MCQ Questions for Class 12 Commerce Maths Integrals Quiz 1

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

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

Explanation

$\displaystyle \because f\left( x \right) =\int _{ 1 }^{ x }{ \frac { x\log { t } }{ 1+t } dt }$ and

$\displaystyle F\left( e \right) =f\left( e \right) +f\left( \frac { 1 }{ e } \right)$

$\displaystyle \Rightarrow F\left( e \right) =\int _{ 1 }^{ e }{ \frac { \log { t } }{ 1+t } dt } +\int _{ 1 }^{ 1/e }{ \frac { x\log { t } }{ 1+t } dt }$

$\displaystyle =\int _{ 1 }^{ e }{ \frac { \log { t } }{ 1+t } dt } +\int _{ 1 }^{ e }{ \frac { \log { t } }{ t\left( 1+t \right) } dt } =\int _{ 1 }^{ e }{ \frac { \log { t } }{ t } dt }$

$\displaystyle ={ \left[ \frac { { \left( \log { t } \right) }^{ 2 } }{ 2 } \right] }_{ 1 }^{ e }=\frac { 1 }{ 2 } \left[ { \left( \log { e } \right) }^{ 2 }-{ \left( \log { 1 } \right) }^{ 2 } \right] =\frac { 1 }{ 2 }$

The following integral

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

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

Explanation

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

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

$cosec x - cot x = e^{-u}$

$\Rightarrow x = \displaystyle \frac{e^u - e^{-u}}{2} (e^u - e^{-u}) dx = -2 cosec x cot x dx$

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

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

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

The value of

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

0%
$\pi$
0%
$2 \pi$
0%
$\displaystyle \frac{\pi}{2}$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

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

$= \int_0^1 \displaystyle \frac{dt}{\sqrt{t - t^2}} = \int_0^1 \frac{dt}{\sqrt{\frac{1}{4} - \left ( t-\frac{1}{2} \right )^2}} = sin^{-1} \left ( \frac{t - \frac{1}{2}}{\frac{1}{2}} \right ) |_0^1$

$= sin^{-1} 1 -sin^{-1} (-1) = \pi$

$\int x log x dx$ is equal to

0%
$\displaystyle\frac{x^2}{4}(2log x-1)+c$
0%
$\displaystyle\frac{x^2}{2}(2log x-1)+c$
0%
$\displaystyle\frac{x^2}{4}(2log x+1)+c$
0%
$\displaystyle\frac{x^2}{2}(2log x+1)$

Explanation

Using ILATE

$\int \underset{II}{x}\underset{I}{log}x dx=logx\cdot \displaystyle\frac{x^2}{2}-\int \frac{1}{x}\cdot \frac{x^2}{2}dx$

$=\displaystyle\frac{x^2}{2}log x-\frac{1}{2}\frac{x^2}{2}+c$

$=\displaystyle\frac{x^2}{4}(2log x-1)+c$

$\int \dfrac {dx}{\sqrt {x^{10} - x^{2}}}; x > 1=$ ______

$+ C$ .

0%
$\dfrac {1}{4}\log |\sqrt {x^{10} - x^{2}} + x^{2}|$
0%
$\dfrac {1}{2}\log |x^{10} - x^{2}|$
0%
$-\dfrac {1}{4}\sec^{-1} (x^{4})$
0%
$\dfrac {1}{4}\sec^{-1} (x^{4})$

Explanation

Let

$I=\displaystyle \int \dfrac {dx}{\sqrt {x^{10} - x^{2}}}$

$=\displaystyle \int \dfrac {4x^{3}dx}{4x^{3} \times x \sqrt {x^{8} - 1}}$

$=\displaystyle \int \dfrac {4x^{3} dx}{4x^{4}\sqrt {x^{8} - 1}}$
Put

$x^{4} = t$

$\Rightarrow 4x^{3}dx = dt$

$\displaystyle \int \dfrac {dt}{4t \sqrt {t^{2} - 1}} = \dfrac {1}{4}\sec^{-1} (x^{4}) + C$ .

Evaluate the integral

$\displaystyle \int_{0}^{1}\frac{1}{1+x^{2}}dx$

0%
$\pi/4$
0%
$\pi$
0%
$\pi/3$
0%
$0$

Explanation

$\displaystyle \int_{0}^{1}\frac{1}{1+x^2}dx$

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

$\displaystyle \int_{0}^{1}\frac{1}{1+x^2}dx=\tan^{-1}x+c\displaystyle \vert_{0}^{1}$

$=(\tan^{-1}(1)+c)-(\tan^{-1}(0)+c)$

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

The value of

$\int { { e }^{ \tan { \theta } } } \left( \sec { \theta } -\sin { \theta } \right) d\theta$ is equal to ?

0%
$-{ e }^{ \tan { \theta } }\sin { \theta } +C$
0%
${ e }^{ \tan { \theta } }\sin { \theta } +C$
0%
${ e }^{ \tan { \theta } }\sec { \theta } +C$
0%
${ e }^{ \tan { \theta } }\cos { \theta } +C$

Explanation

$\displaystyle I = \int e^{\tan x}. \sec x dx - \int e^{\tan x} .\sin x dx$

$\displaystyle I = \int e^{\tan \theta}. \sec\theta. d\theta - \int e^{\tan\theta}. \sin\theta . d\theta$

by using integration by parts.

$\displaystyle I = \int e^{\tan\theta}.\sec\theta \, d\theta + e^{\tan\theta}. \cos\theta - \int e^{\tan\theta}.(\sec^2\theta).\cos\theta.d\theta$

$\displaystyle I = \int e^{\tan\theta}.\sec\theta d\theta - \int e^{\tan\theta}\sec \, \theta.d\theta +e^{\tan\theta}.\cos\theta$

$\displaystyle I = e^{\tan\theta}. \cos\theta + c.$

Find

$\displaystyle\int_0^\dfrac{\pi}{2}{cos^3}{x}\ dx$ =

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

Explanation

$I = \displaystyle\int_0^\dfrac{\pi}{2}cos^3{x}\ dx$

$= \displaystyle\int_0^\dfrac{\pi}{2}cos^2{x}.\cos{x}\ dx$

$=\displaystyle\overset{\pi/2}{\underset{0}{\displaystyle\int}}(1-\sin^2x)\cos xdx\ (cos^2{x}+sin^2{x} = 1)$
Tanking

$\sin x =t$

$\therefore \cos x dx=dt$
If

$x=0$ then

$t=\sin 0$

$\therefore t=0$
And if

$x=\dfrac{\pi}{2}$ then

$t=\sin\dfrac{\pi}{2}$

$\therefore t=1$

$\therefore I=\displaystyle\overset{1}{\underset{0}{\displaystyle\int}}(1-t^2)dx$

$=\left(t-\dfrac{t^3}{3}\right)^1_0$

$=\left(1-\dfrac{1}{3}\right)-(0-0)$

$\therefore I=\dfrac{2}{3}$ .

$\displaystyle \int _{ 0 }^{ 1 }{ \tan ^{ -1 }{ \left( \dfrac { x }{ \sqrt { 1-{ x }^{ 2 } } } \right) } dx }$

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

Explanation

$\int_0^1 {{{\tan }^{ - 1}}\left( {\dfrac{x}{{1 - {x^2}}}} \right)} \,dx$

Putting

$x = \sin \theta$

$dx = \cos \theta \,d\theta$

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

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

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

$= \int_0^{\dfrac{\pi }{2}} {\,\,\theta } \,\,\cos \theta \,d\theta$

$= {\left( {\theta \sin \theta } \right)^{\dfrac{\pi }{2}}} - \int_0^{\dfrac{\pi }{2}} {1.\sin \theta \,d\theta }$

$= {\left[ {\theta \,\,\sin \theta + \cos \theta } \right]_0}^{\dfrac{\pi }{2}}$

$= \left( {\dfrac{\pi }{2} + 0} \right) - \left( 1 \right)$

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

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

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

Explanation

$= \int_0^1 {{e^x}} \left[ {\cfrac{1}{{x + 1}} - \cfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right]dx$

$= {\left[ {\cfrac{{{e^x}}}{{x + 1}}} \right]^1}_0$

$= \cfrac{e}{2} - \cfrac{1}{1}$

$= \cfrac{e}{2} - 1$

Find

$\int\limits_0^{\sqrt 2 } {\sqrt {2 - {x^2}} dx}$

0%
$\frac{\pi }{2}$
0%
8
0%
0
0%
2

Explanation

$\begin{array}{l} \int _{ 0 }^{ \sqrt { 2 } }{ \sqrt { 2-{ x^{ 2 } } } dx }=\int _{ 0 }^{ \sqrt { 2 } }{ \sqrt { { { \left( { \sqrt { 2 } } \right) }^{ 2 } }-{ x^{ 2 } } } dx } \\ =\left[ { \frac { x }{ 2 } \sqrt { 2-{ x^{ 2 } } } +\frac { { { { \left( { \sqrt { 2 } } \right) }^{ 2 } } } }{ 2 } .{ { \sin }^{ -1 } }\frac { x }{ { \sqrt { 2 } } } } \right] _{ 0 }^{ \sqrt { 2 } } \\ =\left[ { 0+{ { \sin }^{ -1 } }\left( 1 \right) } \right] -\left[ { 0+{ { \sin }^{ -1 } }0 } \right] \\ =\frac { \pi }{ 2 } \end{array}$

Hence, the option

$A$ is the correct answer.

Solve

$\int {\dfrac{1}{{\sqrt[{}]{{9 - 25{x^2}}}}}} dx$

0%
$\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{5x}{3} \right)+C$
0%
${{\sin }^{-1}}\left( \dfrac{5x}{3} \right)+C$
0%
$\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{3x}{5} \right)+C$
0%
${{\sin }^{-1}}\left( \dfrac{3x}{5} \right)+C$

Explanation

We have,

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

$I=\int{\dfrac{dx}{5\sqrt{\dfrac{9}{25}-{{x}^{2}}}}}$

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

We know that

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

Therefore,

$I=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{x}{\dfrac{3}{5}} \right)+C$

$I=\dfrac{1}{5}{{\sin }^{-1}}\left( \dfrac{5x}{3} \right)+C$

Hence, this is the correct answer.

Evaluate

$\displaystyle\int^{3}_{2}3^{x}dx$

0%
$\dfrac{1}{\ln 3}$
0%
$\dfrac{8}{\ln 3}$
0%
$\dfrac{18}{\ln 3}$
0%
None of these

Explanation

$I=\displaystyle\int_{2}^{3}3^{x}dx$

$=$

$\left [ \dfrac{3^{x}}{\ln{3}} \right ]_{2}^{3}$

$=\dfrac{3^{3}}{\ln3}-\dfrac{3^{2}}{\ln3}=\dfrac{27-9}{\ln3}=\dfrac{18}{\ln3}$ .

$\int_0^{ \pi /2 } \sin^5 x \cos ^6 x dx =$

0%
$\dfrac {8}{693}$
0%
$\dfrac {32}{693}$
0%
$\dfrac {8}{99}$
0%
$\dfrac {16}{63}$

Explanation

$\begin{array}{l} \int _{ 0 }^{ \pi /2 }{ { { \sin }^{ 5 } }x{ { \cos }^{ 6 } }xdx } \\ put\, \, \cos x=t \\ -\sin xdx=dt \\ \int _{ 0 }^{ \pi /2 }{ { { \sin }^{ 4 } }x{ { \cos }^{ 6 } }x.\sin xdx } \\ =-\int _{ 1 }^{ 0 }{ { { \left( { 1-{ t^{ 2 } } } \right) }^{ 2 } }.{ t^{ 6 } }.dt } \\ =\int _{ 0 }^{ 1 }{ \left( { 1+{ t^{ 4 } }-2{ t^{ 2 } } } \right) { t^{ 6 } }.dt } \\ =\int _{ 0 }^{ 1 }{ \left( { { t^{ 10 } }-2{ t^{ 8 } }+{ t^{ 6 }} } \right) .dt } \\ =\left[ { \dfrac { { { t^{ 11 } } } }{ { 11 } } -\dfrac { 2 }{ 9 } .{ t^{ 9 } }+\dfrac{ t^{ 7 } }{7} } \right] _{ 0 }^{ 1 } \\ =\dfrac { 1 }{ { 11 } } -\dfrac { 2 }{ 9 } +\dfrac { 1 }{ 7 } \\ =\dfrac { { 63-154+99 } }{ { 693 } } \\ =\dfrac { { 162-154 } }{ { 693 } } \\ =\dfrac { { 8 } }{ { 693 } } \end{array}$

The value of

$\int _{ }^{ }{ \cfrac { \log { x } }{ { \left( x+1 \right) }^{ 2 } } } dx$ is

0%
$\cfrac { -\log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C$
0%
$\cfrac { \log { x } }{ x+1 } +\log { x } -\log { \left( x+1 \right) } +C$
0%
$\cfrac { \log { x } }{ x+1 } -\log { x } -\log { \left( x+1 \right) } + C$
0%
$\cfrac { -\log { x } }{ x+1 } -\log { x } -\log { \left( x+1 \right) } +C$

Explanation

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

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

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

$-\dfrac{lnx}{(x+1)}+\int \dfrac{1}{x}\dfrac{1}{(x+1)}dx$

$-\dfrac{lnx}{(x+1)}+\int \dfrac{(x+1)-(x)}{x(x+1)}$

$-\dfrac{lnx}{(x+1)}+\int \dfrac{1}{x}-\dfrac{1}{1+x}dx$

$-\dfrac{lnx}{(1+x)}+lnx-ln(1+x)+C$

$\int {\dfrac{{\cot \sqrt x }}{{2\sqrt x }}dx}$ is equal to

$= \_\_\_\_\_ + C.$

0%
$2\log |\sin \sqrt x |$
0%
$\log |\sin \sqrt x |$
0%
$\dfrac{1}{2}\log |\sin \sqrt x |$
0%
None of these

Explanation

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

Let

$t=\sqrt { x }$

$\dfrac { 1 }{ { 2\sqrt { x } } } dx=dt$

Therefore,

$I=\int { \cot dt }$

$I=\log \sin t+C$

On putting the value of

$t$ , we get

$I=\log \sin \sqrt x+C$

Hence, this is the answer.

$\displaystyle \int_{\pi /4}^{3\pi/4 }\dfrac{dx}{1+\cos x}$ is equal to

0%
$-2$
0%
$2$
0%
$4$
0%
$-1$

$\int \sin ^{-1}(\cos x) d x$ is equal to

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

Explanation

$\text{sin}^{-1}(\cos x)=\dfrac{\pi}{2}-\text{cos}^{-1}(\cos x)=\dfrac{\pi }{2}- x$

$\displaystyle\int \text{sin}^{-1}(\cos x) d x=\int \bigg(\dfrac{\pi}{2}-x\bigg) d x=\dfrac{\pi x-x^2}{2}+C$

Evaluate :

$\int \cos^3 x e^{\log \sin x} dx$

0%
$-\dfrac{\cos ^4x}{4}+C$
0%
$\dfrac{\sin x}{x^2}+C$
0%
$-\dfrac{\sin^3x}{3}+C$
0%
None of these

Explanation

$I=\int \cos^3 x \ e^{\log \sin x} dx$

$I=\int \cos ^3 x \sin x \ dx$

Putting

$\cos x =t$ and

$-\sin x \ dx=dt$

or,

$\sin x \ dx=-dt$ , we get,

$I=-\int t^3 \ dt=-\dfrac{t^4}{4}+C=-\dfrac{\cos ^4x}{4}+C$

What is

$\displaystyle \int \dfrac{dx}{x(1 + ln x)^n}$ equal to

$(n \neq 1)$ ?

0%
$\dfrac{1}{(n - 1)(1 + ln x)^{n - 1}} + c$
0%
$\dfrac{1 - n}{(1 + ln x)^{1- n}} + c$
0%
$\dfrac{n + 1}{(1 + ln x)^{n+1}} + c$
0%
$-\dfrac{1}{(n - 1)(1 + ln x)^{n-1}} + c$

Explanation

Given,

$\int \dfrac{1}{x\left(1+\ln \left(x\right)\right)^n}dx$

apply

$u=1+\ln \left(x\right)$

$=\int \dfrac{1}{u^n}du$

$=\int \:u^{-n}du$

$=\dfrac{u^{-n+1}}{-n+1}$

$=\dfrac{\left(1+\ln \left(x\right)\right)^{-n+1}}{-n+1}$

$=\dfrac{\left(1+\ln \left(x\right)\right)^{-n+1}}{-n+1}+C$

$-\dfrac{1}{(n - 1)(1 + ln x)^{n-1}} + c$

Evaluate

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

0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{4}$
0%
None of these

Explanation

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

$=[\tan^{-1}x]^1_0\\$ .......

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

$=(\tan^{-1}1-\tan^{-1}0)\\$

$=\dfrac{\pi}{4}$ .

Evaluate the integral

$\displaystyle \int_{1}^{e^{3}}\frac{dx}{x\sqrt{1+\ln x}}$

0%
$2$
0%
$2\sqrt{2}$
0%
$\sqrt{2}$
0%
$-2$

Explanation

$\displaystyle \int_{1}^{e^3}\dfrac{dx}{x\sqrt{1+ln x}}=\displaystyle \int_{1}^{e^3}\dfrac{dlnx}{\sqrt{1+ln x}}$

$\displaystyle \int_{1}^{e^3}\dfrac{dlnx}{\sqrt{1+ln x}}=2\sqrt{1+lnx}\displaystyle \displaystyle \vert_{1}^{e^3}$

$=2\sqrt{1+3}-2\sqrt{1}$

$=2\sqrt{4}-2=2(1)$

$=2$

So, $\displaystyle \int_{1}^{e^3}\dfrac{dx}{x\sqrt{1+ln x}}=2$

The integral

$\displaystyle \int_{0}^{\pi_/{3}}\frac{\cos x}{3+4\sin x}d_{X=}$

0%
$\displaystyle \log\left(\frac{3+2\sqrt{3}}{3}\right)$
0%
$\dfrac{1}{4} \log\left(\displaystyle \frac{3+2\sqrt{3}}{3}\right)$
0%
$2\displaystyle \log\left(\frac{3+2\sqrt{3}}{3}\right)$
0%
$\dfrac{1}{2}\log\left(\displaystyle \frac{3+2\sqrt{3}}{2}\right)$

Explanation

$\displaystyle \int_{0}^{\dfrac{\pi}{3}}\dfrac{\cos x }{3+4 \sin x}dx=\int_{0}^{\dfrac{\pi}{3}}\dfrac{d(\sin x)}{3+4 \sin x}$

$\displaystyle =\int_{0}^{\dfrac{\pi}{3}}\dfrac{d (\sin x) }{3+4 \sin x}=\dfrac{1}{4}\left[\log (3+4 \sin x)\right]_{0}^{\dfrac{\pi}{3}}$

$\displaystyle =\dfrac{1}{4}\left[\log \left[3+4\left(\sin \dfrac{\pi}{3}\right)\right]-\log (3+4 \sin (0))\right]$

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

$=\dfrac{1}{4}(\log (3+2\sqrt{3})-\log 3)$

$=\dfrac{1}{4}\log \left ( \dfrac{3+2\sqrt{3}}{3} \right )$

$\therefore \displaystyle \int_{0}^{\dfrac{\pi}{3}}\dfrac{\cos x}{3+4 \sin x}dx=\dfrac{1}{4} \log \left(\dfrac{3+2\sqrt{3}}{3}\right)$

Evaluate the integral

$\displaystyle \int_{0}^{\pi_/{2}}\frac{cosx}{1+sin^{2}x}dx$

0%
$\pi$
0%
$\pi/3$
0%
$\pi/2$
0%
$\pi/4$

Explanation

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

$\displaystyle \int_{0}^{\dfrac{\pi}{2}}\dfrac{d sin x}{1+sin^2 x}=\left [( tan^{-1}(sin x)+ c\right ) \displaystyle ]_{0}^{\dfrac{\pi}{2}}\ (\int\dfrac{1}{1+x^2}=\tan^{-1}x)$

$=\left ( tan^{-1}(sin ( \dfrac{\pi}{2})+ c \right )-\left ( tan^{-1}(sin 0)+ c\right )$

$=tan^{-1}(1)$

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

$\displaystyle \int_{-1}^{3}\left( \tan^{ -1 }\frac { x }{ x^{ 2 }+1 } +\tan ^{ -1 } \frac { x^{ 2 }+1 }{ x } \right) dx=$

0%
$\pi$
0%
$2\pi$
0%
$4\pi$
0%
$3\pi$

Explanation

We need to find the value of $\displaystyle \int_{-1}^{3} \left[ \tan^{-1}\left ( \dfrac{x}{x^2+1} \right) + \tan^{-1}\left ( \dfrac{x^2+1}{x}\right ) \right] dx$

We know $\tan^{-1}p+\cot^{-1}p=\dfrac{\pi}{2}$

$\Rightarrow \tan^{-1}p+\tan^{-1}\dfrac{1}{p}=\dfrac{\pi}{2}$

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

$\displaystyle \int_{-1}^{3}\left [ \tan^{-1} \left ( \dfrac{x}{1+x^2}\right ) + \tan^{-1}\left ( \dfrac{x^2+1}{x}\right ) \right ]dx$

$=\displaystyle \int_{-1}^{3}\dfrac{\pi}{2}dx$

$=\dfrac{\pi}{2}\int_{-1}^{3}1\cdot dx$

$=\dfrac{\pi}{2}[3-(-1)]$

$=\dfrac{\pi}{2}\times 4=2\pi$

$\displaystyle \int_{1/\sqrt{3}}^{k}\dfrac{1}{1+x^{2}}dx=$

$\dfrac{\pi}{6}$

then the upper limit

$k=?$

0%
$\sqrt{3}$
0%
$\displaystyle \frac{1}{\sqrt{3}}$
0%
$1$
0%
$2+\sqrt{3}$

Explanation

$\displaystyle \int_{\frac{1}{\sqrt{3}}}^{k}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{6}$

$\displaystyle \int_{\frac{1}{\sqrt{3}}}^{k}\dfrac{1}{1+x^2}dx=tan^{-1}x\displaystyle \vert _{\frac{1}{\sqrt{3}}}^{k}$

$=[tan^{-1}k-tan^{\dfrac{-1}{\sqrt{3}}}]$

$=tan^{-1}k-tan^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}$

$tan^{-1}\dfrac{1}{\sqrt{3}}=\dfrac{\pi}{6}$ ---- equation (i)

$tan^{-1}k=\dfrac{\pi}{6}+\dfrac{\pi}{6}$ ---- equation (ii)

$tan^{-1}k=\dfrac{\pi}{3}$

$k=\sqrt{3}$

The integral

$\displaystyle \int_{0}^{\pi/4}\frac{\sin^{9}x}{\cos^{11}x}dx=$

0%
$10$
0%
$5$
0%
$\dfrac{1}{10}$
0%
$\dfrac{1}{5}$

Explanation

$\displaystyle I=\int_{0}^{\dfrac{\pi}{4}}\dfrac{\sin ^9 x}{\cos ^{11}x}dx=\int_{0}^{\dfrac{\pi}{4}}\dfrac{\sin ^9 x}{\cos ^{9}x} \sec ^2 dx$

$\displaystyle =\int_{0}^{\dfrac{\pi}{4}} \tan ^9x d(\tan x)$

$=\left[\dfrac{\tan ^{10}x}{10}+c \right]_{0}^{\dfrac{\pi}{4}}$

$=\left [ \left ( \dfrac{\tan ^{10}(\dfrac{\pi}{4})}{10}+c \right ) -\left ( \dfrac{\tan ^{10}(0)}{10} +c \right )\right ]$

$\tan \dfrac{\pi}{4}=1 ; \tan 0=0$

$\Rightarrow I=\dfrac{1}{10}-\dfrac{0}{10}=\dfrac{1}{10}$

So, $\displaystyle \int_{0}^{\dfrac{\pi}{4}}\dfrac{\sin ^9 x}{\cos ^{11}x}dx=\dfrac{1}{10}$

$\displaystyle \int_{0}^{1}\frac{4x^{3}}{\sqrt{1-x^{8}}}dx =?$

0%
$\pi$
0%
$-\pi$
0%
$\pi /2$
0%
$- \pi /2$

Explanation

$\displaystyle \int_{0}^{1}\dfrac{4x^3}{\sqrt{1-x^8}}dx=\int_{0}^{1}\dfrac{d(x^4)}{\sqrt{1-(x^4)^2}}$

$=\displaystyle\int_{0}^{1}\dfrac{d(x^4)}{\sqrt{1-(x^4)^2}}=[sin^{-1}(x^4)+c]_{0}^{1}\ \left(\int\dfrac{1}{\sqrt{1-x^2}}=sin^{-1}x\right)$

$=\displaystyle[(sin^{-1}(1)+c)-(sin^{-1}(0)+c)]$

$=\displaystyle sin^{-1}1 =\dfrac{\pi}{2}$

Evaluate:

$\displaystyle \int_{0}^{1}\frac{\tan^{-1}x}{1+x^{2}}dx$

0%
$\displaystyle \frac{\pi^{2}}{4}$
0%
$\displaystyle \frac{\pi^{2}}{18}$
0%
$\displaystyle \frac{\pi^{2}}{32}$
0%
$\displaystyle \frac{\pi^{2}}{8}-1$

Explanation

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

Let

$z=\tan^{-1}x$

For

$x=0, z=\tan^{-1} 0=0$ and for

$x=1, z= \tan^{-1} 1=\dfrac{\pi}{4}$

$\implies dz=\dfrac{1}{1+x^2} dx$

Hence, integration becomes-

$\displaystyle\int_{0}^{\dfrac{\pi}{4}} zdz$

$= \left[\dfrac{z^2}{2}\right]_0^{\dfrac{\pi}{4}}$

$=\dfrac{1}{2}\left [\dfrac{\pi ^2}{16}-0\right]$

$=\dfrac{\pi ^2}{32}$

Evaluate:

$\displaystyle \int_{1}^{2}\frac{1}{x\sqrt{x^{2}-1}}d{x}$

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

Explanation

We need to find value of $\displaystyle \int_{1}^{2}\dfrac{1}{x\sqrt{x^2-1}}dx$
We know that

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

$\displaystyle \int_{1}^{2}\dfrac{1}{x\sqrt{x^2-1}}dx= \sec^{-1}2-\sec^{-1}1$

$\sec^{-1}2 =\dfrac{\pi}{3}$

and $\sec^{-1}1=0$

So, $\sec^{-1}2-\sec^{-1}1=\dfrac{\pi}{3}$

$\therefore \displaystyle \int_{1}^{2}\dfrac{1}{x\sqrt{x^2-1}}dx=\dfrac{\pi}{3}$

$\displaystyle \int_{0}^{\pi_/{2}}\frac{\cos x}{1+\sin x}d_{X=}$

0%
$\log 2$
0%
$\log \mathrm{e}$
0%
$\dfrac{1}{2}$ log3
0%
$0$

Explanation

$\int_{0}^{\dfrac{\pi}{2}}\dfrac{cos x}{1+sin x}dx$
we know that
$\int_{0}^{\dfrac{\pi}{2}}\dfrac{cos x}{1+sin x}dx=\int_{0}^{\dfrac{\pi}{2}}\dfrac{d(sin x)}{1+sin x}$
$\int_{0}^{\dfrac{\pi}{2}}\dfrac{d (sin x)}{1+sin x}=log (1+sin x)\int_{0}^{\dfrac{\pi}{2}}$
$=log (1+sin \dfrac{\pi}{2}) - log (1+sin 0)$
$=log (2)$
$=log 2$

The value of

$\int_{0}^{\infty} x.e^{-x^{2}}dx_{=}$

0%
$1$
0%
$- 1/2$
0%
$1/2$
0%
$0$

Explanation

$\int_{0}^{\infty}xe^{-x^2}dx=\dfrac{1}{2}\int_{0}^{\infty}e^{-x^2}dx^2$
$=\dfrac{1}{2}\int_{0}^{\infty}e^{-x^2}dx^2=\dfrac{1}{2}\left [ -\dfrac{1}{e^{x^2}}+c \right ] \int_{0}^{\infty}$
$=\dfrac{1}{2}\left [ (0+e)-(-1+e) \right ]$
$=\dfrac{1}{2}$
$\int_{0}^{\infty}xe^{-x^2}dx=\dfrac{1}{2}$

$\displaystyle \int_{0}^{1}\frac{x^{2}}{1+x^{2}}dx$ equals

0%
$1-\displaystyle \frac{\pi}{4}$
0%
$1-\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

$\displaystyle \int_{0}^{1}\dfrac{x^2dx}{1+x^2}$

$\displaystyle \int_{0}^{1}1{dx}-\displaystyle \int_{0}^{1}\dfrac{1}{1+x^2}dx$

$=1-\left [ \tan^{-1}x \right ] \displaystyle \vert_{0}^{1}$

$=1 - \left [ \tan^{-1} 1 \right ]$

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

Thus $\displaystyle \int_{0}^{1}\dfrac{x^2}{1+x^2}dx=1-\dfrac{\pi}{4}$

$\displaystyle \int_{0}^{1}\frac{dx}{e^{x}+e^{-x}}=$

0%
$tan^{-1}e$
0%
$\displaystyle \frac{\pi}{4}$
0%
$ta\displaystyle \mathrm{n}^{-1}\mathrm{e}-\frac{\pi}{4}$
0%
$ta\displaystyle \mathrm{n}^{-1}\mathrm{e}+\frac{\pi}{4}$

Explanation

$\int_{0}^{1}\dfrac{e^x}{1+e^{2x}}dx=\int_{0}^{1}\dfrac{de^x}{1+e^{2x}}$
$=tan^{-1}(e^x)\int_{0}^{1}$
$=tan^{-1}(e)-tan^{-1}(e^{\circ})$
$=tan^{-1}(e)-tan^{-1}(1)$
$=tan^{-1}(e)-\dfrac{\pi}{4}$
So, $\int_{0}^{1}\dfrac{e^x}{1+e^{2x}}dx=tan^{-1}e-\dfrac{\pi}{4}$

$\displaystyle \int_{1}^{2}(\frac{1+x\log x}{x})e^{x}dx_{=}$

0%
$\mathrm{e}^{2}$ $log2$
0%
$elog2$
0%
$\displaystyle \frac{1}{2}$ log2
0%
$\displaystyle \frac{\mathrm{e}^{2}}{2}$ log2

Explanation

$\int_{1}^{2}\left ( \dfrac{1}{x} + \log x \right ) e^xdx$ $\displaystyle \int e^x [f(x)+f'(x) ] dx=e^x f(x)$

$= \int_{1}^{2} \left ( e^x \log x+ c\right )$

$=e^2 \log 2-e \log 1$

$=e^2 \log 2-0$

$\int_{1}^{2}\left ( \dfrac{1+x \log x}{x} \right ) e^x dx=e^2 \log 2$

Evaluate the integral

$\displaystyle \int_{\frac{\sqrt{2}}{3}}^{ \frac{\sqrt{3}}{3}}\displaystyle \frac{dx}{\sqrt{4-9x^{2}}}$

0%
$\displaystyle \frac{\pi}{36}$
0%
$\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{\pi}{4}$
0%
$\displaystyle \frac{7\pi}{30}$

Explanation

$\displaystyle \int_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}} \dfrac{dx}{2\sqrt{1-\dfrac{ax^2}{2}}}=\dfrac{1}{2}\displaystyle \int_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}} \dfrac{dx}{\sqrt{1-(\dfrac{3x}{2})^2}}$

$=\left[\dfrac{2}{3}\times \dfrac{1}{2} sin^{-1}\left(\dfrac{3x}{2}\right)\displaystyle \right]_{\frac{\sqrt{2}}{3}}^{\frac{\sqrt{3}}{3}}$

$=\dfrac{2}{3}\times \dfrac{1}{2} \left [ sin^{-1} \left ( \dfrac{3\sqrt{3}}{3(2)} \right ) -sin^{-1} \left ( \dfrac{3\sqrt{2}}{3\times 2} \right ) \right ]$

$=\dfrac{1}{3} \left [ sin^{-1} \left ( \dfrac{\sqrt{3}}{2} \right ) -sin \left ( \dfrac{1}{\sqrt{2}} \right ) \right ]$

$=\dfrac{1}{3} \left [ \dfrac{\pi}{3}-\dfrac{\pi}{4} \right ]$

$=\dfrac{1}{3} \left ( \dfrac{\pi}{12} \right )$

$=\dfrac{\pi}{36}$

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

0%
$\sqrt{2}l\mathrm{o}\mathrm{g}(\sqrt{2}+1)$
0%
$\sqrt{2}l\mathrm{o}\mathrm{g}(\sqrt{2}-1)$
0%
$\dfrac{1}{\sqrt{2}}l\mathrm{o}\mathrm{g}(\sqrt{2}+1)$
0%
$\dfrac{-1}{\sqrt{2}}l\mathrm{o}\mathrm{g}(\sqrt{2}+1)$

Explanation

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

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

$\displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{2 d (tan \dfrac{x}{2})}{1-(tan\dfrac{x}{2}+1)^2}=2 \displaystyle \int_{0}^{\dfrac{\pi}{2}} \dfrac{d (tan \dfrac{x}{2})}{(\sqrt{2})^2-(tan \dfrac{x}{2}-1)^2}$

$=2\left [ \dfrac{1}{2\sqrt{2}} \right ] log \left | \dfrac{\sqrt{2}+tan\dfrac{x}{2}-1}{\sqrt{2}-(tan\dfrac{x}{2}-1)} \right | _{0}^{\dfrac{\pi}{2}}$

$=\dfrac{1}{\sqrt{2}} log \left ( \dfrac{\sqrt{2}+2}{\sqrt{2}-2} \right )$

$=log \left ( \dfrac{1+\sqrt{2}}{1-\sqrt{2}} \right )$

$=\sqrt{2}log (\sqrt{2}+1)$

Evaluate the integral

$\displaystyle \int_{0}^{\pi_/{2}}\cos^{5}x.\sin 2xdx$

0%
$2/7$
0%
$1/7$
0%
$-1/7$
0%
$3/7$

Explanation

$\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^5x\cdot sin2x dx=\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^5x\cdot 2 sinx cosx dx$

$=2\displaystyle \int_{0}^{\dfrac{\pi}{2}} cos^6 x sinx dx=-2\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^6 x\cdot d(cos x)$

$=-2\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^6 x dcos x=-2\left ( \dfrac{cos^7x}{7} \right )\displaystyle \vert_{0}^{\dfrac{\pi}{2}}$

$=-2\left [ \dfrac{cos^7 x}{7}\displaystyle \vert_{0}^{\frac{\pi}{2}} \right ]$

$=-2\left ( \dfrac{-1}{7} \right)=\dfrac{2}{7}$

Thus $\displaystyle \int_{0}^{\dfrac{\pi}{2}}cos^5 x\times sin 2x dx=\dfrac{2}{7}$

$\displaystyle \int_{0}^{\displaystyle \tfrac{\pi}{4}}\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}dx=$

0%
$\log 2$
0%
$-\log\sqrt{2}$
0%
$2\log 2$
0%
$3\log\sqrt{2}$

Explanation

Consider, $I=\displaystyle \int_{0}^{\dfrac{\pi}{4}}\sqrt{\dfrac{1-sin 2 x}{1+sin 2 x}}\cdot dx$

$\displaystyle I= \int_{0}^{\dfrac{\pi}{4}}\sqrt{\dfrac{(sinx - cos x)^2}{(sin x + cos x)^2}}\cdot dx$

$\displaystyle I=\int_{0}^{\dfrac{\pi}{4}}{\dfrac{(sinx - cos x)}{sin x + cos x}}\cdot dx$

$\displaystyle I=-\int_{0}^{\dfrac{\pi}{4}}{\dfrac{cos x- sin x}{sin x + cos x}}dx$

$I=-\left[log |sin x + cos x|\right]_0^{\frac{\pi}{4}}$

$I=-\left[log (sin {\frac{\pi}{4}} + cos {\frac{\pi}{4}}) -\log (\sin 0+\cos 0))\right]$

$I=-\left[log \left (\dfrac{1}{\sqrt2}+\dfrac{1}{\sqrt 2}\right)-\log 1\right]$

$I=-log \left ( \dfrac{1}{\sqrt 2} +\dfrac{1}{\sqrt 2} \right ) =-log \sqrt{2}$

So, $\displaystyle \int_{0}^{\dfrac{\pi}{4}}\sqrt{\dfrac{1- sin 2 x}{1+sin 2 x}}x=-log (\sqrt{2})$

$\int_{\pi /4}^{\pi /2} Cotx.dx_{=}$

0%
2 log 2
0%
$\displaystyle \frac{\pi}{2}$ log2
0%
$\log\sqrt{2}$
0%
$\log 2$

Explanation

$\displaystyle \int_{\frac{\pi}{4}} ^{\frac{\pi}{2}} cotx dx$

$=[ln|sinx|]_{{\pi}/{4} }^{{\pi}/{2}}$

$=ln\left |sin\left(\dfrac{\pi}{2}\right)\right |-ln\left |sin\left(\dfrac{\pi}{4}\right)\right |$

$=ln(1)-ln(\dfrac{1}{\sqrt{2}})$

$=0-(-log(\sqrt{2}))$

$=log(\sqrt{2})$ .

Evaluate the integral

$\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}$

0%
$0$
0%
$-1$
0%
$\pi/2$
0%
$-\pi/2$

Explanation

$\displaystyle \int_{0}^{1}\dfrac{dx}{\sqrt{1-x^2}}$

$\displaystyle \int \dfrac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+c$

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

$=(\sin^{-1}1+c)-(\sin^{-1}0+c)$

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

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

Evaluate the integral

$\displaystyle \int_{0}^{1}\frac{(\sin^{-1} {x})^{2}}{\sqrt{1-x^{2}}}dx$

0%
$\displaystyle \frac{\pi^{3}}{24}$
0%
$\pi^{2}$
0%
$-\pi^{2}$
0%
$0$

Explanation

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

$\displaystyle \int \dfrac{(sin^{-1}x)^2}{\sqrt{1-x^2}} \cdot dx=\displaystyle \int (sin^{-1}x)d(sin^{-1}x)$

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

$\displaystyle \int_{0}^{1}\dfrac{(sin^{-1}x)^2}{\sqrt{1-x^2}} \cdot dx=\left[\dfrac{(sin^{-1}x)^3}{3}+c\right]_{0}^{1}$

$=\left ( \dfrac{(sin^{-1}1)^3}{3}+c\right )-\left ( \dfrac{(sin^{-1}0)^3}{3}+c\right )$

$=\dfrac{1}{3}\left ( \dfrac{\pi}{2} \right )^3$

$=\dfrac{\pi^3}{24}$

Evaluate the integral

$\displaystyle \int_{0}^{a}\sqrt{a^{2}-x^{2}}dx$

0%
$\displaystyle \frac{a^{2}}{4}$
0%
$\pi {a}^{2}$
0%
$\displaystyle \frac{\pi a^{2}}{2}$
0%
$\displaystyle \frac{\pi a^{2}}{4}$

Explanation

$\displaystyle \int_{0}^{a}\sqrt{a^2-x^2} dx$

$\displaystyle \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1} (\frac{x}{a})+c$

$\displaystyle \int_{0}^{a}\sqrt{a^2-x^2}\cdot dx=\left [ \frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left ( \frac{x}{a} \right ) +c \right ]_{0}^{a}$

$=\left [ \dfrac{a}{2} \sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\left ( \dfrac{a}{a} \right ) +c \right ]-\left [ \dfrac{x}{2} \sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\left ( \dfrac{0}{a} \right ) +c \right ]$

$=\left [ \dfrac{a^2}{2} \sin^{-1}(1)+c \right]- \left [ \sqrt{a^2}+c \right ]$

$=\dfrac{a^2}{2} \left ( \dfrac{\pi}{2} \right )$

$=\dfrac{\pi a^2}{4}$

Evaluate the integral

$\displaystyle \int_{1/2}^{1}\frac{1}{\sqrt{1-x^{2}}}dx$

0%
$\pi$
0%
$\pi/2$
0%
$\pi/3$
0%
$\pi/4$

Explanation

$\displaystyle \int_{\frac{1}{2}}^{1}\dfrac{1}{\sqrt{1-x^2}}dx$

We know that $\displaystyle \int \frac{1}{\sqrt{1-x^2}}dx=(\sin^{-1}x+c)$

$\displaystyle \int_{\frac{1}{2}}^{1}\dfrac{1}{\sqrt{1-x^2}}dx=\left[(\sin^{-1}x+c)\displaystyle \right]_{\frac{1}{2}}^{1}$

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

$=\sin^{-1}1-\sin^{-1}\dfrac{1}{2}$

$=\dfrac{\pi}{2}-\dfrac{\pi}{6}$

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

$\displaystyle \int_{\frac{1}{2}}^{1}\dfrac{1}{\sqrt{1-x^2}}dx=\dfrac{\pi}{3}$

$\displaystyle \int_{0}^{1}\sqrt{1-x^{2}}dx_{=}$

0%
$1-\displaystyle \frac{\pi}{4}$
0%
$1-\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{\pi}{4}$

Explanation

$\int_{0}^{1}\sqrt{1-x^2}\cdot dx$
$\int \sqrt{1-x^2}dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2} sin^{-1}\left ( \dfrac{x}{a} \right )+ c$
$\int_{0}^{1}\sqrt{1-x^2}\cdot dx= \left ( \dfrac{1}{2} \sqrt{1^2-x^2}+\dfrac{a^2}{2}sin^{-1} \left ( \dfrac{x}{a} \right )+c \right )\int_{0}^{1}$
$=\left ( \dfrac{1}{2} (0)+ \dfrac{1}{2} sin^{-1}(1)+c \right ) - (0+0+c)$
$=\dfrac{1}{2}\dfrac{\pi}{2}$
$=\dfrac{\pi}{4}$
$\int_{0}^{1}\sqrt{1-x^2}dx=\dfrac{\pi}{4}$

$\displaystyle \int_{0}^{a}\frac{1}{a^{2}+x^{2}}dx_{=}$

0%
$\pi/2$
0%
$\pi/3$
0%
$\pi/4$
0%
$\pi/4a$

Explanation

$\int_{0}^{a}\dfrac{1}{a^2+x^2}dx$

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

$\int \dfrac{d(\dfrac{x}{a})}{1+(\dfrac{x}{a})^2}=\dfrac{1}{a}\int \dfrac{d(\dfrac{x}{a})}{1+(\dfrac{x}{a})^2}$

$=\dfrac{1}{a}tan^{-1}(\dfrac{x}{a})+c$

$\int_{0}^{a}\dfrac{1}{a^2+x^2}dx=\left [ \dfrac{1}{a} tan^{-1}\left ( \dfrac{x}{a} \right ) +c \right ] \int_{0}^{a}$

$=\left [ \dfrac{1}{a}tan^{-1}\left ( \dfrac{a}{a} \right ) +c \right ] -(0+c)$
$=\dfrac{1}{a}tan^{-1}(1)$

$=\dfrac{1}{a}\cdot \dfrac{\pi}{4}$

$=\dfrac{\pi}{4a}$

Evaluate the integral

$\displaystyle \int_{\frac{a}{2}}^{a}\frac{1}{\sqrt{a^{2}-x^{2}}}dx$

0%
$\dfrac{\pi}{2}$
0%
$\pi a$
0%
$\pi-1$
0%
$\dfrac{\pi}{3}$

Explanation

$\displaystyle \int_{a}^{\frac{a}{2}}\dfrac{1}{\sqrt{a^2-x^2}}\cdot dx$

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

$=\displaystyle \int \dfrac{d\left ( \dfrac{x}{a} \right )}{\sqrt{1-\left ( \dfrac{x}{a} \right )^2}}$

$=\sin^{-1}(\dfrac{x}{a})+c$

$\displaystyle \int_{\frac{a}{2}}^{a}\dfrac{1}{\sqrt{a^2-x^2}}\cdot =\left [ \sin^{-1}(\dfrac{x}{a})+c \right ] _{\frac{a}{2}}^{a}$

$=(\sin^{-1}(\dfrac{a}{a})+c)-(\sin^{-1}(\dfrac{\frac{a}{2}}{a})+c)$

$=\left [ (\sin (1)+c ) -(\sin^{-1}(\dfrac{1}{2})+c) \right ]$

$=\dfrac{\pi}{2}-\dfrac{\pi}{6}$

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

Evaluate the integral

$\displaystyle \int_{0}^{1}\frac{x}{1+x^{2}}dx$

0%
$\log 2$
0%
$\displaystyle \frac{1}{2} \log2$
0%
$2$
0%
$\log 4$

Explanation

$\displaystyle \int_{0}^{1}\dfrac{x}{a+x^2}dx$

$\displaystyle \int \dfrac{x}{1+x^2}dx=\dfrac{1}{2}\displaystyle \int \dfrac{dx^2}{1+x^2}=\dfrac{1}{2} \log (1+x^2)+c$

$=\displaystyle \int_{0}^{1}\dfrac{x}{1+x^2}dx=\left [ \dfrac{1}{2} \log (1+x^2)+c \right ]_{0}^{1}$

$=\left ( \dfrac{1}{2} \log (2)+c \right ) - \left ( \dfrac{1}{2} \log (1)+c\right )$

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

$\displaystyle \int_{0}^{1}\dfrac{x}{1+x^2}dx=\dfrac{1}{2} \log 2$

The integral

$\displaystyle \int_{0}^{1}\frac{x^{3}}{1+x^{8}}dx=$

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

Explanation

Let $\displaystyle I=\int_{0}^{1}\dfrac{x^3}{1+x^8} dx$

Now, $\displaystyle \int \dfrac{x^3}{1+x^8 }dx =\dfrac{1}{4}\int \dfrac{d(x^4)}{1+x^8}=\dfrac{1}{4}\tan^{-1}(x^4)+c$

$I=\displaystyle \int_{0}^{1}\dfrac{x^3}{1+x^8}dx=\left(\dfrac{1}{4}\tan^{-1}(x^4)+c\right)_{0}^{1}$

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

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

$=\dfrac{1}{4} \cdot \dfrac{\pi}{4}=\dfrac{\pi}{16}$

Hence, $\displaystyle \int_{0}^{1}\dfrac{x^3}{1+x^8}dx =\dfrac{\pi}{16}$

$\displaystyle \int_{0}^{1}\frac{1}{1+x}dx_{=}$

0%
$\log 2$
0%
$\displaystyle \frac{1}{2} \log2$
0%
$2$
0%
$\log 3$

Explanation

$\int_{0}^{1}\dfrac{1}{1+x}dx=\ln(1+x)+c$

$\int_{0}^{1}\dfrac{1}{1+x}dx =\left [ \ln (1+x)+ c \right ]\int_{0}^{1}$

$=(\ln(2)+c)-(\ln(1)+c)$

$=\ln2$

$\int_{0}^{1}\dfrac{1}{1+x}dx=ln2$

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 1 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Integrals Quiz 1 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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