MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 2

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

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

Explanation

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

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

What is

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

$(n \neq 1)$ ?

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

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

$\displaystyle \int xsecx.tanxdx=$

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$x\sec x+\log|\tan(\pi/2+x/2)|+c$
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$x\sec x-\log$ $|\tan(\pi/4+x/2)|+c$
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$x\sec x-\log |\tan(\pi/4+x)|+c$
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$x\sec x+\log|\tan x/2|+c$

Explanation

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

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

$-\int1\int(\sec x\ \tan x)dx$ ....... [Using integration parts]

$=x \sec x-\int \sec x\ dx$

$=x\sec x-\log\left | \sec x+\tan \ x \right |+c$

$=x\sec x-\log\left | \tan (\pi/4+x/2) \right |+c$ .....

$[\because \sec x+\tan x=\tan (\pi/4+x/2)]$

$\therefore \int x\sec x \tan x\ dx=x\sec x-\log\left | \tan (\pi/4+x/2) \right |+c$

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

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$\displaystyle \frac{x}{(\log x)^{2}+1}+c$
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$\displaystyle \frac{xe^{x}}{1+x^{2}}+c$
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$^{\dfrac{x}{x^{2}+1}+c}$
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$\displaystyle \frac{\log x}{(\log x)^{2}+1}+c$

Explanation

$\int [\dfrac {log x-1}{1+(log x)^2}]^2dx$

$\int \dfrac {(log x-1)^2}{[1+(log x)^2]^2}=\int \dfrac {(log x)^2+1-2 log x)}{[1+(log x)^2]^2}$

$=\int (\dfrac {1}{[(1+(log x)^2]}-\dfrac {2 log x}{[1+log^2x]^2})dx$

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

$=\dfrac {1}{(1+log^2x)}\int 1 dx+\int [\dfrac {2 log x}{(1+log^2x)^2}\times\dfrac {1}{x}\int 1\cdot dx]dx-\int \dfrac {2 log x}{(1+log^2x)^2}dx$

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

$=\dfrac {x}{1+log^2x}+c$

$\displaystyle \int \sec^{2}x.\text{cosec}^{2}xdx=$

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$\tan x-\cot x +c$
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$\tan x +\cot x +c$
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$-\tan x + \cot x +c$
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$\sec x \tan x +c$

Explanation

$\displaystyle \int \sec^{2}x. \text{cosec}^{2}xdx$

$\displaystyle =\int \dfrac{1}{\cos^{2}x.\sin^{2}x}dx$

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

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

$\displaystyle =\int \dfrac{1}{\sin^{2}x}+\dfrac{1}{\cos^{2}x}dx$

$\displaystyle =\int (\text{cosec}^{2}x+\sec^{2}x) dx$

$=-\cot x+\tan x +c$

$=\tan x-\cot x+c$

$\displaystyle \int\sqrt{\frac{x}{x-1}}dx, x\in(0,\pi/2)$ equals

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$\sqrt{x(x-1)}+\log(\sqrt{x}+\sqrt{x-1})+c$
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$\sqrt{x(x-1)}-\log(\sqrt{x}+\sqrt{x-1})+c$
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$\sqrt{x(x-1)}+\log(\sqrt{x}-\sqrt{x-1})+c$
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$\sqrt{x(x+1)}-\log(\sqrt{x}-\sqrt{x+1})+c$

Explanation

$x=sec^{2}\theta$

$dx =2 sec^{2}\theta \cdot tan \theta \cdot d\theta$

$\int \sqrt{\dfrac{sec^{2}\theta}{tan^{2}\theta}} \cdot2 sec^{2}\theta \ tan\ \theta d\theta$

$\int \dfrac{1}{sin \theta} \cdot 2 sec^{2}\theta \cdot \dfrac{sin \theta}{cos \theta} \cdot d\theta$

$=2\int sec^{3}\theta \cdot d\theta \cdot$

$=2\int (sec \theta)(sec^{2}\theta)d\theta \cdot$

$=2[sec\theta tan \theta -\int (sec \theta tan \theta \cdot tan \theta)d\theta \cdot ]$

$2\int sec^{3}\theta ^{d\theta}=2[sec\theta tan \theta]-2\int sec \theta(sec^{2}\theta-1)d\theta$

$2\int sec^{3}\theta d\theta =sec\theta tan \theta +\int sec \theta d\theta$

$2\int sec^{3} \theta d\theta = sec\theta tan\theta \rightarrow log |sec\theta+tan \theta |$

and

$=\sqrt{x(x-1)}+log |\sqrt{x}+\sqrt{x-1}|+c$

$\int \dfrac{x}{\sqrt{x-1}} dx = \sqrt{x(x-1)} +log |\sqrt{x}+\sqrt{x-1}|+c \cdot$

$\displaystyle \int xe^{2x}(1+x)dx$ equal to

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$\displaystyle \frac{xe^{x}}{2}+c$
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$\displaystyle \frac{(e^{x})^{2}}{2}r$
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$\displaystyle \frac{(1+x)^{2}}{2}+c$
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$\displaystyle \frac{(xe^x)^2}{2}$

Explanation

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

$\displaystyle =\int xe^{2x}dx+\int x^{2}e^{2x}dx$

$\displaystyle =e^{2x} \frac{x^{2}}{2}-\int x^{2}e^{2x}dx+\int x^{2}e^{2x}dx$

$\displaystyle I=\frac{(xe^{x})^{2}}{2}$

The value of

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

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$\displaystyle \frac{x^{2}2^{3x}}{4\log 2}-\frac{x.2^{3x+1}}{9.(\log 2)^3}+\frac{2^{3x+1}}{27(\log 2)}+c$
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$\displaystyle \frac{x^{2}2^{3x}}{3\log 2}-\frac{x.2^{3x+1}}{9.(\log 2)}+\frac{2^{3x+1}}{27(\log 2)}+c$
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$\displaystyle \frac{x^{2}2^{3x}}{3\log 2}-\frac{2^{3x+1}.x}{3^{2}(\log 2)^{2}}+\frac{2^{3x+1}}{3^{3}(\log 2)^{3}}+c$
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$\displaystyle \frac{x^{2}2^{3x}}{\log 2}-\frac{2^{3x+1}.x}{3^{2}(\log 2)^{2}}+\frac{2^{3x+1}}{(\log 2)^{3}}+c$

Explanation

$\int x^2\cdot 2^{3x} dx\cdot$

$=x^2\int 2^{3x}dx+\int 2x\int 2^{3x}\cdot dx\cdot$

$=x^2\dfrac {2^{3x}}{log 8}-\int 2x\dfrac {2^{3x}}{log 8^{dx}}$

$=\dfrac {x^22^{3x}}{log 8}-\dfrac {2}{log 8}[\dfrac {x2^{3x.}}{log 8}-\int \dfrac {2^{3x}}{log 8^{.dx}}]$

$=\dfrac {x^22^{3x}}{log 8}-\dfrac {2}{log 8}[\dfrac {x2^{3x}}{log 8}-\dfrac {1}{log 8}\dfrac {2^{3x}}{log 8}]+c.$

$=\dfrac {x^{2}2^{3x}}{log 8}-\dfrac {2^{3x+1}\cdot x}{3^2(log 2)^2}+\dfrac {2^{3x+1}}{3^3(log 2)^{3}}+c$

$\displaystyle \int\log{x}d{x}=$

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$x(\log x-x)+c$
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$x(\log x)+c$
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$x(\log x-1)+c$
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$x(\log x+x)+c$

Explanation

$\int\ log\ x\ dx= \int\ log\ x-1\ dx$

$= log\ x\int\ 1-dx-\ \int(^1/_x.\sqrt{1.dx})dx$

$= log\ x(x)-x+c$

$= x(log\ x-1)+c$

$\because \int \ log\ x\ dx= x(log\ x-1)+c$

Evaluate :

$\displaystyle \int\frac{\cot^{2}x}{(co\sec^{2}x+co\sec x)}d{x}$

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$x-\sin x +C$
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$x + \cos x +C$
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$\sin x - x+C$
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$2\tan(\displaystyle \frac{ax}{2})+c$

Explanation

$\displaystyle \int\frac{\cot^{2}x}{(co\sec^{2}x+co\sec x)}d{x}$

$\displaystyle \int \dfrac{cos ec^{2}x-1}{cos ecx(cosecx+1)}dx$

$\displaystyle \int \dfrac{(cosecx-1)(cosecx+1)}{cos ecx(cosecx+1)}dx$

$\displaystyle \int \dfrac{cosecx-1}{cosecx}dx=\int (1-sin x)dx$

$=x+cos x+c$

$\because -\int sin x.dx=cos x+c$

$\displaystyle \int[\frac{cosx}{x}-\sin{x}\log x]dx=$

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$(\log x) sinx+c$
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$(\log x) (cos x) +c$
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$2(\log x) (cos x) +c$
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$-(\log x) sin x+c$

Explanation

$\int\left [ \displaystyle \dfrac{cos\ x}{x}-sinx\ log\ x \right ]dx$

$\int \displaystyle \dfrac{cos\ x}{x}dx-\ \int\ sin\ x\ log\ x\ dx$

$cos\ x\ \int \displaystyle \dfrac{1}{x}\ dx+\ \int\ sin\ x\ log\ x.\ dx$

$-\int\ sin\ x\ log\ x\ dx.$

$= (\ cos\ x)( log\ x)+c$

Hence,

$\int \left [ \displaystyle \dfrac{cos x}{x}-(sin\ x)(log\ x) \right ]dx=(\ cos\ x)(\ log\ x)+c$

Evaluate

$\displaystyle \int x^{2}e^{x}dx=$

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$e^{x}(x^{2}-2x+2)+c$
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$e^{x}(x^{2}+2x+2)+c$
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$x^{2}+ex+c$
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$e^{x}(x^{2}+x+2)+c$

Explanation

Let

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

$= x^{2} \displaystyle \int e^{x}dx$

$-\displaystyle \int \left (\frac{dx^{2}}{dx}\cdot \int e^{x} dx\right).dx$

$=x^{2} e^{x} -\displaystyle \int (2xe^{x}) dx\cdot$

$=x^{2}e^{x}-2\displaystyle \int xe^{x}\cdot dx\cdot$
Taking

$\displaystyle \int xe^{x}= xe^{x}-e^{x}\cdot$ ....(by using by parts)
Therefore,

$I=x^{2}e^{x}-2[xe^{x}-e^{x}]+c$

$=e^{x}[x^{2}-2x+2]+c$

$\displaystyle \int e^{\log x}\cos xdx=$

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$xsinx-cosx+c$
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$\dfrac{x}{2}sinx+cos^{2}x+c$
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$xsinx+cosx+c$
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$xsinx+sinx+c$

Explanation

$\int e^{logx}cosx\ dx=\int x\ cosx\ dx.$

$=x\int cosx\ dx-\ \int \left ( 1\int cosxdx \right )dx$

$=x\ sinx+cosx+c$

$\int e^{logx}cosx\ dx=\ x\ sinx+cosx+c.$

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

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

Explanation

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

$=\displaystyle\int \frac{\cos x}{\log x} dx - \int\frac{\sin x}{x(\log\ x)^{2}}dx$ Using Integration by parts in the first expression, we get

$=\displaystyle\frac{1}{\log\ x}\int\cos x dx -\int\left(\frac{d\left (\dfrac{1}{\log x}\right )}{dx}\cdot\int\cos x dx\right)dx-\int \frac{\sin x}{x(\log x)^{2}}dx$

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

$=\dfrac{\sin x}{\log x}+c$
where,

$c$ is a constant of integration.

The value of

$\int x(cosecx\ cotx)dx=$ is

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$xcosec x-log |tanx/2|+c$
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$2-xcosec x+\log| tan \displaystyle \frac{x}{2}|+c$
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$xcosec x-2 \log|tanx/2|+c$
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$xcot x-log |tan{\displaystyle \frac{x}{2}}|+c$

Explanation

$\int x(cosecx\ cot)dx=x\int (cosecx\ cotx)dx$

$-\int 1\left ( \int cosecx\ cot\ xdx \right )dx$

$=x\left [ -cosecx \right ]+c+\int (cosecx)dx$

$=\ -x\ cosecx+c+log\left | cosecx-cotx \right |$

$=-x\ cosecx+c+\ log\left | \dfrac{1-cosx}{sinx} \right |$

$=2-x\ cosecx+c_{1}+log\left | tan\ x/2 \right |$

$\int x\ cosecx\ cotx\ dx$

$=2-x\ cosecx+c_{1}+log\ \left | tan\ \dfrac{x}{2} \right |$

Solve

$\displaystyle \int \tan^{-1}xdx$

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$x\tan x -\log|1+x^{2}|+c$
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$x$ $\tan^{-1} x -\displaystyle \frac{1}{2}\log|1+x^{2}|+c$
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$x$ $\tan^{-1} x+\log\sqrt{1+x^{2}}+c$
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$x\tan x+\log|1+x^{2}|+c$

Explanation

$\displaystyle \int \tan^{-1}x \ .1 \ \ dx$

Let us assume here

$\tan^{-1}x$ is the first function and constant 1 is the second function. Then the integral of the second function is x.

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

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

Now by integrating the functions by parts, we get

$\displaystyle =\tan^{-1}x\int 1\ dx$

$\displaystyle -\int \left ( d\frac{\ tan^{-1}x}{dx}\int 1\ dx \right )dx$

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

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

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

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

Evaluate

$\displaystyle \int\frac{\log(x/e)}{(\log x)^{2}}dx$

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$\displaystyle \frac{\log x}{x}+c$
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$\displaystyle \frac{x}{\log x}+c$
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$^{\dfrac{x}{log(x)^{2}}+c}$
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$\displaystyle \frac{(\log x)^{2}}{x}+c$

Explanation

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

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

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

$=\displaystyle \dfrac{1}{(\log \ x)}\int\ 1.dx+\ \int\ \dfrac{1}{(\log \ x)^{2}}dx-\int\ \dfrac{1}{(\log \ x)^{2}}dx$ ..... [Using integration by parts]

$=\ \dfrac{x}{\log \ x}+c$

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

$\displaystyle \int x^{2} {\it cosh} 4xdx=$

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$\displaystyle \frac{x^{2}}{4} {\it sinh}4x-\displaystyle \frac{x}{8}{\it cosh} 4x+\displaystyle \frac{\sinh 4x}{32}+c$
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$\displaystyle \frac{x^{2}}{4}sinh4x+\frac{x}{8}$ cosh $4{x-}\displaystyle \frac{\sinh 4x}{32}+c$
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$\displaystyle \frac{x^{2}}{4}sinh4x-\frac{x}{8}$ cosh $4x-\displaystyle \frac{\sinh 4x}{32}+c$
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$\displaystyle \frac{x^{2}}{4}sinh4x+\frac{x}{8}$ cosh $4x-\displaystyle \frac{\sin h4x}{64}+$

Explanation

$\int x^{2}\ cosh\ 4x.dx$

$x^{2}\int cosh\ 4x\ dx-\int 2x\left [ \left ( \int cosh\ 4x \right )dx \right ]dx$

$=\dfrac{x^{2}}{4}sin\ h4x-\ 1/2\ \int (x\ sinh4x)dx\ (1)$

$\int x\ sinh\ 4x\ dx=\dfrac{x}{4}\ cos\ h4x-\int \dfrac{cosh}{4}4x\ dx$

$=\left ( \dfrac{x}{4}cos\ h\ 4x-\dfrac{sin}{16} 4x\right )\ -(2)$

Using (2) in (1),

$=\dfrac{x^{2}}{4}sin\ h4x-\dfrac{x}{8}cosh\ 4x+\dfrac{sinh4x}{32}+c$

$\int x^{2}\ cosh\ 4x\ dx=\dfrac{x^{2}}{4}sin\ h4x-x/8\ cosh\ 4x+\dfrac{sinh\ 4x}{32}+c$

$\displaystyle \int cot^{-1}(\frac{x-1}{x+1})dx=$

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

Explanation

$cosec\theta =\sqrt{1+x^{2}}$

$sin \theta =\dfrac{1}{\sqrt{1+x^{2}}}$

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

$x=cot\theta$

$\int cot^{-1}(\dfrac{cot\theta -1}{cot\theta +1})[cosec^{2}\theta ]d\theta$

$-\int cot^{-1}[cot(\pi /4+\theta )](cosec^{2}\theta )d\theta$

$=-\int (\pi /4+\theta (cosec^{2}\theta ))d\theta$

$=-[\int \pi /4cosec^{\gamma }\theta d\theta +\int \theta cosec^{\gamma \theta d\theta }]$

$=-[-\pi /4cot\theta +[\theta (-cot\theta )+\int cot\theta d\theta ]$

$=\pi /4cot\theta +cot\theta -log|sin\theta |+c$

$=\pi /4x+x cot^{-1}x-1/2log(\sqrt{1+x^{2}})+c$

$\int cot^{-1}(\dfrac{x-1}{x+1})dx=\pi /4{x}+x cot^{-1}x -1/2 log (\sqrt{1+x^{2}})+c$

$\displaystyle \int e^{x}cosec x(1-\cot x)dx=$

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$e^{x}\cot x+c$
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$e^{x}cosec x\cot x+c$
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$e^{x}cosec x+c$
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$-e^{x}\cot x+c$

Explanation

$\int e^{x}cosec\ x(1-cot\ x)dx$

$\int e^{x}\left ( cosec\ x-cose\ x\ cot\ x \right )dx$

$\int e^{x}cosec\ x\ dx-\int e^{x}\left ( cosec\ x.cot\ x \right )dx$

$\int e^{x}\left ( cosec\ x\ cot\ x \right )dx=ex\int cosec\ x\ cot\ x\ dx+\int e^{x}cosec\ x.dx$

$=e^{x}cosec\ x+\int cosec\ x\ dx.$

$\int e^{x}cosec\ x\ dx-\left ( -e^{x}cosec\ x+\int cosec\ x\ dx \right )+c$

$=\int e^{x}cosec\ x\ dx+e^{x}\ cosec\ x-\int e^{x}\ cosec\ x\ dx$

$=\ e^{x}cosec\ x+c$

$\displaystyle \int x^{2}a^{x}dx=$

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$a^{x}[\displaystyle \frac{x^{2}}{\log a}-\frac{2x}{(\log a)^{2}}+\frac{2}{(\log a)^{3}}]+c$
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$a^{x}[\displaystyle \frac{x^{2}}{\log a}+\frac{2x}{(\log a)^{2}}-\frac{2}{(\log a)^{3}}]+c$
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$a^{x}[\displaystyle \frac{x^{2}}{\log a}+\frac{2x}{(\log a)^{2}}+\frac{2}{(\log a)^{3}}]+c$
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$a^{x}(\displaystyle \frac{x}{\log a}-\frac{1}{(\log a)^{2}})+c$

Explanation

$\int x^{2}a^{x}dx = x^{2}\int a^{x}dx-\int \left [ \dfrac{dx^{2}}{dx}\int a^{x}\ dx \right ]dx$

$\dfrac{x^{2}a^{x}}{log\ a}-\int \left [ 2x\int a^{x}dx \right ]dx$

$\dfrac{x^{2}a^{x}}{log\ a}-2\int \dfrac{x\ a^{x}}{log^{a}}dx$

$\dfrac{x^{2}a^{x}}{log\ a}-\dfrac{2}{log^{a}}\int [xa^{x}]dx$

$\int xa^{x}dx=\dfrac{xa^{a}}{ ^{a}}-\int 1\dfrac{a^{x}}{(log^{a})}dx$

$=\dfrac{xa^{x}}{log^{a}}-\dfrac{a^{x}}{(log^{a})^{2}}$

$=\dfrac{x^{2}a^{x}}{log^{a}}-\dfrac{2}{log^{a}}\left [ \dfrac{xa^{x}}{log^{a}}-\dfrac{a^{x}}{(log^{a})^{2}} \right ]$

$a^{x}\left [ \dfrac{x^{2}}{log{a}}-\dfrac{2x}{(loga)^{2}}+\dfrac{2}{(log\ a)^{3}} \right ]+c$

lf polynomials

$P$ and

$Q$ satisfy

$\displaystyle{\int((3x-1)\cos x+(1-2x)\sin x)dv}=P\cos x+Q\sin x+R$ (ignoring the constant of integration) then

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$P=3x-2$
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$Q=2+x$
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$P=3(x-1)$
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$Q=3(x-1)$

Explanation

Let

$I=\int((3x-1)\cos x+(1-2x)\sin x)dx$

Using integration by parts, we evaluate the 2 expressions individually,

$\int (3x-1) \cos x = (3x - 1) \sin x - \int 3\sin x dx$

$= (3x - 1) \sin x + 3 \cos x + c$

Similarly,

$\int (1-2x)\sin x dx = -(1-2x)\cos x - \int (-2)(-\cos x) dx$

$= (2x - 1) \cos x - 2\sin x + c$

Adding both the expressions, we get

$I= 3(x - 1)\sin x + 2(x+1)\cos x$ Ignoring the constant

Comparing it with the RHS, we get

$P = 2(x+1)$ and

$Q = 3(x-1)$

Hence, option 'D' is correct.

Evaluate

$\displaystyle \int\frac{\arcsin\sqrt{x}}{\sqrt{1-x}}dx$

$=$

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$2[\sqrt{x}-\sqrt{1-x}$ arc $\sin\sqrt{x}]+c$
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$2[\sqrt{x}+\sqrt{1-x}$ arc $\sin\sqrt{x}]+c$
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$2[\sqrt{x}+\sqrt{1-x}$ arc $\cos\sqrt{x}]+c$
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$2[\sqrt{x}-\sqrt{1-x}$ arc $\cos\sqrt{x}]+c$

Explanation

$\displaystyle \int \dfrac {\sin ^{-1}\sqrt x}{\sqrt {1-x}}\cdot dx.$

Put

$x=\sin ^2\theta$

$\Rightarrow dx=2 \sin \theta \cos \theta d\theta.$

$\therefore \displaystyle \int \dfrac {\sin ^{-1}\sqrt x}{\sqrt {1-x}}\cdot dx.$

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

$\displaystyle =2\int \theta \sin \theta d\theta$

$=2[\theta(-\cos \theta)+\int \cos \theta d\theta.]$

$=2[\theta(-\cos \theta)+\sin \theta]+c$

$=2[\sin \theta-\theta \cos \theta]+c$

$=2[\sqrt x-(\sin ^{-1}\sqrt x)\sqrt {1-x}]+c$

$\therefore \displaystyle \int \dfrac {\sin ^{-1}\sqrt x}{\sqrt {1-x}}dx=2[\sqrt x-(\sqrt {1-x})\sin ^{-1}\sqrt x]+c$

$\displaystyle \int e^{x}(x^{2}-5x+8)$ dx

$=e^{x}f(x)+c$ then

$f(x)$

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$x^{2}-5x+12$
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$x^{2}+7x+15$
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$x^{2}-7x-15$
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$x^{2}-7x+15$

Explanation

$\int e^{x}(x^{2}-5x+8)dx$

$=\int e^{x}x^{2}dx-5\int e^{x}.xdx+8\int e^{x}\ dx$

$=x^{2}e^{x}=2\int xe^{x}\ dx-5\int e^{x}\ x+8\int e^{x}dx$

$x^{2}e^{x}-7[xe^{x}-e^{x}]+8e^{x}$

$=x^{2}e^{x}-7xe^{x}+15\ e\ ^{x}$

$=e^{n}[x^{2}-7x+15]+c$

$so,\ f(x)-\ x^{2}-7x+15$

Evaluate

$\displaystyle \int x\frac{(\sec 2x-1)}{(\sec 2x+1)}dx$

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$x \tan x-\log |\displaystyle \sec x|+\frac{x^{2}}{2}+c$
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$x \tan x-\log |\displaystyle \sec x|-\frac{x^{2}}{2}+c$
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$x \tan x-\log |\sec\frac{x}{2}|+c$
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$x \tan x-\log |\displaystyle \sec\frac{x}{2}|+\frac{x^{2}}{2}+c$

Explanation

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

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

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

$\displaystyle =\int x \sec ^{2}x \ {dx}-\int xdx$

We know,

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

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

Now by integrating the functions by parts, for the first part taking

$x$ as

$u$ and

$sec^2 x$ as

$v$ .

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

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

$=x \tan x-\log |\sec x|-\dfrac{x^{2}}{2}+c$

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

$\displaystyle \int e^{x}2^{3\log_{2}x}dx=e^{x}f(x)+c$ , then

$f(x)=$

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${ x }^{ 3 }-3{ x }^{ 2 }+6x-6$
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$x^{3}-3x^{2}-6x-3$
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$x^{3}-3x^{2}+6x+6$
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$x^{3}+3x^{2}+6x+6$

Explanation

Given,

$\displaystyle \int e^{x}2^{3\log_{2}x}dx=e^{x}f(x)+c$ ......(A)
Consider,

$\int e^{ x }2^{ 3\log _{ 2 }{ x } }dx=\int e^{ x }.x^{ 3 }dx$

$\int e^{x} \times x^{3}dx=x^{3}\int e^{x}dx-\int 3x^{2}e^x\ dx$

$=x^{3}e^{x}-3 \int x^{2}e^{x}\ dx\$ ....(1)

$\int x^{2}e^{x}=x^{2}e^{x}-2\int xe^{x}\ dx\$ .....(2)

$\int xe^{x}=\ xe^{x}-e^{x}\$ .....(3)
Using (3) in (2),

$\int x^{2}e^{x}=x^{2}e^{x}-2(xe^{x}-e^{x})\$ ....(4)
Using (4) in (1)

$\displaystyle \int e^{ x }.x^{ 3 }dx= x^{3}e^{x}-3\left [ x^{2}e^{x}-2(xe^{x}-e^{x}) \right ]+c$

$=x^{3}e^{x}-3x^{2}e^{x}+6xe^{x}-6e^{x}+c$

$=e^{x}\left [ x^{3}-3x^{2}+6x-6 \right ]+c$
On comparing with eqn (A), we get

$f(x)=x^{3}-3x^{2}+6x-6$

$\displaystyle \int\frac{x+\sin x}{1+\cos x}dx=$

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$xtan \displaystyle \frac{x}{2}+c$
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$xcot \displaystyle \frac{x}{2}+c$
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$xsin \displaystyle \frac{x}{2}+c$
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$xcos \displaystyle \frac{x}{2}+c$

Explanation

$\displaystyle \int\frac{x+\sin x}{1+\cos x}dx$

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

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

$\displaystyle=\frac { 1 }{ 2 } [x\frac { \tan { (x/2) } }{ 1/2 } -\int \frac { \tan { (x/2) } }{ 1/2 } dx]+\int \tan { (x/2) } dx+C$

$=x\tan { (x/2) } +C$

$\displaystyle \int\cos^{-1}(\frac{1}{x})dx$ equal to

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$x\sec^{-1}x+\cosh^{-1}x+c$
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$x\sec^{-1}x-\cosh^{-1}x+c$
0%
$x\sec^{-1}x-\sin^{-1}x+c$
0%
$x\sec^{-1}x+\sin^{-1}x+c$

Explanation

$x=sec\theta$

$dx=sec\theta \cdot tan\theta d\theta.$

$\int cox^{-1}(cos\theta)\cdot (sec\theta \cdot tan\theta)d\theta.$

$\int \theta (sec\theta \cdot tan\theta)d\theta \cdot \theta \int (sec\theta \cdot tan\theta)d\theta -\int (\int(sec\theta \cdot tan\theta)d\theta)d\theta$

$\int sec\theta tan\theta\cdot d\theta=sec\theta+c$

$=\theta sec\theta-\int sec\theta d\theta$

$=\theta sec\theta+(log \mid sec\theta+tan\theta \mid)+1.$

$=x sec^{-1}x.-log \mid x+\sqrt {x^2-1}\mid +c$

$=x sec^{-1}x-cos h^{-1}x+c.$

$\int cos^{-1}(\dfrac {1}{x})=x sec^{-1}{x}-cos h^{-1}x+c.$

Evaluate:

$\displaystyle \int\log_{10}xdx$

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$(x-1)\log_{e}x+c$
0%
$\displaystyle \log_{e}10.x\log_{e}(\frac{x}{e})+c$
0%
$\displaystyle \log_{10}e.x\log_{e}(\frac{x}{e})+c$
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$\cfrac{1}{x}+c$

Explanation

We know that,

$\int { uvdx } =u'\int { vdx } -\int { u'(\int { vdx } )dx } +C$ (Integration by parts rule)

$\therefore$ we can write,

$I=\int { \log _{ 10 }{ x } dx } =\log _{ 10 }{ e } \int { \log _{ e }{ x } \cdot (1)dx }$ ...as

$\log _{ 10 }{ x } =\cfrac { \log _{ e }{ x } }{ \log _{ e }{ 10 } } =\log _{ 10 }{ e } \cdot \log _{ e }{ x }$

Applying integration by parts;

$u=\log _{ e }{ x } v=1\\ u'=\cfrac { 1 }{ x } \int { vdx } =x\\$

$\therefore I=\log _{ 10 }{ e } ((\log _{ e }{ x } )x-\int { \cfrac { 1 }{ x } \cdot xdx } )+C=\log _{ 10 }{ e } (x\log _{ e }{ x } -\int { dx } )+C\\ I=\log _{ 10 }{ e } \cdot x(\log _{ e }{ x } -1)+C=\log _{ 10 }{ e } \cdot x(\log _{ e }{ x } -\log _{ e }{ e } )+C\\ \therefore I=\log _{ 10 }{ e } \cdot x(\log _{ e }{ \cfrac { x }{ e } } )+C$

Evaluate

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

0%
$\displaystyle e^{x}[\dfrac{1}{x+1}]+c$
0%
$\displaystyle e^{x}[\dfrac{-1}{x+1}]+c$
0%
$e^{x}x+c$
0%
$\displaystyle e^{x}[\dfrac{-1}{(x+1)^{2}}]+c$

Explanation

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

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

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

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

We know,

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

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

Now by integrating the functions by parts, for the first part taking

$e^x$ as

$u$ and

$\dfrac{1}{x+1}$ as

$v$

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

$\displaystyle I=\frac{e^{x}}{x+1}+C$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 2 - 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 2 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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