MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 10

$\int {{{(\sin x)}^{99}}{{(\cos x)}^{ - 101}}dx = \_\_\_\_\_\_\_ + C.}$

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$\dfrac{{{{(\tan x)}^{100}}}}{{100}}$
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$\dfrac{{{{(\tan x)}^2}}}{2}$
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$\dfrac{{{{(\tan x)}^{98}}}}{{98}}$
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$\dfrac{{{{(\tan x)}^{97}}}}{{97}}$

Explanation

$\begin{array}{l}I= { \int { { { \left( { \sin x } \right) }^{ 99 } }.\left( { \cos x } \right) } ^{ -101 } }dx. & \\I= \int { \dfrac { { { { \left( { \sin x } \right) }^{ 99 } } } }{ { { { \left( { \cos x } \right) }^{ 101 } } } } dx } & \\I= \int { { { \left( { \dfrac { { \sin x } }{ { \cos x } } } \right) }^{ 99 } }.\dfrac { 1 }{ { { { \cos }^{ 2 } }x } } dx } & \\I= \int { { { \left( { \tan x } \right) }^{ 99 } }.{ { \sec }^{ 2 } }xdx } & Put\, \, \tan x=t.\, \, { \sec ^{ 2 } }xdx=dt \\I= \int { { t^{ 99 } }dt } & \\I= \dfrac { { { t^{ 100 } } } }{ { 100 } } +c & \\ I=\dfrac { { { { \left( { \tan x } \right) }^{ 100 } } } }{ { 100 } } +c & \end{array}$

Hence, this is the answer.

Evaluate

$\int {{x^2}\log x\,dx\, }$ .

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$\cfrac{{{x^2}}}{2}\log x - \cfrac{1}{{9}}{x^2}+c$
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$\cfrac{{{x^3}}}{3}\log x - \cfrac{1}{{9}}{x^2}+c$
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$\cfrac{{{x^3}}}{3}\log x - \cfrac{1}{9}{x^3}+c$
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$\cfrac{{{x^3}}}{3}\log x + \cfrac{1}{9}{x^3}+c$

Explanation

We have,

$I=\displaystyle \int {x^2}\log x\,dx$

We have,

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

$\displaystyle I=\log x\int x^2\ dx -\int \left(\dfrac{d}{dx}\log x \int x^2\ dx\right)\ dx$

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

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

$\displaystyle I=\dfrac{x^3\log x}{3} - \dfrac{x^3}{9}+C$

Hence, this is the answer.

The value of

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

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${ e }^{ x }\left( \dfrac { x-1 }{ x+1 } \right) +C$
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${ e }^{ x }\left( \dfrac { x+1 }{ x-1 } \right) +C$
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${ e }^{ x }.x+C$
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None of these

Explanation

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

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

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

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

$=\displaystyle\int e^xdx-\displaystyle\int e^x\cdot \dfrac{2x}{(x+1)^2}dx$ .

$=e^x-I_1+e$

Let

$I_1=\displaystyle\int e^x\cdot \dfrac{2x}{(x+1)^2}dx$

Here

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

$2x=A(x+1)+B$

$x=-1$ ,

$2(-1)=B$

$\Rightarrow B=-2$

$x=1$ ,

$2=2A+(-2)$

$2A=2+2$

$A=\dfrac{4}{2}=2$

$\therefore \dfrac{2x}{(x+1)^2}=\dfrac{2}{x+1}-\dfrac{2}{(x+1)^2}$

$\therefore I_1=\displaystyle\int e^x\left\{\dfrac{2}{x+1}-\dfrac{2}{(x+1)^2}\right\}dx$

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

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

Which is of the form

$\displaystyle\int e^x[f(x)+f'(x)]dx$

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

$[\because\displaystyle\int e^x[f(x)+f'(x)]dx=e^xf(x)+c]$

$=\dfrac{2e^x}{x+1}$

$\therefore I=e^x-\dfrac{2e^x}{x+1}+c$

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

$=e^x\left\{\dfrac{x+1-2}{x+1}\right\}+c$

$\therefore I=e^x\left(\dfrac{x-1}{x+1}\right)+c$ .

1191721_1333860_ans_ec09c6a3e14c4a65ac69fa4c85fc1009.jpg

$\int {{e^x}(\log \sin x + \cot x)\,dx = } \_\_\_\_\_\_ + C.$

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${e^x}\cot x$
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${e^x}\log \sin x$
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${e^x}\tan x$
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None of these

Explanation

$\int {{e^x}\left( {\log \sin x + \cot x} \right)dx}$

As we know

$\begin{array}{l} \int { { e^{ x } }\left[ { f\left( x \right) +f'\left( x \right) } \right] dx } ={ e^{ x } }f\left( x \right) +c \\ f\left( x \right) =\log \sin x \\ f'\left( x \right) =\dfrac { 1 }{ { \sin x } } .\cos x=\cot x \\ \int { { e^{ x } }\left( { \log \sin x+\cot x } \right) dx. } ={ e^{ x } }\left( { \log \sin x } \right) +c \end{array}$

Hence, this is the answer.

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

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

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.

The integral

$\int \cos \left( \log _ { e } x \right) d x$ is equal to :
(where

$C$ is a constant of integration)

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$\dfrac { { x } }{ 2 } \left[ \sin \left( \log _{ { { e } } }{ x } \right) -\cos \left( \log _{ { { e } } }{ x } \right) \right] +{ C }$
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$\dfrac { { x } }{ 2 } \left[ \cos \left( \log _{ { { e } } }{ x } \right) +\sin \left( \log _{ { { e } } }{ x } \right) \right] +{ C }$
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${ x }\left[ \cos \left( \log _{ { { e } } }{ x } \right) +\sin \left( \log _{ { { e } } }{ x } \right) \right] +{ C }$
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$x \left[ \cos \left( \log _ { e } x \right) - \sin \left( \log _ { c } x \right) \right] + C$

Explanation

${ I }=\int \cos (\ell { n }{ x }){ d }{ x }$

${ I }=\cos (\ln { } { x })\cdot { x }+\int \sin (\ell { n }{ x }){ d }{ x }$

$\cos ( \ell n x ) x + \left[ \sin ( \ell n x ) \cdot x - \int \cos ( \ell n x ) d x \right]$

${ I }=\dfrac { { x } }{ 2 } [\sin (\ell { n }{ x })+\cos (\ell { n }{ x })]+{ C }$

Evaluate:

$\displaystyle\int { \dfrac { { x }^{ 4 }+1 }{ 1+{ x }^{ 6 } } } dx$

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${ tan }^{ -1 }(x)-{ tan }^{ -1 }({ x }^{ 3 })+c$
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${ tan }^{ -1 }(x)-\frac { 1 }{ 3 } { tan }^{ -1 }({ x }^{ 3 })+c$
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${ tan }^{ -1 }(x)+{ tan }^{ -1 }({ x }^{ 3 })+c$
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${ tan }^{ -1 }(x)+\frac { 1 }{ 3 } { tan }^{ -1 }({ x }^{ 3 })+c$

Explanation

Given,

$\int \dfrac{x^4+1}{x^6+1}dx$

$=\int \dfrac{x^4+1}{x^6+1}\times \dfrac{x^2+1}{x^2+1}dx$

$=\int \dfrac{(x^6+1)+x^2(x^2+1)}{(x^6+1)(x^2+1)}dx$

$=\int \dfrac{dx}{x^2+1}+\dfrac{1}{3}\int \dfrac{3x^2}{x^6+1}dx$

$=\tan ^{-1}x+\dfrac{1}{3}3\cdot \int \dfrac{x^2}{x^6+1}dx$

substitute

$u=x^3$

$=\tan ^{-1}x+\dfrac{1}{3}3\cdot \int \dfrac{1}{3\left(u^2+1\right)}du$

$=\tan ^{-1}x+\dfrac{1}{3}3\cdot \dfrac{1}{3}\cdot \int \dfrac{1}{u^2+1}du$

$=\tan ^{-1}x+\dfrac{1}{3}3\cdot \dfrac{1}{3}\cdot \tan ^{-1}u$

$=\tan ^{-1}x+\dfrac{1}{3}3\cdot \dfrac{1}{3}\cdot \tan ^{-1}x^3$

$=\tan ^{-1}x+\dfrac{1}{3} \tan ^{-1}x^3+C$

$\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 4 }+1 } } dx=$

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$\dfrac { 1 }{ \sqrt { 2 } } { \tan }^{ -1 }(\dfrac { { x }^{ 2 }+1 }{ \sqrt { 2x } } )+c$
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${ \tan }^{ -1 }(\dfrac { { x }^{ 2 }+1 }{ \sqrt { 2x } } )+c$
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$\dfrac { 1 }{ \sqrt { 2 } } {\tan }^{ -1 }(\dfrac { { x }^{ 2 }-1 }{ \sqrt { 2x } } )+c$
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${ \tan }^{ -1 }(\dfrac { { x }^{ 2 }-1 }{ \sqrt { 2x } } )+c$

Explanation

Given,

$\int \dfrac{x^2+1}{x^4+1}dx$

$=\int \dfrac{\dfrac{1}{x^2}+1}{\dfrac{1}{x^2}+x^2}dx$ ..............dividing by

${x^2}$

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

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

$x-\dfrac{1}{x}=t\rightarrow dt=\dfrac{1}{x^2}+1 dx$

$=\int \dfrac{1}{t^2+2}dt$

$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{t}{\sqrt 2}+c$

$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{x-\dfrac{1}{x}}{\sqrt 2}+c$

$=\dfrac{1}{\sqrt{2}}\tan ^{-1}\dfrac{x^2-1}{\sqrt 2x}+c$

$\int { \frac { dx }{ \sqrt [ 4 ]{ { (x+1) }^{ 5 }{ (x+2) }^{ 3 } } } }$ is equal to :

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$4{ (\frac { x+1 }{ x+2 } ) }^{ 1/4 }+c$
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$-4{ (\frac { x+1 }{ x+2 } ) }^{ -1/4 }+c$
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$-4{ (\frac { x+2 }{ x+1 } ) }^{ 1/4 }+c$
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None of these

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

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

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

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

$\int { \cfrac { \cos { x } -1 }{ \sin { x } +1 } } { e }^{ x }dx$ is equal to:

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$\cfrac { { e }^{ x }\cos { x } }{ 1+\sin { x } } +c$
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$c-\cfrac { { e }^{ x }\sin { x } }{ 1+\sin { x } }$
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$c-\cfrac { { e }^{ x }}{ 1+\sin { x } }$
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$c-\cfrac { { e }^{ x }\cos { x } }{ 1+\sin { x } }$

$\int { \cfrac { 1 }{ \left( 1+x \right) \sqrt { x } } } dx=f\left( x \right) +A$ , where A is any arbitary constant, then the function f(x) is

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$2\tan^{-1}{x}$
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$2\tan^{-1}{\sqrt{x}}$
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$2\cot^{-1}{\sqrt{x}}$
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$log_{e}{(1+x)}$

Explanation

Given,

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

put

$u=\sqrt{x}$

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

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

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

$=f(x)+A$

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

$\int { \dfrac { dx }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+4) } } =$

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$\frac { 1 }{ 3 } \tan ^{ -1 }{ x } -\frac { 1 }{ 3 } \tan ^{ -1 }{ \frac { x }{ 2 } } +c$
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$\frac { 1 }{ 3 } \tan ^{ -1 }{ x } +\frac { 1 }{ 3 } \tan ^{ -1 }{ \frac { x }{ 2 } } +c$
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$\frac { 1 }{ 3 } \tan ^{ -1 }{ x } -\frac { 1 }{ 6 } \tan ^{ -1 }{ \frac { x }{ 2 } } +c$
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$\tan ^{ -1 }{ x } -2\tan ^{ -1 }{ \frac { x }{ 2 } } +c$

Explanation

Given,

$\int \dfrac{dx}{\left(x^2+1\right)\left(x^2+4\right)}$

$=\dfrac 13 \int \dfrac{3 \, dx}{\left(x^2+1\right)\left(x^2+4\right)}$

$=\dfrac 13 \int \dfrac{x^2+4-(x^2+1) \, }{\left(x^2+1\right)\left(x^2+4\right)}dx$

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

$=\dfrac{1}{3}\tan ^{-1}\left(x\right)-\dfrac{1}{6}\tan ^{-1}\left(\dfrac{x}{2}\right)+C$

Integrate:

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

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

Explanation

Given,

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

let

$u=\sqrt{x+4}$

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

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

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

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

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

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

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

The angle made by the tangent line at (1, 3) on the curve

$y=4x-{ x }^{ 2 }$ with

$\overline{ OX }$ is

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$\tan ^{-1}2$
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$\tan ^{-1}(1/2)$
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$\tan ^{-1}-2$
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None of these

Explanation

Given,

$y=4x-x^2$

$\dfrac{dy}{dx}=4-2x$

$\dfrac{dy}{dx}_{(1,3)}=4-2(1)=2$

Therefore, angle made by tangent,

$\tan \theta =2$

$\therefore \theta =\tan ^{-1}2$

$\displaystyle \int { \left( u\cfrac { dv }{ dx } \right) } dx=uv-\int { wdx }$ , then

$w=$

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$\cfrac { du }{ dx } \cfrac { dv }{ dx }$
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$v\cfrac { du }{ dx }$
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$\cfrac { d }{ dx } (uv)$
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$u\cfrac { dv }{ dx }$

Explanation

Given,

$\displaystyle \int \left ( u\dfrac{dv}{dx} \right )dx=uv-\int wdx$

it is called as, product rule integration,

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

$\displaystyle \int \left ( u\dfrac{dv}{dx} \right )dx$

$\displaystyle =uv-\int v\dfrac{du}{dx}dx$

$\displaystyle \therefore w=v\dfrac{du}{dx}$

$\displaystyle \int { \cfrac { x\tan ^{ -1 }{ x } }{ { \left( 1+x^{ 2 } \right) }^{ 3/2 } } } dx=$

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$\cfrac { x+\tan ^{ -1 }{ x } }{ { \sqrt { \left( 1+x^{ 2 } \right) } }^{ } } +c$
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$\cfrac { x-\tan ^{ -1 }{ x } }{ { \sqrt { \left( 1+x^{ 2 } \right) } }^{ } } +c$
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$\cfrac { \tan ^{ -1 }{ x }-x }{ { \sqrt { \left( 1+x^{ 2 } \right) } }^{ } } +c$
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None of these

Explanation

Let

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

$x = \tan \theta$

$dx = \sec^2 \theta d \theta$

$I = \displaystyle \int \dfrac{\tan \theta . \theta . \sec^2 \theta}{\sec^3 \theta} d \theta = \int \dfrac{\theta \tan \theta }{\sec \theta } d \theta$

$I = \displaystyle \int \theta. \sin \theta \, d \theta = - \theta \cos \theta + \int \cos \theta$

$= - \theta \cos \theta + \sin \theta$

$= \dfrac{-\tan^{-1} x}{\sqrt{1 + x^2}} + \dfrac{x}{\sqrt{1 + x^2}} = \dfrac{x - \tan^{-1} x}{\sqrt{1 + x^2}} + C$

$\displaystyle \int \dfrac{dx}{9x^2 + 1} =$ _____.

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

Explanation

Consider the given integral.

$I=\displaystyle \int \dfrac{1}{(9x^2+1)}dx$

$I=\displaystyle \int \dfrac{1}{9(x^2+\dfrac{1}{9})}dx$

$I=\dfrac{1}{9}\displaystyle \int \dfrac{1}{x^2+(\dfrac{1}{3})^2}dx$

$I=\dfrac{1}{9}\times \dfrac{1}{1/3}\tan^{-1}(\dfrac{x}{1/3})+C$

$I=\dfrac{1}{3} \tan ^{-1}{3x}+C$

Hence, this is the answer.

Evaluate

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

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$e^x(\log{x}+\cfrac{1}{x^2})$
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$e^x(\log{x}+\cfrac{1}{x})$
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$e^x(\log{x}-\cfrac{1}{x^2})$
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$e^x(\log{x}-\cfrac{1}{x})$

Explanation

Given,

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

$\displaystyle =\int \:e^x\log \left(x\right)dx+\int \dfrac{e^x}{x^2}dx$

We know,

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

$\displaystyle =\left (e^x\log \left(x\right)-\text{Ei}\left(x\right) \right )-\dfrac{e^x}{x}+\text{Ei}\left(x\right)$ where

$\displaystyle\text{Ei}(x)=\int \dfrac{e^x}{x} dx$

$\displaystyle =e^x\log \left(x\right)-\dfrac{e^x}{x}+C$

$=e^x(\log x-\dfrac{1}{x})$

$\int { \left( 2+\log { x } \right) { \left( ex \right) }^{ x }dx } =.....+C;x>1$

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${ \left( ex \right) }^{ x }$
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${ x }^{ x }$
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${ \left( ex \right) }^{ -x }$
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${ e }^{ { x }^{ x } }$

Explanation

Let

${ \left( ex \right) }^{ x }=t$

$x\ln { ex } =\ln { t }$

$\Rightarrow x(1+\ln { x } )=\ln { t }$

$\Rightarrow \left( x.\cfrac { 1 }{ x } +\left( 1+\ln { x } \right) \right) dx=\cfrac { 1 }{ t } dt$

$\Rightarrow { \left( ex \right) }^{ x }\left( 2+\ln { x } \right) dx=dt$

now

integrate on both sides

$\int { dt } =t+c\Rightarrow { \left( ex \right) }^{ x }+c$

$\displaystyle \int { e^ { \sec} } (\sec x\, \tan x \, f(x)+ \sec x\, \tan x+ \tan^2x)dx=e^{\sec x}f(x)+c. Then f(x) is$

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$\sec x+x\ tan x+\dfrac{1}{2}$
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$x \sec x+x \tan x+\dfrac{1}{2}$
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$x \sec x+x^2 \tan x+\dfrac{1}{2}$
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$\sec x+ \tan x+\dfrac{1}{2}$

Explanation

$\displaystyle \int { e^ { secx} } (\sec x\, \tan x \, f(x)+ \sec x \, \tan x+ \sec^2x)dx=e^{sec x}f(x)$
Differentiate both side

$\int { e^ { secx} } (\sec x\, \tan x\, f(x)+\sec x\,\tan x+\sec^2x)dx=e^{secx}f'(x)+e^{secx}\,tan x\,secx\,f(x)$

$\therefore \sec x\,tanx+\sec^2x=f'(x)$

$\therefore f(x)=\sec x+\tan x+d$

$\int _{ a }^{ b }{ \dfrac{dx}{2 + 3x} } =$

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$\dfrac{1}{3} log_e (2 + 3b)$
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$\dfrac{1}{3} log_e (2 + 2a)$
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$\dfrac{1}{3} log_e \left(\dfrac{2 + 3b}{2 + 3a}\right)$
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$\dfrac{1}{3} log_e \left(\dfrac{2 + 3a}{2 + 3b}\right)$

$\displaystyle\int \dfrac{2x^3-1}{x^4+x}dx$ is equal to?

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$ln\left|\dfrac{x^3+1}{x}\right|+c$
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$ln\left|\dfrac{x^3+1}{x^2}\right|+c$
0%
$\dfrac{1}{2}\ln\left|\dfrac{x^3+1}{x^2}\right|+c$
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$\dfrac{1}{2}ln\left|\dfrac{x^3+1}{x}\right|+c$

Explanation

$\displaystyle\int \dfrac{2x^3-1}{x^4+x}dx=\displaystyle\int \dfrac{2x-x^{-2}}{x^2+x^{-1}}dx=ln(x^2+x^{-1})+c=ln(x^3+1)-lnx+c$ .

$\displaystyle\int 2^{2x}\cdot 2^xdx=A\cdot 2^{2^x}+c$ , then

$A=?$

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

Explanation

$2^x=z\Rightarrow 2^xdx=\dfrac{dz}{ln 2}$

$\Rightarrow \dfrac{1}{ln 2}\displaystyle\int 2^zdz=\dfrac{1}{(ln 2)^2}2^{2^x}+c$ .

$\displaystyle \int \dfrac{dx}{x^3(1+x^6)^{2/3}} = xf(x) (1 + x^6)^{\frac{1}{3}} + C$
where

$C$ is a constant of integration, then the function

$f(x)$ is equal to -

0%
$-\dfrac{1}{6x^3}$
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$\dfrac{3}{x^2}$
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$-\dfrac{1}{2x^2}$
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$-\dfrac{1}{2x^3}$

Explanation

$\displaystyle \int \dfrac{dx}{x^3(1+x^6)^{2/3}} = xf(x) ( 1 +x^6)^{1/3} + C$

$\displaystyle \int \dfrac{dx}{x^7\left(\dfrac{1}{x^6} + 1\right)^{2/3}} = xf(x) (1 + x^6)^{1/3} + C$
Let

$t = \dfrac{1}{x^6} + 1$

$dt = \dfrac{-6}{x^7} dx$

$\displaystyle -\dfrac{1}{6} \int \dfrac{dt}{t^{2/3}} = - \dfrac{1}{2} t^{1/3}$

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

$\therefore f(x) = -\dfrac{1}{2x^3}$

Evaluate :

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

0%
$tan x+ sec x+ C$
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$tan x -sec x + C$
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$-tan x + sec x + C$
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$-tan x - sec x + C$

What is

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

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

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

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

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

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

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

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

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

$\displaystyle \int\frac{x\cos x\log x-\sin x}{x(\log x)^{2}}dx=$

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$\displaystyle \frac { \sin { x } }{ \log { x } } +C$
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$\displaystyle \frac { \cos { x } }{ \log { x } } +C$
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$\displaystyle \frac { \log { x } }{ \sin { x } } +C$
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$\displaystyle \frac { \log { x } }{ \cos { x } } +C$

Explanation

$\displaystyle\int { \frac { x\cos(x)\log(x)-\sin(x) }{ x{ (\log(x)) }^{ 2 } } dx }$

$\displaystyle=\int { \frac {\cos(x) }{\log(x) } -\frac {\sin(x) }{ x{ (\log(x)) }^{ 2 } } } dx$

$\displaystyle=\int { \frac { \cos(x) }{ \log(x) } dx } -\int { \frac { \sin(x) }{ x{ (\log(x)) }^{ 2 } } dx }$

$\displaystyle=\frac { 1 }{ \log(x) } \sin(x)-\int { \frac { 1 }{ x{ (\log(x)) }^{ 2 } } (-\cos(x)) } dx-\int { \frac { \sin(x) }{ x{ (\log(x)) }^{ 2 } } dx } +C$

$\displaystyle=\frac { \sin(x) }{ \log(x) } +\int { \frac {\ sin(x) }{ x{ (\log(x)) }^{ 2 } } dx } -\int { \frac { \sin(x) }{ x{ (\log(x)) }^{ 2 } } dx } +C$

$\displaystyle=\frac { \sin(x) }{ \log(x) } +C\\$

$\int { x\sin { x } } dx=-x\cos { x } +\alpha$ , then

$\alpha$ is equal to

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$\sin x+C$
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$\cos x+C$
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$C$
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None of these

Explanation

$I=\displaystyle\int{x\sin{x}dx}$

Integrating by parts,

Let

$u=x\Rightarrow\,du=dx$

$dv=\sin{x}dx\Rightarrow\,v=-\cos{x}$

$\int u.v dx=u \int vdx-\int \left [\int vdx. \dfrac{du}{dx}.dx \right ]$ ......by parts formula

$I=-x\cos{x}+\displaystyle\int{\cos{x}dx}$

$I=-x\cos{x}+\sin{x}+c$ ......where

$c$ is the constant of integration.

We have

$I=\displaystyle\int{x\sin{x}dx}=-x\cos{x}+\alpha$

Comparing with

$I=-x\cos{x}+\sin{x}+c$ we have

$\alpha=\sin{x}+c$

Evaluate :

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

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$- cot x -tan x +C$
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$-cot x + tan x + C$
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$cot x -tan x +C$
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$cot x + tan x + C$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 10 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 10 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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