentiating both sides)

now, when

$x=0\Rightarrow t=\tan 0=0$

& when

$x=\pi/4\Rightarrow t=\tan \pi/4=1$

$I=\displaystyle \int_{0}^{1}{(1+t^2)dt}=\displaystyle \int_{0}^{1}{dt}+\displaystyle \int_{0}^{1}{t^2 dt}$

$=[t]_{0}^{1}+\left[\dfrac{t^3}{3}\right]_{0}^{1}$

$=(1-0)+(1/3-0)$

$I=1+\dfrac{1}{3}=\dfrac{4}{3}$

For

$x>0$ , let

$f(x)=\displaystyle\int_{1}^{x}\dfrac {\log t}{1+t}dt$ . Then

$f(x)+f\left(\dfrac {1}{x}\right)$ is equal to:

0%
$\dfrac {1}{4}(\log x)^{2}$
0%
$\dfrac {1}{2}(\log x)^{2}$
0%
$\log x$
0%
$\dfrac {1}{4}\log x^{2}$

Explanation

$f\left( x \right) = \displaystyle\int_1^x {\dfrac{{\log t}}{{1 + t}}} - \left( 1 \right)$

$Now$

$f {\left( {\dfrac{1}{x}} \right)} = \displaystyle\int_1^{1/x} {\dfrac{{\log t}}{{1 + t}}}$

$Let\,t = \dfrac{1}{y}\Rightarrow dt = \dfrac{{ - 1}}{{{y^2}}}dy$

$f {\left( {\dfrac{1}{x}} \right)} = \displaystyle\int_1^x {\dfrac{{\log \dfrac{1}{y}}}{{1 + \dfrac{1}{y}}} \times - \dfrac{1}{{{y^2}}}dy}$ (replacing

$t$ with

$\dfrac{1}{y}$ )

$as, \ {t=\dfrac{1}{y}\Rightarrow\dfrac{1}{x}=\dfrac{1}{y}}\Rightarrow x=y$ (upper limit)

$= \displaystyle\int_1^x {\dfrac{{ - \log y}}{{1 + y}}} \times y \times - \dfrac{1}{{{y^2}}}dy$

$= \displaystyle\int_1^x {\dfrac{{\log y}}{{1 + y}}} \times \dfrac{1}{y}dy$

$f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = \displaystyle\int_1^x {\dfrac{{\log t}}{{\left( {1 + t} \right)}}\left( {1 + \dfrac{1}{t}} \right)dt}$

$= \displaystyle\int_1^x {\dfrac{{\log t}}{{\left( {t + y} \right)}}\left( {\dfrac{{t + 1}}{t}} \right)dt}$

$= \displaystyle\int_1^x {\dfrac{{\log t}}{t}dt}$

$putting\,\log t = z$

$\dfrac{1}{t}dt = dz$

$\displaystyle\int_0^{\log x} {zdz}$ ......

$(upper \ limit \ becomes \ logx \ similarly)$

$= {\left[ {\dfrac{{{z^2}}}{2}} \right]_0}^{\log x} = \dfrac{1}{2}{\left( {\log x} \right)^2}$

Evaluate

$\displaystyle\int^1_0\dfrac{x^3}{(1+x^8)}dx$

0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{8}$
0%
$\dfrac{\pi}{16}$

Explanation

Let

$I=\displaystyle\int^1_0\dfrac{x^3}{(1+x^8)}dx$

Put

$x^4=t$ and

$4x^3dx=dt$ .

$[x=0\Rightarrow t=0]$ and

$[x=1\Rightarrow t=1]$

$\therefore I=\dfrac{1}{4}\displaystyle\int^1_0\dfrac{dt}{(1+t^2)}\\$

$=\dfrac{1}{4}[\tan^{-1}t]^1_0$ ...........

$\because\displaystyle \int \dfrac {dx}{ {x^2+a^2} }=\tan^{-1}\dfrac xa+C\\$

$=\dfrac{1}{4}[\tan^{-1}1-\tan^{-1}0]\\$

$=\left(\dfrac{1}{4}\times \dfrac{\pi}{4}\right)=\dfrac{\pi}{16}$ .

The value of

$\int _{ 0 }^{ \dfrac { \pi }{ 2 } }{ \log { \left( \dfrac { 4+3\sin { x } }{ 4+3\cos { x } } \right) dx } }$ is

0%
$2$
0%
$\dfrac{3}{4}$
0%
$0$
0%
$-2$

Explanation

As,

$\begin{array}{l}\int_0^a {f\left( x \right)} dx = \int_0^a {f\left( {x - a} \right)} dx\\So,\\I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \\ = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin \left( {\frac{\pi }{2} - x} \right)}}{{4 + 3\cos \left( {\frac{\pi }{2} - x} \right)}}} \right)} \\ = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \end{array}$

Adding the above two integral,

$\begin{array}{l}I + I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} dx + \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} dx\\2I = \int_0^{\frac{\pi }{2}} {\left[ {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) + \log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \right]dx} \\As,\\\log a + \log b = \log ab\\So,\\2I = \int_0^{\frac{\pi }{2}} {\log \left[ {\left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) \times \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \right]dx} \\ = \int_0^{\frac{\pi }{2}} {\log \left( 1 \right)dx} \\ = \int_0^{\frac{\pi }{2}} {0dx} \\2I = 0\\I = 0\end{array}$

Thus Option C

$\displaystyle \int _{ 0 }^{ \pi /4 }{ \tan ^{ 2 }{ x } dx= }$

0%
$1-\dfrac {\pi}{4}$
0%
$1+\dfrac {\pi}{4}$
0%
$\dfrac {-\pi}{4}-1$
0%
$\dfrac {\pi}{4}-1$

Explanation

$tan^2x = sec^2x - 1$

$\Rightarrow$

$\int_0^{\pi/4} -1 dx$ +

$\int_0^{\pi/4} (sec^2x) dx$

$-\pi/4$ + [

$tanx$ }

$_0^{\pi/4}$

= 1 -

$\pi/4$

So, the correct option is 'A'.

The value of

$\displaystyle \underset{0}{\overset{x}{\int}} \dfrac{(t - |t|)^2}{(1 + t^2)} dt$ is equal to

0%
$4(x - \tan^{-1} x), \, \ \text{if} \, x < 0$
0%
$0 \, if \, x > 0$
0%
$\ln ( 1 + x^3) \, \ \text{if} \, x > 0$
0%
$4(x + \tan^{-1} x ) \, \text{if} \, x < 0$

The integral

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

0%
$\cfrac { 15 }{ 128 }$
0%
$\cfrac { 15 }{ 64 }$
0%
$\cfrac { 13 }{ 32 }$
0%
$\cfrac { 13 }{ 256 }$

Explanation

$\int _{ { \pi }/{ 12 } }^{ { \pi }/{ 4 } }{ \cfrac { 8\cos { 2x } }{ { \left( \dfrac { \sin { x } }{ \cos { x } } +\dfrac { \cos { x } }{ \sin { x } } \right) }^{ 3 } } dx }$

$\Rightarrow \int _{ { \pi }/{ 12 } }^{ { \pi }/{ 4 } }{ \dfrac { 8\cos { 2x } }{ { \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x } \right) }^{ 3 } } { \left( \sin { x } \cos { x } \right) }^{ 3 }dx }$

$\Rightarrow \int _{ { \pi }/{ 12 } }^{ { \pi }/{ 4 } }{ 8\cos { 2x } { \left( 2\sin { x } \cos { x } \right) }^{ 3 }\times \dfrac { 1 }{ { 2 }^{ 3 } } dx }$

$\Rightarrow \int _{ { \pi }/{ 12 } }^{ { \pi }/{ 4 } }{ \cos { 2x } \sin ^{ 3 }{ 2x } dx }$

$\sin { 2x } =t$

$2\cos { 2x } dx=dt$

$\Rightarrow \int _{ 1/2 }^{ 1 }{ \dfrac { { t }^{ 3 } }{ 2 } dt }$

$\Rightarrow \cfrac { 1 }{ 2 } \dfrac { { \left[ { t }^{ 4 } \right] }_{ 1/2 }^{ 1 } }{ 4 }$

$\Rightarrow \cfrac { 1 }{ 8 } \left[ 1-\cfrac { 1 }{ 16 } \right]$

$=\cfrac { 1 }{ 8 } \times \dfrac { 15 }{ 16 } =\cfrac { 15 }{ 128 }$

$\int _{ 0 }^{ 1 }{ \cfrac { \tan ^{ -1 }{ x } }{ x } } dx=k\int _{ 0 }^{ \pi /2 }{ \cfrac { x }{ \sin { x } } } dx$ then the value of

$k$ is

0%
$1$
0%
$\cfrac{1}{4}$
0%
$4$
0%
$\cfrac{1}{2}$

Explanation

Put

$x=\tan a \implies\displaystyle \int_{0}^{\frac{\pi}{4}}\text{sec}^{2}a\frac{a}{\tan a}da;\ put\ \ t=2a\implies \int_{0}^{\frac{\pi}{2}}\frac{t}{\sin t}dt\implies k=1$

The solution for x of the equation

$\displaystyle\int^x_{\sqrt{2}}\dfrac{dt}{t\sqrt{t^2-1}}=\dfrac{\pi}{2}$ is?

0%
$- \sqrt2$
0%
$\pi$
0%
$\sqrt{3}/2$
0%
$2\sqrt{2}$

Explanation

$\displaystyle \int_{\sqrt 2}^{x} \dfrac {dt}{t\sqrt {t^2 -1}}=\dfrac {\pi}{2}$

Let

$t=\sec \theta$

$dt=\sec \theta \tan \theta \ d\theta$

$\displaystyle \int_{\pi /4}^{\sec^{-1}x} \dfrac {\sec \theta \tan \theta d\theta}{\sec \theta \sqrt {\sec^2 \theta -1}} \ \Rightarrow \displaystyle \int_{\pi /4}^{\sec^{-1}\theta} \dfrac {\sec \theta \tan \theta d\theta }{\sec \theta \tan \theta }=\dfrac {\pi}{2}$

$\displaystyle \int_{\pi /4}^{\sec^{-1}x} d\theta =\dfrac {\pi}{2}$

$\sec^{-1}x-\dfrac {\pi}{4}=\dfrac {\pi}{2}$

$\sec^{-1}x=\dfrac {\pi}{2}+\dfrac {\pi}{4}\ \Rightarrow x=\sec \left (\dfrac {3\pi}{4}\right)$

$\boxed {x=-\sqrt 2}$

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

0%
$\dfrac{32}{35}$
0%
$\dfrac{64}{35}$
0%
$\dfrac{256}{35}$
0%
$\dfrac{512}{35}$

Solve

$\int\limits_l^e {\dfrac{{dx}}{{\ln \left( {{x^x} \cdot {e^x}} \right)}}}$

0%
$\ln 2$
0%
$2\ln 2$
0%
$-\ln 2$
0%
None of these

Explanation

Consider the given integral.

$I=\int_{1}^{e}{\dfrac{dx}{\ln \left( {{x}^{x}}\cdot {{e}^{x}} \right)}}$

$I=\int_{1}^{e}{\dfrac{dx}{\ln {{x}^{x}}+\ln {{e}^{x}}}}$

$I=\int_{1}^{e}{\dfrac{dx}{x\ln x+x\ln e}}$

$I=\int_{1}^{e}{\dfrac{dx}{x\ln x+x}}$

$I=\int_{1}^{e}{\dfrac{dx}{x\left( \ln x+1 \right)}}$

Let $t=\ln x+1$

$\dfrac{dt}{dx}=\dfrac{1}{x}+0$

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

Therefore,

$I=\int_{1}^{2}{\dfrac{dt}{t}}$

$I=\left[ \ln \left( t \right) \right]_{1}^{2}$

$I=\ln \left( 2 \right)-\ln \left( 1 \right)$

$I=\ln 2$

Hence, this is the answer.

The value of

$\int _{ 0 }^{ \infty }{ \cfrac { \log { x } }{ 1+{ x }^{ 2 } } } dx$ , equals

0%
$\cfrac{\pi}{2}$ $\log{2}$
0%
$-\cfrac{\pi}{2}$ $\log{2}$
0%
$0$
0%
$\cfrac{\pi}{4}$ $\log{2}$

Evaluate:

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

0%
$0$
0%
$1$
0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{4}$

Evaluate

$\displaystyle\int^1_0\dfrac{xe^x}{(1+x)^2}dx$

0%
$\left(\dfrac{e}{2}-1\right)$
0%
$(e-1)$
0%
$e(e-1)$
0%
None of these

Explanation

Let

$I=\displaystyle\int^1_0\dfrac{xe^x}{(1+x)^2}dx$

$=\displaystyle\int^1_0e^x\left\{\dfrac{(1+x)-1}{(1+x)^2}\right\}dx$

$=\displaystyle\int^1_0e^x\left\{\dfrac{1}{(1+x)}-\dfrac{1}{(1+x)^2}\right\}dx$

$=\displaystyle\int^1_0e^x\{f(x)+f'(x)dx,$ where

$f(x)=\dfrac{1}{(1+x)}$

$=[e^xf(x)]^1_0$

$=\left[e^x\cdot \dfrac{1}{(1+x)}\right]^1_0$

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

Evaluate

$\displaystyle\int^{\pi/2}_{\pi/3} cosec xdx$

0%
$\dfrac{1}{2} log 2$
0%
$\dfrac{1}{2} log 3$
0%
$-log 2$
0%
None of these

$\displaystyle \overset{\ln \pi}{\underset{\ln\pi - \ln2}{\int}} \dfrac{e^x}{1 - \cos \left(\tfrac{2}{3}e^x\right)}dx$ is equal to

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

Explanation

Now,

$\displaystyle \overset{\ln \pi}{\underset{\ln\pi - \ln2}{\int}} \dfrac{e^x}{1 - \cos \left(\dfrac{2}{3}e^x\right)}dx$

$=\displaystyle \overset{\ln \pi}{\underset{\ln\tfrac{\pi}{2}}{\int}} \dfrac{e^x}{1 - \cos \left(\dfrac{2}{3}e^x\right)}dx$

Put

$\dfrac{2e^x}{3}=z$ .

or,

$e^x dx=\dfrac{3}{2}dz$ , again when

$x=\ln{\pi}$ then

$z=\dfrac{2}{3}{\pi}$ and

$x=\ln\frac{\pi}{2}$ then

$z=\dfrac{\pi}{3}$ .

So the given integration becomes,

$=\displaystyle \overset{\tfrac{2\pi}{3} }{\underset{\tfrac{\pi}{3}}{\int}} \frac{\dfrac{3}{2}}{1 - \cos \left(z\right)}dz$

$=\dfrac{3}{4}\displaystyle \overset{ \tfrac{2\pi}{3}}{\underset{\tfrac{\pi}{3}}{\int}}\cos ec^2 \dfrac{z}{2}dz$

$=\dfrac{3}{4}\times 2\left[-\cot \dfrac{z}{2}\right]_{\tfrac{\pi}{3}}^{\tfrac{2\pi}{3}}$

$=\dfrac{3}{2}\times\left[\sqrt3-\dfrac{1}{\sqrt 3}\right]$

$=\sqrt 3$ .

Solve

$\displaystyle\int\limits_0^{\dfrac{{\ln 3}}{2}} {\dfrac{{{e^x} + 1}}{{{e^{2x}} + 1}}dx}$

0%
$\dfrac{\pi }{{2}} + \dfrac{1}{2}ln\left( {\dfrac{3}{2}} \right)$
0%
$\dfrac{\pi }{{12}} + \dfrac{1}{2}ln\left( {\dfrac{3}{2}} \right)$
0%
$\dfrac{\pi }{{12}}-+ \dfrac{1}{2}ln\left( {\dfrac{3}{2}} \right)$
0%
None of these

Explanation

$\begin{array}{l}\displaystyle\int\limits_0^{\dfrac{{\ln 3}}{2}} {\dfrac{{{e^x} + 1}}{{{e^{2x}} + 1}}dx} \\\left[ \begin{array}{l}{e^x} = t\\{e^x}dx = dt\,\,\,\,\,\,\, \Rightarrow dx = \dfrac{{dt}}{t}\\x = 0\,\, \Rightarrow t = 1\\x = \dfrac{1}{2}\ln 3\,\, \Rightarrow t = {e^{\dfrac{1}{2}\ln 3}}\\t = \sqrt 3 \\ \end{array} \right]\\ = \displaystyle\int\limits_1^{\sqrt 3 } {\dfrac{{t + 1}}{{\left( {1 + {t^2}} \right)t}}} dt\\ = \displaystyle\int\limits_1^{\sqrt 3 } {\dfrac{{\left( {1 + \dfrac{1}{t}} \right)t}}{{\left( {1 + \dfrac{1}{{{t^2}}}} \right){t^2}}}\dfrac{{dt}}{t}} \\let\,\,\,\dfrac{1}{t} = u\\ - \dfrac{1}{{{t^2}}}dt = du\\ = - \displaystyle\int\limits_1^{\dfrac{1}{{\sqrt 3 }}} {\dfrac{{1 + u}}{{1 + {u^2}}}du} \\ = \displaystyle\int\limits_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{1}{{1 + {u^2}}}du} + \displaystyle\int\limits_{\dfrac{1}{{\sqrt 3 }}}^1 {\dfrac{u}{{1 + {u^2}}}du} \\ = \left( {{{\tan }^{ - 1}}\left( u \right)} \right)_{\dfrac{1}{{\sqrt 3 }}}^1 + \dfrac{1}{2}\left( {\ln \left( {{u^2} + 1} \right)} \right)_{\dfrac{1}{{\sqrt 3 }}}^1\\ = \dfrac{\pi }{{12}} + \dfrac{1}{2}\left( {\ln 2 - \ln \dfrac{4}{3}} \right)\\ = \dfrac{\pi }{{12}} + \dfrac{1}{2}ln\left( {\dfrac{3}{2}} \right)\end{array}$

$\int (1 + 2x + 3x^2 + 4x^3 + ...)dx =$

0%
$(1 + x)^{-1} + c$
0%
$(1 - x)^{-1} + c$
0%
$(1 - x)^{-1} - 1 + c$
0%
None of these

Explanation

$\displaystyle \int(1+2x+3x^2+4x^3+..)dx$

$=\displaystyle \int (1-x)^2dx\quad [(1-x)^2=1+2x+3x^2]$

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

$=(1-x)^{-1}+c$

$\therefore \boxed{\displaystyle \int(1+2x+3x^2+.....)dx}=(1-x)^{-1}+c$

Evaluate

$\displaystyle\int^{\pi}_0\dfrac{x}{1+\sin x}dx$ .

0%
$x\tan x - ln|\cos x| - x\sec x -ln|\sec x - \tan x| + C$
0%
$x\tan x + ln|\cos x| - x\sec x -ln|\sec x - \tan x| + C$
0%
$\tan x + ln|\cos x| - x\sec x -ln|\sec x - \tan x| + C$
0%
None of these

Explanation

$\displaystyle \int \cfrac{x(1-\sin x)}{1-\sin^2x}dx$

$\displaystyle \int \cfrac{x(1-\sin x)}{\cos^2x}dx$

$\displaystyle \int (x \sec^2x - x \tan x \sec x )dx$

Applying byparts,we get

$x\tan x + ln|\cos x| - x\sec x -ln|\sec x - \tan x| + C$

Evaluate

$\displaystyle\int^{\pi/3}_{\pi/6}\dfrac{dx}{1+\sqrt{\tan x}}$ .

0%
$\cfrac{\pi}{12}$
0%
$\cfrac{7\pi}{12}$
0%
$\cfrac{5\pi}{12}$
0%
None of these

Explanation

$I=\displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{dx}{1+\sqrt{\tan x}}$

$\displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{dx}{1+\sqrt{\tan(\cfrac{\pi}{3}+\cfrac{\pi}{6} - x)}}$

$\displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{dx}{1+\sqrt{\cot x}}$

$\displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{\sqrt{\tan x}dx}{1+\sqrt{\tan x}}$

Adding another I in the above equantion, we get

$2I = \displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{\sqrt{\tan x}dx}{1+\sqrt{\tan x}} + \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} \cfrac{dx}{1+\sqrt{\tan x}}$

$2I = \displaystyle \int_\cfrac{\pi}{6}^\cfrac{\pi}{3} dx$

$2I = \cfrac{\pi}{6}$

$I = \cfrac{\pi}{12}$

Obtain

$\displaystyle \int _{ 0 }^{ \pi }{ \sqrt { 1+\cos { 2x } } dx }$

0%
$2$
0%
$12$
0%
$0$
0%
None of these

Explanation

$\displaystyle\int^{\pi}_0\sqrt{1+\cos^2x}dx$

$\cos 2x=2\cos^2x-1$

$\displaystyle\int^{\pi}_0\sqrt{1+2\cos^2x-1}dx$

$\displaystyle\int^{\pi}_0\sqrt{2}\cos xdx$

$=\sqrt{2}[\sin x]^{\pi}_0$

$=\sqrt{2}[0-0]$

$=0$ .

$\displaystyle \underset{0}{\overset{\pi}{\int}} \dfrac{x \, dx}{1 + \sin \, x} =$

0%
$\dfrac{\pi}{6}$
0%
$\pi$
0%
$\dfrac{\pi}{3}$
0%
none of these

Explanation

$\int \dfrac{x}{1+\sin x}dx$

$\int \dfrac{x \sec x}{\tan x+\sec x}dx$

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

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

$=-\dfrac{x}{\tan x+\sec x}+\int \dfrac{\cos x}{1+\sin x}dx$

--------------------------------------------------------

substitute

$u=\sin x\rightarrow du=\cos x dx$

$=\int \dfrac{1}{u+1}du$

$=\ln (u+1)$

$=\ln (\sin x+1)$

-----------------------------------------------------------------

$=-\dfrac{x}{\tan x+\sec x}+\ln (\sin x+1)+C$

$\int_{0}^{\pi} \dfrac{x}{1+\sin x}dx$

$=\left [ -\dfrac{x}{\tan x+\sec x}+\ln (\sin x+1)\right ]_0^{\pi}$

$=\pi -0$

$=\pi$

$\displaystyle\int^1_0\dfrac{2x}{\sqrt{1-x^4}}dx$ is equal to?

0%
$\pi$
0%
$\dfrac{\pi}{2}$
0%
$2\pi$
0%
$0$

Explanation

We have,

$I=\displaystyle\int^1_0\dfrac{2x}{\sqrt{1-x^4}}dx$

Let

$t=x^2$

$dt=2x\ dx$

Therefore,

$I=\displaystyle\int^1_0\dfrac{dt}{\sqrt{1-t^2}}dt$

$I=\left[\sin^{-1}t\right]_0^1$

$I=\left[\sin^{-1}1-\sin^{-1} 0\right]$

$I=\left[\sin^{-1} \sin \dfrac{\pi}{2}-\sin^{-1} \sin 0\right]$

$I=\dfrac{\pi}{2}$

Hence, this is the answer.

Evaluate

$\displaystyle\int^{\pi/2}_{\pi/4}\cot xdx$

0%
$log 2$
0%
$2 log 2$
0%
$\dfrac{1}{2}log 2$
0%
None of these

The value of the integral

$\displaystyle\int^{1}_{\dfrac{1}{3}}\dfrac{(x-x^3)^{\dfrac{1}{3}}}{x^4}dx$ is?

0%
$6$
0%
$0$
0%
$3$
0%
$4$

Explanation

$\begin{array}{l} \int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { { { (x-{ x^{ 3 } }) }^{ \frac { 1 }{ 3 } } } } }{ { { x^{ 4 } } } } }dx \\ =\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { { x^{ 3 } }\times \frac { 1 }{ 3 } { { (\frac { 1 }{ { { x^{ 2 } } } } -1) }^{ \frac { 1 }{ 3 } } } } }{ { { x^{ 4 } } } } }dx \\ =\int _{ \frac { 1 }{ 3 } }^{ 1 }{ x\frac { { { { \left( { \frac { 1 }{ { { x^{ 2 } } } } -1 } \right) }^{ \frac { 1 }{ 3 } } } } }{ { { x^{ 4 } } } } }dx \\ =\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { 1 }{ { { x^{ 3 } } } } \left( { \frac { 1 }{ { { x^{ 2 } } } } -1 } \right) }dx \\ let{ \left( { \frac { 1 }{ { { x^{ 2 } }-1 } } } \right) ^{ \frac { 1 }{ 3 } } }=t \\ \frac { 1 }{ 3 } { \left( { \frac { 1 }{ { { x^{ 2 } } } } -1 } \right) ^{ \frac { { -2 } }{ 3 } } }\cdot \left( { -\frac { 1 }{ { { x^{ 3 } } } } } \right) dx=d \\ =-3\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { 1 }{ { { x^{ 3 } } } } }t\cdot { x^{ 3 } }{ \left( { \frac { 1 }{ { { x^{ 2 } } } } -1 } \right) ^{ \frac { 2 }{ 3 } } }dt \\ =-3\int _{ \frac { 1 }{ 3 } }^{ 1 }{ t\cdot { t^{ 2 } }dt } \\ =-3\int _{ \frac { 1 }{ 3 } }^{ 1 }{ { t^{ 3 } }dt } \\ =-3\left[ { \frac { { { t^{ 4 } } } }{ 4 } } \right] _{ \frac { 1 }{ 3 } }^{ 1 } \\ =-3\left[ { { { \left( { \frac { 1 }{ { { x^{ 2 } }-1 } } } \right) }^{ \frac { 1 }{ 3 } } } } \right] _{ \frac { 1 }{ 3 } }^{ 1 } \\ =3\left[ { 0-2 } \right] \\ =6 \end{array}$

$\int {\frac{{{{\sin }^{ - 1}}x}}{{{{\left( {1 - {x^2}} \right)}^{\frac{3}{2}}}}}dx}$

0%
$\frac{{x\left( {{{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - {x^2}} }} + \frac{1}{2}\log \left| {\left( {1 - {x^2}} \right)} \right| + C.$
0%
$\frac{1}{2}\log \left| {\left( {1 - {x^2}} \right)} \right| + C$
0%
$\frac{{x\left( {{{\sin }^{ - 1}}x} \right)}}{{\sqrt {1 - {x^2}} }}+C$
0%
$4+\frac { \pi }{ 2 }$

Explanation

$\begin{array}{l} Put\, x=\sin t\, \, so,\, that\, dx={ { costdt } }\, and\, t=si{ n^{ -1 } }x \\ \therefore \int { \frac { { { { \sin }^{ -1 } }x } }{ { { { \left( { 1-{ x^{ 2 } } } \right) }^{ \frac { 3 }{ 2 } } } } } dx=\int { \frac { { t\cos t } }{ { { { \left( { 1-{ { \sin }^{ 2 } }t } \right) }^{ \frac { 3 }{ 2 } } } } } dt } =\int { \frac { { t{ { cost } } } }{ { { { \cos }^{ 3 } }t } } dt } } \\ =\int { t{ { \sec }^{ 2 } }tdt } \\ =t.\left( { \tan t } \right) -\int { 1.\tan tdt } \\ =t\left( { \tan t } \right) +\log \left| { \cos t } \right| +C \\ ={ \sin ^{ -1 } }x\frac { x }{ { \sqrt { 1-{ x^{ 2 } } } } } +\log \left| { \sqrt { 1-{ x^{ 2 } } } } \right| +C \\ =\frac { { x\left( { { { \sin }^{ -1 } }x } \right) } }{ { \sqrt { 1-{ x^{ 2 } } } } } +\frac { 1 }{ 2 } \log \left| { \left( { 1-{ x^{ 2 } } } \right) } \right| +C. \\ hence,\, the\, \, option\, \, A\, is\, correct\, \, answer. \end{array}$

$\displaystyle\int \limits_{ -1 }^{ 1 }|x| dx = a$ then -

0%
$a=1$
0%
$a=2$
0%
$a=3$
0%
$a=4$

Explanation

Given,

$\displaystyle\int \limits_{ -1 }^{ 1 }|x| dx$

$=2\displaystyle\int \limits_{ 0 }^{ 1 }x dx$ [ Since

$|x|$ is an even function]

$=2\times \left[\dfrac{x^2}{2}\right]_{0}^{1}$

$=2\times \dfrac{1}{2}$

$=1$ .

What is the value of

$\int_{0}^{a}\dfrac{x-a}{x+a}\ dx$ ?

0%
$a+2a\log 2$
0%
$a-2a\log 2$
0%
$2a\log 2-a$
0%
$2a\log 2$

Explanation

$I= _{0}^{a}\int \dfrac{x-a}{x+a} dx$

Let

$t =x+a \Rightarrow dt =dx$

When

$x= 0, t=a$

$x=a, t= 2a$

Substituting,

$I= _{a}^{2a} \int \dfrac{t-2a}{t} dt$

$= [t-2a \log |t|]^{2a}_{a}$

$=2a-2a \log 2a- a+2a \log a$

$=a-2a\log 2$

$\therefore$ Option

$B$ is correct.

The value of

$\int\limits_0^{\pi /2} {\dfrac{{x\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx\,is}$ is

0%
$\dfrac{{{\pi ^2}}}{4}$
0%
$\dfrac{{{\pi ^2}}}{8}$
0%
$\dfrac{{{\pi ^2}}}{{16}}$
0%
$\dfrac{{3{\pi ^2}}}{{16}}$

Explanation

$\displaystyle I = \int_{0}^{\pi/2} \frac{x sinx \, cosx}{sin^4x + cos^4 x}dx$

using

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

$\displaystyle I= \int_{0}^{\pi/2} \frac{(\pi/2 -\pi) sin (\pi /2 -\pi) cos (\pi/2 - \pi)}{sin^4 (\pi/2 -\pi)+ cos^4 (\pi /2 -\pi)}$

$\displaystyle = \int_{0}^{\pi/2} \frac{(\pi/2-\pi)cosx sinx}{cos^4x + sin^4 x} dx$

$\displaystyle I = \int_{0}^{\pi/2} \frac{cos sinx}{cos^4 x+ sin^4 x}dx - \int_{0}^{\pi/2} \frac{+xcosx \, sinx}{cos^4x+sin^4x}dx$

$\displaystyle \therefore 2I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{sinx cosx}{sin^4 x + cos^4x} dx - I$

$\displaystyle \therefore I = \dfrac{\pi}{2}\int_{0}^{1} \frac{dt/2}{t^2+(1-t)^2}$

$x = \pi / 2,$ then t = 1

and

$x = 0 ,$ then

$t =0$

$\displaystyle = \frac{\pi}{8} \int_{0}^{1} \dfrac{dt}{t^2+t^2+1-2t}$

$\displaystyle = \frac{\pi}{8} \int_{0}^{1} \dfrac{dt}{2t^2-2t+1}$

$\displaystyle = \dfrac{\pi}{16} \int_{0}^{1} \dfrac{dt}{(t-\dfrac{1}{2})^2 + (\dfrac{1}{2})^2}$

$\displaystyle = \dfrac{\pi}{16} \times \frac{1}{1/2} [ tan^{-1} (\frac{t-1/2}{1/2})]$

$\displaystyle \because \int \frac{1}{x^2+a^2}dx = \frac{1}{a} tan^{-1}x/a$

$\displaystyle = [tan^{-1} 2(1-1/2)-tan^{-1}2(0-1/2)]$

$\displaystyle = \dfrac{\pi}{8} [tan^{-1}1-tan^{-1}(-1)]$

$\displaystyle \because tan^{-1}(-1) = -tan^{-1}1 = -\pi/4$

$\displaystyle I = \frac{\pi}{8} [ \frac{\pi}{4}+\frac{\pi}{4}] = \frac{\pi^2}{16}$

1067940_1138807_ans_ded56a648ed94f06a86b8d7916aeab9b.JPG

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

0%
$e^{-1}-1$
0%
$2\left (1-\dfrac {1}{e}\right)$
0%
$1-\dfrac {1}{e}$
0%
$e-1$

Explanation

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

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

0%
$\frac {\pi}{4}+\log {\sqrt {2}}$
0%
$\frac {\pi}{4}-\log {\sqrt {2}}$
0%
$1+\log {\sqrt {2}}$
0%
$1-\frac{1}{2}\log {2}$

$\displaystyle \underset{0}{\overset{\pi}{\int}} \dfrac{x \, dx}{ 1 + \sin \, x}$ =

0%
$\dfrac{\pi}{6}$
0%
$\pi$
0%
$\dfrac{\pi}{3}$
0%
Cannot be valued

Explanation

$\int \dfrac{x}{1+\sin x}dx$

$=\int \dfrac{x}{\frac{\cos ^2x}{1-\sin x}}dx$

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

$=x \tan x +\ln|\cos x|-x\sin x \tan x-x \cos x+ \ln| \tan x +\sec x |+C$

$\int_{0}^{\pi}\dfrac{x}{1+\sin x}dx = \left [ x \tan x +\ln|\cos x|-x\sin x \tan x-x \cos x+ \ln| \tan x +\sec x | \right ]_0^{\pi}$

$\int_{0}^{\pi}\dfrac{x}{1+\sin x}dx = \pi$

$\displaystyle\int {\dfrac{{\ln \left( {1 + {x}} \right)}}{{1 + {x}}}} dx\,equals$

0%
$\dfrac {(\ln (1+x))^2}2$
0%
$- \pi \ln (1+x)$
0%
$\frac{\pi }{2}\ln (1+x)$
0%
$- \frac{\pi }{2}\ln (1+x)$

Explanation

Given

$\displaystyle \int \dfrac {\ln (1+x)}{1+x}dx$

let

$1+x=t$

$\implies dx=dt$

$\implies \displaystyle \int \dfrac {\ln t}{t}dt$

Let

$\ln t=p$

$\implies \dfrac 1tdt =dp$

$\implies \displaystyle \int pdp$

$\implies \dfrac{p^2}{2}$

$\implies \dfrac {(\ln t)^2}2$

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

$\int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \cfrac { 2x-1 }{ 1+x-{ x }^{ 2 } } \right) } } dx$ is equal to

0%
$0$
0%
$1$
0%
$-1$
0%
None of these

Explanation

$\displaystyle \int_{0}^{1} tan^{-1} (\dfrac{2x-1}{1-x-x^{2}})dx$

$\displaystyle \int_{0}^{1} tan^{-1} (\dfrac{x-(1-x)}{1+x(1-x)})dx$

$\displaystyle \int_{0}^{1} tan ^{-1}(x)-tan^{-1}(1-x)dx$

$\displaystyle \int_{0}^{1} tan ^{-1}(x)dx - \int_{0}^{1}tan^{-1}(1-x)dx$ ___ (1)

$\displaystyle \int_{0}^{1} tan ^{-1}x\,dx$ =

$\displaystyle [x\,tan^{-1}x]_{0}^{1}-\int_{0}^{1}\dfrac{x}{1+x^{2}}dx$

$\displaystyle 1+x^{2} = t$

$\displaystyle 2x\,dx = dt$

$\displaystyle [1 tan ^{-1}]-\dfrac{1}{2}\int_{1}^{2} \dfrac{dt}{t}$

$\displaystyle\dfrac{\pi }{4}-\dfrac{1}{2} ln2$

$\displaystyle \int_{0}^{1} tan^{-1}(1-x)dx$

$\displaystyle 1-x = t$

$\displaystyle -dx = dt$

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

$[t\ tan^{-1}t]_{0}^{1}-\int_{0}^{1}\frac{t}{1+t^{2}}dt$

$= \dfrac{\pi }{4} - \dfrac{1}{2}ln2$

$\displaystyle eq^n-(1)\Rightarrow$

$\displaystyle \dfrac{\pi }{4}-\dfrac{1}{2}ln-\dfrac{\pi }{4}+\dfrac{1}{2}ln2$

$\displaystyle 0$

1089857_1176941_ans_6cf2eb1b67044690914eb3575083e949.png

$\frac{d}{dx}(\phi (x))=f(x)$ , then

$\int_1^2 {f\left( x \right)}$ is equal to.

0%
$f(1)-f(2)$
0%
$\phi(1)-\phi(2)$
0%
$f(2)-f(1)$
0%
$\phi(2)-\phi(1)$

Explanation

$\dfrac {d}{dx} (\phi (x))=f(x)$

$\Rightarrow \ \displaystyle \int_{1}^{2} d (\phi (x)) \displaystyle \int_{1}^{2} f(x) dx$

$\Rightarrow \ \phi (x) \displaystyle \int_{1}^{2}= \displaystyle \int_{1}^{2} f(x) dx$

$\Rightarrow \ \displaystyle \int_{1}^{2} f(x) dx=\phi (2)-\phi (1)$

State whether the statement is ture/false.

$\displaystyle \int _ { - \pi / 2 } ^ { \pi / 2 } \left( \frac { \sin x } { 1 - \cos x } \right) d x$ =0

0%
True
0%
False

Explanation

$\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\dfrac{\sin{x}}{1-\cos{x}}dx}$

$=\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\dfrac{2\sin{\dfrac{x}{2}}\cos{\dfrac{x}{2}}}{2{\sin}^{2}{\dfrac{x}{2}}}dx}$

$=\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\dfrac{\cos{\dfrac{x}{2}}}{\sin{\dfrac{x}{2}}}dx}$

$=\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\cot{\dfrac{x}{2}}dx}$

$=2\left[\ln{\left|\sin{\dfrac{x}{2}}\right|}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$=2\left[\ln{\left|\sin{\dfrac{\pi}{4}}\right|}-\ln{\left|\sin{\dfrac{\pi}{4}}\right|}\right]=0$

Evaluate:

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

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

The value of the integral

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

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

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

$\displaystyle\int \frac{ (\sec \theta)}{(\tan^2 \theta)} d \theta$

0%
$\dfrac{1}{cos \theta}+c$
0%
$-\dfrac{1}{sin \theta}+c$
0%
$\dfrac{1}{tan \theta}+c$
0%
$\dfrac{1}{cosec \theta}+c$

Explanation

$\displaystyle\int{\dfrac{\sec{\theta}}{{\tan}^{2}{\theta}}d\theta}$

$=\displaystyle\int{\sec{\theta}{\cot}^{2}{\theta}d\theta}$

$=\displaystyle\int{\dfrac{1}{\cos{\theta}}\dfrac{{\cos}^{2}{\theta}}{{\sin}^{2}{\theta}}d\theta}$

$=\displaystyle\int{\cot{\theta}\csc{\theta}d\theta}$

$=-\csc{\theta}+c$

$=-\dfrac{1}{\sin{\theta}}+c$

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 6 - MCQExams.com

0:0:2

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Definite Integrals Quiz 6 - MCQExams.com

0:0:2

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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