If $$\displaystyle I_{n}=\int (\ln x)^{n} dx,$$ then $$I_{n}+nI_{n-1}=$$

$$\displaystyle \frac{\left ( \ln x \right )^{n}}{x}+C$$

MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 11

Evaluate :

$\displaystyle \int \frac { cot x}{( cosec x - cot x)} dx$

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$-cosec x -cot x- x +C$
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$cosec x - cot x -x +C$
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$-cosec x +cot x -x + C$
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$-sin x + x+ C$

Let

$\displaystyle\dfrac{x^{1/2}}{\sqrt{1-x^{3}}}dx=\dfrac{2}{3}gof(x)+c$ then

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$f(x)=\sqrt{x}$
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$f(x)=x^{3/2}$
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$f(x)=x^{2/3}$
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$g(x)=\sin^{-1}x$

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

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

Evaluate :

$\displaystyle \int \dfrac {( cos 2x - cos 2 \alpha)}{( cos x- cos \alpha)} dx$

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$2 sin x +x cos \alpha + C$
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$2 sin x - 2x cos \alpha + C$
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$-2 sin x +2x cos \alpha + C$
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$-2 sin x -2x cos \alpha + C$

Evaluate :

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

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$tan x + sec x+ C$
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$tan x - sec x + C$
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$-tan x + sec x + C$
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$-tan x -sec x + C$

The value of

$\int_{1}^{e} \dfrac{1+x^{2} \ln x}{x+x^{2} \ln x} d x$ is

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$e$
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$\ln (1+e)$
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$e+\ln (1+e)$
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$e-\ln (1+e)$

Explanation

$\int_{1}^{e} \dfrac{1+x^{2} \ln x}{x+x^{2} \ln x} d x$

$=\int_{1}^{e} \dfrac{1+x-x+x^{2} \ln x}{x(1+x \ln x)} d x$

Splitting them we get,

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

$=[\ln x+x]_{1}^{e}-[\ln (1+x \ln x)]_{1}^{e} \\$

$=e-\ln (1+e)$

$\int_{1}^{2} e^{x^{2}} d x=a,$ then

$\int_{e}^{e^{4}} \sqrt{\ln x} d x$ is equal to

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$2 e^{4}-2 e-a$
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$2 e^{4}-e-a$
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$2 e^{4}-e-2 a$
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$e^{4}-e-a$

Explanation

$I_{1}=\int_{e}^{e^{4}} \sqrt{\ln x} d x,$ putting

$t=\sqrt{\ln x},$ i.e.,

$d t=\dfrac{d x}{2 x \sqrt{\ln x}}\\$

$\Rightarrow d x=2 t e^{t^{2}} d t \\$

$\Rightarrow \int_{e}^{e^{4} \sqrt{\ln x} d x} \\$

$=\int_{1}^{2} 2 t^{2} e^{t^{2}} d t \\$

using

$uv$ integration formula , taking

$2t^2$ as first function and

$e^{t^2}$ as second function, we get

$=\left.t \cdot e^{t^{2}}\right|_{1} ^{2}-\int_{1}^{2} e^{t^{2}} d t=2 e^{4}-e-a$

The value of

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

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$e^{x}\sqrt{\dfrac{1+x}{1-x}}+c$
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$e^{x}\sqrt{1+x}+c$
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$e^{x}\sqrt{1-x}+c$
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$e^{x}\sqrt{\dfrac{1-x}{1+x}}+c$

Explanation

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

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

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

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

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

The second term is the derivative of the first term.
So,

$=\displaystyle e^{x}\sqrt{\frac{1+x}{1-x}}$

Evaluate

$\displaystyle \int 3^{x}\cos4xdx=$

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$\displaystyle \dfrac{3^{x}}{(\log 3)^{2}+16}[(\log 3)\cos4x+4\sin4x]+c$
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$\displaystyle \dfrac{3^{x}}{9+(\log 4)^{2}}[\cos4x+(\log 3)\cdot \sin4x]+c$
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$\displaystyle \dfrac{3^{x}}{(\log 3)^{2}+(\log 4)^{2}}[(\log 3)\sin4x+(\log 4)\cos3x]+c$
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$\displaystyle \dfrac{3^{x}}{(\log 3)^{2}+16}[(\log 3)\sin4x+(\log 4)\cos3x]+c$

Explanation

$L=\int 3^x \cos 4x dx.$ ......

$(i)$

$=\displaystyle \cos 4x\int 3^x dx + \int 4 \sin 4x\dfrac {3^x}{\left(\log 3\right)}dx.$

$\displaystyle =\dfrac {3^{x}\cos 4x}{\log \left(3\right)}+\dfrac {4}{\log 3}\int 3^x \sin 4x\ dx$ ......

$(ii)$

Now,

$\displaystyle \int 3^x \sin 4x dx=\dfrac {\sin 4x 3^x}{\log 3}-\int 4 \cos 4x\cdot \dfrac {3^x}{\log 3}$

$\displaystyle =\dfrac {3^x \sin 4x}{\log 3}-\dfrac {4}{\log 3}\int 3^x \cos 4x dx$

$\displaystyle =\dfrac {3^x \sin 4x}{\log 3}-\dfrac {4}{\log 3}L$ ..... From

$(i)$

$\therefore \displaystyle L=\dfrac {3^x \cos 4x}{\log 3}+\dfrac {4}{\log 3}\left[\dfrac {3^x \sin 4x}{\log 3}-\dfrac {4}{\log 3}\times L\right]$ ..... From

$(ii)$

$\displaystyle L\left(\dfrac {\left(\log 3\right)^2+4^2}{\left(\log 3\right)^2}\right)=\dfrac {3^x}{\left(\log 3\right)^2}[\cos 4x {\left(\log 3\right)}+4\sin 4x]$

$\displaystyle L=\dfrac {3^{x}}{\left(\log 3\right)^2+4^2}[\left(\log 3\right)\cos 4x + 4\sin 4x]+c$

$\displaystyle \int\frac{\log x-2008}{(\log x+1)^{2010}}dx=$

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$\displaystyle \frac{1}{(\log x+1)^{2008}}+c$
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$\displaystyle \frac{x}{(\log x+1)^{2010}}+c$
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$\displaystyle \frac{1}{(\log x+1)^{2009}}+c$
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$\displaystyle \frac{x}{(\log x+1)^{2009}}+c$

Explanation

$\int [\dfrac{(logx+1)-2009}{(log x+1)^{2010}}]dx$

$\int [\dfrac{1}{(log x+1)^{2009}}-\dfrac{2009}{(log x+1)^{2010}}]dx --(1)$

$\int \dfrac{1}{(log x+1)^{2009}}dx- \int \dfrac{2009}{(log x+1)^{2010}}dx$

$\int \dfrac{1}{(log x+1)}^{2009}1 dx=\dfrac{x}{(log x+1)^{2009}}-\int (\dfrac{-2009}{(log x+1)^{2010}}-\dfrac{1}{x}\int 1.dx)dx$

$=\dfrac{x}{(log x+1)^{2009}}+\int \dfrac{2009}{(log x+1)^{2010}}dx --(2)$

putting (2) in (1)

$\dfrac{x}{(log x+1)^{2009}}+\int \dfrac{2009}{(log x+1)^{2010}}dx-\int \dfrac{2009}{(log x+1)^{2010}}dx+c$

$=\dfrac{x}{(log x+1)^{2009}}+c$

Let f be a function defined for every x, such that f'' = -f ,f(0)=0, f' (0) = 1, then f(x) is equal to

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tanx
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$e^{x}-1$
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sinx
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2sinx

Explanation

$\int f^{''}(x)f^{'}(x)=-\int f(x)f^{'}(x)$

$(f^{'}(x))^{2}=f^{2}(x)+c$

$c=1$

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

$\int \frac{df(x)}{\sqrt{1-f^{2}(x)}}=\int dx$

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

$f(x)=sin x$

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

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$\displaystyle \frac{4}{\pi}[x\sin^{-1}x+\sqrt{1-x^{2}}]-x+c$
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$\log[\sin^{-1}x+\cos^{-1}x]+c$
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$\dfrac{4}{\pi}[x\sin^{-1}x+\sqrt{1-x^{2}}]+c$
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$\dfrac{4}{\pi}[x\sin^{-1}x-\sqrt{1-x^{2}}]+C$

Explanation

$sin^{-1}(x)=cos^{-1}(\sqrt{1-x^{2}})$

$x=sin\theta$

$\displaystyle = \int \dfrac{\theta -cos^{-1}[cos\pi /2-\theta ]}{\theta +cos^{-1}[cos\pi /2-\theta ]}. d\theta$

$=\int \dfrac{\theta -(\pi /2-\theta )}{\theta +(\pi /2-\theta )}=cos\theta (d\theta)$

$=\int [\dfrac{2\theta -\pi /2}{\pi /2}]cos\theta (d\theta)$

$\dfrac{4}{\pi }\int \theta cos\theta (d\theta) -\int cos\theta (d\theta)$

$\displaystyle = \dfrac{4}{\pi }[\theta sin\theta +cos\theta ]-sin\theta +c$

$\theta =sin^{-1}x$

$\dfrac{4}{\pi }[sin^{-1}x.x+\sqrt{1-x}^{2}]-x+c$

$=\dfrac{4}{\pi} [xsin^{-1}x+\sqrt{1-x^{2}]}-x+c$

$\int \dfrac{sin^{-1}x-cos^{-1}x}{sin^{-1}x+cos^{1}x}dx=\dfrac{4}{\pi }[xsin^{-1}x+\sqrt{1-x^{2}}]-x+c$

$\displaystyle\int xe^{-5x^{2}}\sin4x^{2}dx=e^{-5x^{2}} (A \sin4x^{2}+B\cos4x^{2})+C$ , then

$A+B$

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$-\dfrac{1}{66}$
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$\dfrac{1}{66}$
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$-\dfrac{9}{66}$
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$-\dfrac{7}{66}$

Explanation

Given,

$\displaystyle \int xe^{-5x^{2}}\sin4x^{2}dx=Ke^{-5x^{2}} (A \sin4x^{2}+B\cos4x^{2})+C$ .
Let

$\displaystyle I=\int xe^{ -5x^2}\sin 4x^{ 2 }dx$ .
Put

$x^{ 2 }=t$ .

$\Rightarrow xdx=\dfrac { dt }{ 2 }$

$\Rightarrow I=\displaystyle \frac { 1 }{ 2 } \int e^{ -5t }\sin {4t}dt$
Using Integration by parts, we get

$=\displaystyle \frac { 1 }{ 2 } \left[\sin { 4t } \frac { e^{ -5t } }{ -5 } -\int { 4\cos { 4t } } \frac { e^{ -5t } }{ -5 } dt \right]$

$\displaystyle =-\frac { 1 }{ 10 } e^{ -5t }\sin { 4t } +\frac { 4 }{ 10 } \int e^{ -5t }\cos { 4t } dt$

$\displaystyle =-\frac { 1 }{ 10 } e^{ -5t }\sin { 4t } +\frac { 4 }{ 10 } \left [\cos { 4t } \frac { e^{ -5t } }{ -5 } -\int (-4)\sin { 4t } \frac { e^{ -5t } }{ -5 }.dt \right]$

$\displaystyle \Rightarrow I=-\frac { 1 }{ 10 } e^{ -5t }\sin { 4t } -\frac { 2 }{ 25 } \cos { 4t }. e^{ -5t }-\frac { 8 }{ 25 } I+C$

$\displaystyle \Rightarrow \frac { 33 }{ 25 } I=-\frac { 1 }{ 10 } e^{ -5t }\sin { 4t } -\frac { 2 }{ 25 } \cos { 4t }. e^{ -5t }+C$

$\displaystyle I=\frac { -5 }{ 66 } e^{ -5x^{ 2 } }\sin { 4{ x }^{ 2 } } -\frac { 2 }{ 33 } e^{ -5x^{ 2 } }\cos { 4{ x }^{ 2 } }+C$

$\displaystyle I=e^{ -5{ x }^{ 2 } }\left[-\dfrac{5}{66}\sin { 4{ x }^{ 2 } } -\dfrac{2}{33}\cos { 4{ x }^{ 2 } } \right]+C$

On comparing with given expression,

$\displaystyle A=-\dfrac{5}{66}, B=-\dfrac{2}{33}$

$\therefore A+B = -\dfrac{9}{66}$

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

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$\displaystyle \frac{e^{x}\cos x}{1+\sin x}+c$
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$c-\displaystyle \frac{e^{x}\sin x}{1+\sin x}$
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$c-\displaystyle \frac{e^{x}}{1+\sin x}$
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$c-\displaystyle \frac{e^{x}cosx}{1+\sin x}$

Explanation

Let

$u$ be

$\dfrac{cosx-1}{sinx+1}$
and

$dv$ be

$e^{x}$ then

$v=e^{x}$
we know that

$\int udv=uv-\int vdu$

$\Rightarrow \int (\dfrac{cosx-1}{sinx+1})e^{x}$ =

$(\dfrac{cosx-1}{sinx+1})e^{x}-\int e^{x}\dfrac{(cosx-1)(cosx)-(sinx+1)(-sinx)}{(sinx+1)^{2}}$ \
we have to do the by parts again taking

$u=\dfrac{(cosx-1)(cosx)-(sinx+1)(-sinx)}{(sinx+1)^{2}}$ and

$v=e^{x}$

$\Rightarrow \int (\dfrac{1-cosx+sinx}{(1+sinx)^{2}})e^{x}\\=(\dfrac{1-cosx+sinx}{(1+sinx)^{2}})e^{x}-\int e^{x}(\dfrac{(1-cosx+sinx)2(1+sinx)(cosx)-(1+sinx)^{2}(sinx+cosx)}{(1+sinx)^{4}})$

$\Rightarrow \int (\dfrac{1-cosx+sinx}{(1+sinx)^{2}})e^{x}=-(\dfrac{e^{x}}{1+sinx})$ on back substitution we get

$\Rightarrow \int (\dfrac{cosx-1}{sinx+1})e^{x}$ =

$(\dfrac{cosx-1}{sinx+1})e^{x}-(-(\dfrac{e^{x}}{1+sinx}))$

$\Rightarrow \int (\dfrac{cosx-1}{sinx+1})e^{x}$ =

$\dfrac{e^{x}(cosx)}{1+sinx}$
Hence, option, 'A' is correct.

$\displaystyle \int(\frac{x+2}{x+4})^{2}e^{x}dx$ is equal to

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$e^{x}(\displaystyle \frac{x}{x+4})+c$
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$e^{x}(\displaystyle \frac{x+2}{x+4})+c$
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$e^{x}(\displaystyle \frac{x-2}{x+4})+c$
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$(\displaystyle \frac{2xe^{x}}{x+4})+c$

Explanation

$\displaystyle \int (\frac { x+2 }{ x+4 } )^{ 2 }e^{ x }dx$

$\displaystyle =\int (1-\frac { 2 }{ x+4 } )^{ 2 }e^{ x }dx$

$\displaystyle =\int (1+\frac { 4 }{ (x+4)^{ 2 } } -\frac { 4 }{ x+4 } )e^{ x }dx$

$\displaystyle =\int (1+\frac { 4 }{ (x+4)^{ 2 } } )e^{ x }dx-\int { \frac { 4 }{ x+4 } e^{ x }dx }$

$\displaystyle =e^{ x }+4[e^{ x }(\frac { -1 }{ x+4} )-\int { e^{ x } } \frac { -1 }{ x+4 } dx]-\int { \frac { 4 }{ x+4 } e^{ x }dx }$

$\displaystyle =(\frac { x }{ x+4 } )e^{ x }+C$

The value of

$\displaystyle \int(3x^{2}\tan\frac{1}{x}-x\sec^{2}\frac{1}{x})dx {\it}$ is

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

Explanation

$\displaystyle \int( 3x^{2} \tan\cfrac{1}{x}- x \sec^{2}\cfrac{1}{x}) dx$

$\displaystyle \int (3x^{2} \tan \frac{1}{x}) dx- \int x \sec^{2}\frac{1}{x} dx$

$\displaystyle \tan \frac{1}{x} \int 3x^{2}dx- \int \sec^{2}\frac{1}{x}(\frac{-1}{x^2})x^{3} dx$

$\displaystyle \int x \sec^{2}\frac{1}{x} dx$

$=\displaystyle \tan \frac{1}{x} x^{3} + \int sec^{2} \frac{1}{x} -x dx$

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

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

$\displaystyle \int (3x^{2} \tan \frac{1}{x}-x sec^{2}\frac{1}{x})dx=x^{3}(\tan \frac{1}{x})+c$

$\displaystyle \int e^{2x} (\sqrt{3}cosx-sinx)dx=$

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$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)cosx -(\sqrt{3}-2)sinx]+c$
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$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)sinx+(\sqrt{3}-2)\cos x]+c$
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$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)sinx-(\sqrt{3}-2)cosx]+c$
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$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)cosx+(\sqrt{3}-2)sin x]+c$

Explanation

$=\displaystyle \int e^{2x}(\sqrt{3}cosx-sinx)dx$

$=\displaystyle \int e^{2x}\sqrt{3}cosx.dx- \int e^{2x} sinx.dx$

$\displaystyle \int e^{2x}\sqrt{3}cosx.dx=e^{2x}\sqrt{3}sinx-\int 2e^{2x}\sqrt{3}sinx$

$\displaystyle \int e^{2x}\sqrt{3}cosx.dx=e^{2x}\sqrt{3}sinx+2\sqrt{3}e^{2x}cosx-\int 4e^{2x}\sqrt{3}cosx$

$\displaystyle5 \int e^{2x}\sqrt{3}cosx.dx=e^{2x}\sqrt{3}sinx+2\sqrt{3}e^{2x}cosx$
Similarly,

$\displaystyle \int e^{2x}sinx.dx=-e^{2x}cosx+\int 2e^{2x}cosx.dx$

$=\displaystyle-e^{2x}cosx+2e^{2x}sinx-\int 4e^{2x}sinxdx +c$

$\displaystyle 5\int e^{2x} sinx. dx= -e^{2x}cosx+2e^{2x}sinx$

So,

$\displaystyle \int e^{2x}(\sqrt{3}cosx-sinx)dx$

$=\displaystyle \frac{1}{5}(e^{2x}\sqrt{3}sinx+2\sqrt{3}e^{2x}cosx)-\frac{1}{5}(-e^{2x}cosx+2e^{2x}sinx)+c$

$=\displaystyle\frac{1}{5}e^{2x}(\sqrt{3}sinx+2\sqrt{3}cosx+cosx-2sinx)+c$

$=\displaystyle\frac{1}{5}e^{2x}((2\sqrt{3}+1)cosx+(\sqrt{3}-2)sinx)+c$
Hence, option 'D' is correct.

$\displaystyle \int\frac{e^{4x}-1}{e^{2x}}\log(\frac{e^{2x}+1}{e^{2x}-1})ck=\frac{t^{2}}{2}\log t-\frac{t^{2}}{4}-\frac{u^{2}}{2} \log u + \frac{u^{2}}{4}+c$ then

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$\displaystyle t=e^{-x}-e^{x},u=e^{x}+e^{-x}$
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$\displaystyle t=e^{x}-e^{-x},u=e^{x}+e^{-x}$
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$\displaystyle t=e^{x}+e^{-x},u=e^{x}-e^{-x}$
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$\displaystyle u=e^{-x}-e^{x},t=e^{x}+e^{-x}$

Explanation

$\displaystyle \int { \frac { { e }^{ 4x }-1 }{ { e }^{ 2x } } } \log { \left( \frac { { e }^{ 2x }+1 }{ { e }^{ 2x }-1 } \right) } dx$

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

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

$\displaystyle =\frac { { t }^{ 2 } }{ 2 } \log { t } -\frac { { t }^{ 4 } }{ 4 } -\frac { { u }^{ 2 } }{ 2 } \log { u } +\frac { { u }^{ 2 } }{ 4 } +c$
Where

$t={ e }^{ x }+{ e }^{ -x }$ and

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

$\displaystyle \int \cos x\log(\tan \frac{x}{2})dx=$

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

Explanation

$\int\ cosx\ log\ (tanx/2)dx$

$=\ log(tanx/2)\ \int\ cosx\ dx$

$-\int\ 1/2\ \dfrac{sec\ ^{2}\ x/_2}{tan\ x/2}\ \int\ cosx\ dx$

$=log\ (tan\ x/2)sinx-\ \int\ cosecx\ sinx\ dx$

$=\ log(tan\ x/2)\ sinx-\ x+c$

$=\ log(tan\ x/2)sinx-x+c=\ \int\ cosx\ log\ (tan\ x/2)^{dx}$

$=sinx log|tan\dfrac{x}{2}|-x+c$
Hence, option 'D' is correct.

$\displaystyle \int \cos^{-1}\sqrt{\frac{1-x}{2}}dx=$

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$\displaystyle \frac{\pi}{2}x-\frac{1}{2}xcos^{-1}x+c$
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$\dfrac{\pi}{2}x-\dfrac{1}{2}xcos^{-1}x+\dfrac{1}{2}\sqrt{1-x^{2}}+c$
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$\dfrac{\pi}{2}x-\dfrac{1}{2}xcos^{-1}x-\sqrt{1-x^{2}}+c$
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$\displaystyle \dfrac{\pi}{2}x+\dfrac{1}{2}xcos^{-1}x+c$

Explanation

$\displaystyle \int \cos^{-1}\sqrt{\dfrac{1-x}{2}}dx$

Put

$x=\cos {\theta}$

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

So,

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

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

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

$\displaystyle =-\int (\dfrac { \pi }{ 2 } -\dfrac { \theta }{ 2 } )\sin { \theta } d\theta$

$\displaystyle =-\dfrac { \pi }{ 2 } \int \sin { \theta } d\theta +\dfrac { 1 }{ 2 } \int \theta \sin { \theta } d\theta$

$\displaystyle =-\dfrac { \pi }{ 2 } (-\cos { \theta } )+\dfrac { 1 }{ 2 } [-\theta \cos { \theta } +\int \cos { \theta } d\theta ]$ (Integrating by Parts)

$\displaystyle =\dfrac { \pi }{ 2 } \cos { \theta } -\dfrac { 1 }{ 2 } \theta \cos { \theta } +\sin { \theta } +C$

Re-substituting the value of

$x$

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

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

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$\displaystyle \frac{1}{2}e^{x^{2}}(x^{2}-1)+c$
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$e^{x^{2}}(x^{2}-1)+c$
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$\dfrac{1}{2}e^{x^{2}}(x^{2}+1)+c$
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$e^{x^{2}}(x^{2}+1)+c$

Explanation

$\displaystyle \int x^{3}e^{x^{2}}dx$
Put

$x^{2}=t$

$xdx=\displaystyle \frac{dt}{2}$

$\displaystyle =\frac { 1 }{ 2 } \int t{ e }^{ t }dt$

$\displaystyle =\frac { 1 }{ 2 } \left[t{ e }^{ t }-\int { e } ^{ t }dt \right]$

$\displaystyle = \frac { 1 }{ 2 } t{ e }^{ t }-\frac { 1 }{ 2 } { e }^{ t }+c$

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

$\displaystyle \int\dfrac{e^{(x^{2}+4\ln x)}-x^{3}e^{x^{2}}}{x-1}dx$ is equal to

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$(\displaystyle \dfrac{e^{3\ln x}-e^{\ln x}}{2x})e^{x^{2}}+c$
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$\displaystyle \dfrac{(x-1)xe^{x^{2}}}{2}+c$
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$\displaystyle \dfrac{(x-1)xe^{x^{2}}}{2x}+c$
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$\dfrac{1}{2}(x^{2}-1)e^{x^{2}}+c$

Explanation

$\displaystyle I=\int\frac{e^{(x^{2}+4\ln x)}-x^{3}e^{x^{2}}}{x-1}dx$

$\displaystyle =\int \frac { e^{ x^{ 2 } }{ e }^{ \ln { { x }^{ 4 } } }-x^{ 3 }e^{ x^{ 2 } } }{ x-1 } dx$

$\displaystyle=\int \frac { x^{ 4 }e^{ x^{ 2 } }-x^{ 3 }e^{ x^{ 2 } } }{ x-1 } dx$

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

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

Put

${ x }^{ 2 }=t$

$\displaystyle xdx=\frac { dt }{ 2 }$

$\displaystyle I=\frac { 1 }{ 2 } \int te^{ t }dt$

$\displaystyle =\frac { 1 }{ 2 } [te^{ t }-\int e^{ t }dt]$

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

option

$A$ and

$D$ are similiar

$f(x)$ is a polynomial of nth degree then

$\displaystyle \int e^{x}f(x)dx=$

0%
$e^{x}[f(x)-f^{'}(x)+f^{''}(x)-f^{'''}(x)+\ldots\ldots+(-1)^{n}f^{n}(x)]$ Where $f^{n}(x)$ denotes nth order derivative of w.r.t. $x$
0%
$e^{x}[f(x)+f(x)+f^{'}(x)+f^{''}(x)+\ldots\ldots+(-1)^{n}J^{n}(x)]$
0%
$e^{x}[f(x)+f^{'}(x)+f^{'}(x)+f^{'''}(x)+\ldots\ldots+(-1)^{n}f^{2n}(x)]$
0%
$d[f(x)+f(x)+f^{'}(x)+f^{'''}(x)+\ldots\ldots+\{-1)^{n}f^{3n}(x)]$

Explanation

$\int e^{x}f(x)dx$

$=f(x).e^{x}-\int f'(x)e^{x}$

$=f(x).e^{x}-f'(x)e^{x}+\int f''(x)e^{x}$

$=f(x).e^{x}-f'(x)e^{x}+f''(x)e^{x}-\int f'''(x)e^{x}$

$=e^{x}[f(x)-f'(x)+f''(x).......(-1)^{n}f^{n}(x)]$

$\displaystyle \int_{0}^{\pi }\left \{ f(x) +f^{''}(x)\right \}\sin\, x\, dx=5,f(\pi )=$ Then

$f(0)$ equals

0%
$2$
0%
$3$
0%
$5$
0%
$4$

Explanation

$\displaystyle \int f''(x) \sin x dx$

$\displaystyle = \sin x f'(x) - \int \cos x f'(x) dx$

$\displaystyle = \sin x f'(x) - \left[ \cos x f(x)+ \int \sin x f(x) dx \right]$

$\displaystyle \Rightarrow \int \left[ f(x)+f''(x) \right] \sin x dx= \sin x f'(x)dx- \cos x f(x)$

$\displaystyle \Rightarrow \int_0^\pi \left[ f(x)+f''(x) \right] \sin x dx=\left[ \sin x f'(x)dx- \cos x f(x) \right]_0^\pi$

$\Rightarrow 5= f(\pi)+f(0)$

$\Rightarrow f(0)=5-2=3$

$\displaystyle \int \frac{\mathrm{cosec} ^{2}x-2005}{\cos ^{2005}x}dx$ is equal to

0%
$\displaystyle \frac{\cot x}{(\cos x)^{2005}}+c$
0%
$\displaystyle \frac{\tan x}{(\cos x)^{2005}}+c$
0%
$\displaystyle \frac{(-\tan x)}{(\cos x)^{2005}}+c$
0%
none of these

Explanation

$\displaystyle \int \frac{\csc^{2}x-2005}{\cos ^{2005}x}dx$

$=\displaystyle \int (\cos ^{-2005}x)\csc^{2}xdx-2005\int \frac{dx}{\cos ^{2005}x}$

$=\displaystyle (\cos x)^{-2005}(-\cot x)-\int (-2005)(\cos x)^{-2006}(-\sin x)(-\cot x)dx-2005\int \frac{dx}{\cos ^{2005}x}$ ...... [Using integration by parts]

$=-\displaystyle \frac{\cot x}{(\cos x)^{2005}}+C$

Evaluate:

$\displaystyle \int x\left ( \frac{\ln a^{a^{x/2}}}{3a^{5x/2}b^{3x}}+\frac{\ln b^{b^{x}}}{2a^{2x}b^{4x}} \right )$

0%
$\displaystyle \frac{1}{6\ln a^{2}b^{3}}a^{2x}b^{3x}\ln \frac{a^{2x}b^{3x}}{e}+k$
0%
$-\displaystyle \frac{1}{6\ln a^{2}b^{3}}\frac{1}{a^{2x}b^{3x}}\ln \frac{1}{ea^{2x}b^{3x}}+k$
0%
$\displaystyle \frac{1}{6\ln a^{2}b^{3}}\frac{1}{a^{2x}b^{3x}}\ln (a^{2x}b^{3x})+k$
0%
$\displaystyle \frac{1}{6\ln a^{2}b^{3}}\frac{1}{a^{2x}b^{3x}}\ln (a^{2x}b^{4x})+k$

Explanation

$I=\int { \left( \cfrac { \ln { { a }^{ { a }^{ x/2 } } } }{ 3{ a }^{ 5x/2 }{ b }^{ 3x } } +\cfrac { \ln { { b }^{ { b }^{ x } } } }{ 2{ a }^{ 2x }{ b }^{ 4x } } \right) dx } \\ I=\int { \left( \cfrac { { a }^{ x/2 }\ln { { a } } }{ 3{ a }^{ 5x/2 }{ b }^{ 3x } } +\cfrac { { b }^{ x }\ln { { b } } }{ 2{ a }^{ 2x }{ b }^{ 4x } } \right) dx } \\ I=\int { \left( \cfrac { \ln { { a } } }{ 3{ a }^{ 2x }{ b }^{ 3x } } +\cfrac { \ln { { b } } }{ 2{ a }^{ 2x }{ b }^{ 3x } } \right) dx } \\ I=\int { \left( \cfrac { 2\ln { { a } } }{ 6{ a }^{ 2x }{ b }^{ 3x } } +\cfrac { 3\ln { { b } } }{ 6{ a }^{ 2x }{ b }^{ 3x } } \right) dx } \\ I=\int { \left( \cfrac { 2\ln { { a } } +3\ln { b } }{ 6{ \left( { a }^{ 2 }{ b }^{ 3 } \right) }^{ x } } \right) dx } \\ I=\int { \cfrac { \ln { \left( { a }^{ 2 }{ b }^{ 3 } \right) } }{ 6{ \left( { a }^{ 2 }{ b }^{ 3 } \right) }^{ x } } dx } \\$

Let

$\left( { a }^{ 2 }{ b }^{ 3 } \right) =t$

$I=\int { \cfrac { \ln { \left( t \right) } }{ 6{ t }^{ x } } dx } \\ I=\cfrac { \ln { \left( t \right) } }{ 6 } \int { { t }^{ -x }dx } \\ I=\cfrac { \ln { \left( t \right) } }{ 6 } \left[ -\cfrac { { t }^{ -x } }{ \ln { \left( t \right) } } \right] +C\\ I=-\cfrac { 1 }{ 6{ \left( { a }^{ 2 }{ b }^{ 3 } \right) }^{ x } } +C$

Solve the above options to match with the final answer.

$\displaystyle I_{n}=\int (\ln x)^{n} dx,$ then

$I_{n}+nI_{n-1}=$

0%
$\displaystyle \frac{\left ( \ln x \right )^{n}}{x}+C$
0%
$x(\ln x)^{n-1}+C$
0%
$x(\ln x)^{n}+C$
0%
none of these

Explanation

$\displaystyle I_{ n }=\int (\ln { x } )^{ n }dx$

$\displaystyle=x(\ln { x } )^{ n }-\int \frac { x(n)(\ln { x } )^{ n-1 } }{ x } dx$

$\displaystyle =x(\ln { x } )^{ n }-n\: I_{ (n-1) }$

$\displaystyle\Rightarrow I_{ n }-n\: I_{ n-1 }=x(\ln { x } )^{ n }$

What is the value of a + b ?

0%
2
0%
1
0%
$\frac{3}{2}$
0%
$\frac{5}{2}$

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

0%
$\tan x-\cot x+c$
0%
$\tan x-x+1$
0%
$\tan x-x$
0%
$\tan x+x$

Explanation

$\displaystyle \int \dfrac{4dx}{4\sin ^{2}x\cos ^{2}x}$

$=\displaystyle \int \dfrac{4dx}{\sin ^{2}2x}$

$=4\displaystyle \int \text{cosec}^{2}2x dx$

$=-2\cot 2x+C$

$=\dfrac{-2(\cos 2x)}{\sin 2x}+C$

$=\dfrac{2(\sin ^{2}x-\cos ^{2}x)}{2\sin x.\cos x}+C$

$=\dfrac{\sin ^{2}x}{\sin x.\cos x}-\dfrac{\cos ^{2}x}{\sin x.\cos x}+C$

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

$\displaystyle \int {\displaystyle \frac { \sec^{ 2 }x-2010 }{ \sin^{ 2010 }x } dx=\displaystyle \frac { P(x) }{ (\sin x)^{ 2010 } } +c}$ , where

$c$ is arbitrary constant then value of

$P\left( \displaystyle \frac { \pi }{ 3 } \right)$

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

Explanation

$\displaystyle \int {\displaystyle \frac { {\sec }^{ 2 }x-2010 }{ {\sin }^{ 2010 }x } dx=\displaystyle \frac { P(x) }{ (\sin x)^{ 2010 } } +c}$

Let

$I=\displaystyle \int { \frac { \sec ^{ 2 }{ x } -2010 }{ \sin ^{ 2010 }{ x } } } dx$

$\Rightarrow I=\displaystyle \int { \sec } ^{ 2 }x.(\sin { x } )^{ -2010 }dx-2010\int { (\sin { x } )^{ -2010 }dx }$ ......(i)

Let

${ I }_{ 1 }=\displaystyle \int { \sec } ^{ 2 }x.(\sin { x } )^{ -2010 }dx$

Integrating by parts,

$\Rightarrow I_{1}=\displaystyle \frac { \tan x }{ (\sin x)^{ 2010 } } +2010\int { \displaystyle \frac { \tan x\quad \cos x }{ (\sin x)^{ 2011 } } dx } +c$

$=\displaystyle \frac { \tan x }{ (\sin x)^{ 2010 } } +2010\int { \displaystyle \frac { dx }{ (\sin x)^{ 2010 } } } +c$

Putting this value in (i), we get

$\Rightarrow I=\displaystyle \frac { \tan\quad x }{ (\sin x)^{ 2010 } } +c$

$\Rightarrow \displaystyle \frac { \tan\quad x }{ (\sin x)^{ 2010 } } +c=\displaystyle \frac { P(x) }{ (\sin x)^{ 2010 } } +c$

$P(x) = \tan x$

$\therefore \quad P(\dfrac {\pi }3)=\tan\displaystyle \frac { \pi }{ 3 } =\sqrt { 3 }$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 11 - 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 11 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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