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

Evaluate : $$\displaystyle \int \frac { cot x}{( cosec x - cot x)} dx$$
  • $$ -cosec x -cot x- x +C $$
  • $$ cosec x - cot x -x +C $$
  • $$ -cosec x +cot x -x + C $$
  • $$ -sin x + x+ C $$
Let $$\displaystyle\dfrac{x^{1/2}}{\sqrt{1-x^{3}}}dx=\dfrac{2}{3}gof(x)+c$$ then
  • $$f(x)=\sqrt{x}$$
  • $$f(x)=x^{3/2}$$
  • $$f(x)=x^{2/3}$$
  • $$g(x)=\sin^{-1}x$$
$$\displaystyle\int \dfrac{xe^{x}}{(1+x)^{2}}dx$$ is equal to
  • $$\dfrac{e^{x}}{x+1}+c$$
  • $$e^{x}(x+1)+c$$
  • $$-\dfrac{e^{x}}{(x+1)}+c$$
  • $$\dfrac{e^{x}}{1+x^{2}}+c$$
Evaluate : $$\displaystyle \int \dfrac {( cos 2x - cos 2 \alpha)}{( cos x- cos \alpha)} dx$$
  • $$2 sin x +x cos \alpha + C $$
  • $$ 2 sin x - 2x cos \alpha + C $$
  • $$ -2 sin x +2x cos \alpha + C $$
  • $$ -2 sin x -2x cos \alpha + C $$
Evaluate : $$\displaystyle \int \frac {(1- sin x) }{cos^2 x} dx  $$
  • $$ tan x + sec x+ C $$
  • $$ tan x - sec x + C $$
  • $$ -tan x + sec x + C $$
  • $$ -tan x -sec x + C $$
The value of $$\int_{1}^{e} \dfrac{1+x^{2} \ln x}{x+x^{2} \ln x} d x$$ is
  • $$e$$
  • $$\ln (1+e)$$
  • $$e+\ln (1+e)$$
  • $$e-\ln (1+e)$$
If $$\int_{1}^{2} e^{x^{2}} d x=a,$$ then $$\int_{e}^{e^{4}} \sqrt{\ln x} d x$$ is equal to
  • $$2 e^{4}-2 e-a$$
  • $$2 e^{4}-e-a$$
  • $$2 e^{4}-e-2 a$$
  • $$e^{4}-e-a$$
The value of $$\displaystyle \int\frac{e^{x}(2-x^{2})}{(1-x)\sqrt{1-x^{2}}}dx$$ is equal to
  • $$e^{x}\sqrt{\dfrac{1+x}{1-x}}+c$$
  • $$e^{x}\sqrt{1+x}+c$$
  • $$e^{x}\sqrt{1-x}+c$$
  • $$e^{x}\sqrt{\dfrac{1-x}{1+x}}+c$$
Evaluate $$\displaystyle \int 3^{x}\cos4xdx=$$
  • $$\displaystyle \dfrac{3^{x}}{(\log 3)^{2}+16}[(\log 3)\cos4x+4\sin4x]+c$$
  • $$\displaystyle \dfrac{3^{x}}{9+(\log 4)^{2}}[\cos4x+(\log 3)\cdot \sin4x]+c$$
  • $$\displaystyle \dfrac{3^{x}}{(\log 3)^{2}+(\log 4)^{2}}[(\log 3)\sin4x+(\log 4)\cos3x]+c$$
  • $$\displaystyle \dfrac{3^{x}}{(\log 3)^{2}+16}[(\log 3)\sin4x+(\log 4)\cos3x]+c$$
$$\displaystyle \int\frac{\log x-2008}{(\log x+1)^{2010}}dx=$$
  • $$\displaystyle \frac{1}{(\log x+1)^{2008}}+c$$
  • $$\displaystyle \frac{x}{(\log x+1)^{2010}}+c$$
  • $$\displaystyle \frac{1}{(\log x+1)^{2009}}+c$$
  • $$\displaystyle \frac{x}{(\log x+1)^{2009}}+c$$
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
  • tanx
  • $$e^{x}-1$$
  • sinx
  • 2sinx
$$\displaystyle \int\frac{\sin^{-1}x-\cos^{-1}x}{\sin^{-1}x+\cos^{-1}x}dx=$$
  • $$\displaystyle \frac{4}{\pi}[x\sin^{-1}x+\sqrt{1-x^{2}}]-x+c$$
  • $$\log[\sin^{-1}x+\cos^{-1}x]+c$$
  • $$\dfrac{4}{\pi}[x\sin^{-1}x+\sqrt{1-x^{2}}]+c$$
  • $$\dfrac{4}{\pi}[x\sin^{-1}x-\sqrt{1-x^{2}}]+C$$
If $$\displaystyle\int xe^{-5x^{2}}\sin4x^{2}dx=e^{-5x^{2}} (A \sin4x^{2}+B\cos4x^{2})+C$$, then $$A+B$$
  • $$-\dfrac{1}{66}$$
  • $$\dfrac{1}{66}$$
  • $$-\dfrac{9}{66}$$
  • $$-\dfrac{7}{66}$$
$$\displaystyle \int\frac{\cos x-1}{\sin x+1}e^{x}dx$$ is equal to
  • $$\displaystyle \frac{e^{x}\cos x}{1+\sin x}+c$$
  • $$c-\displaystyle \frac{e^{x}\sin x}{1+\sin x}$$
  • $$c-\displaystyle \frac{e^{x}}{1+\sin x}$$
  • $$c-\displaystyle \frac{e^{x}cosx}{1+\sin x}$$
$$\displaystyle \int(\frac{x+2}{x+4})^{2}e^{x}dx$$ is equal to
  • $$e^{x}(\displaystyle \frac{x}{x+4})+c$$
  • $$e^{x}(\displaystyle \frac{x+2}{x+4})+c$$
  • $$e^{x}(\displaystyle \frac{x-2}{x+4})+c$$
  • $$(\displaystyle \frac{2xe^{x}}{x+4})+c$$
The value of $$\displaystyle \int(3x^{2}\tan\frac{1}{x}-x\sec^{2}\frac{1}{x})dx {\it}$$ is
  • $$ x^{3}\displaystyle \tan\frac{1}{x}+c$$
  • $$ x^{2}\displaystyle \tan\frac{1}{x}+c$$
  • $$ x\displaystyle \tan\frac{1}{x}+c$$
  • $$ \displaystyle \tan\frac{1}{x}+c$$
$$\displaystyle \int e^{2x} (\sqrt{3}cosx-sinx)dx=$$
  • $$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)cosx -(\sqrt{3}-2)sinx]+c$$
  • $$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)sinx+(\sqrt{3}-2)\cos x]+c$$
  • $$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)sinx-(\sqrt{3}-2)cosx]+c$$
  • $$\displaystyle \frac{e^{2x}}{5}[(2\sqrt{3}+1)cosx+(\sqrt{3}-2)sin x]+c$$
If $$\displaystyle \int\frac{e^{4x}-1}{e^{2x}}\log(\frac{e^{2x}+1}{e^{2x}-1})ck=\frac{t^{2}}{2}\log t-\frac{t^{2}}{4}-\frac{u^{2}}{2} \log u + \frac{u^{2}}{4}+c$$ then
  • $$\displaystyle t=e^{-x}-e^{x},u=e^{x}+e^{-x}$$
  • $$\displaystyle t=e^{x}-e^{-x},u=e^{x}+e^{-x}$$
  • $$\displaystyle t=e^{x}+e^{-x},u=e^{x}-e^{-x}$$
  • $$\displaystyle u=e^{-x}-e^{x},t=e^{x}+e^{-x}$$
$$\displaystyle \int \cos x\log(\tan \frac{x}{2})dx=$$
  • $$\displaystyle \sin x \log|\tan x|-x+c$$
  • $$\displaystyle -\sin x \log|\tan \frac{x}{2}|+x+c$$
  • $$\displaystyle -\sin x \log|\tan \frac{x}{2}|-x+c$$
  • $$\displaystyle \sin x \log|\tan \frac{x}{2}|-x+c$$
$$\displaystyle \int \cos^{-1}\sqrt{\frac{1-x}{2}}dx=$$
  • $$\displaystyle \frac{\pi}{2}x-\frac{1}{2}xcos^{-1}x+c$$
  • $$\dfrac{\pi}{2}x-\dfrac{1}{2}xcos^{-1}x+\dfrac{1}{2}\sqrt{1-x^{2}}+c$$
  • $$\dfrac{\pi}{2}x-\dfrac{1}{2}xcos^{-1}x-\sqrt{1-x^{2}}+c$$
  • $$\displaystyle \dfrac{\pi}{2}x+\dfrac{1}{2}xcos^{-1}x+c$$
$$\displaystyle \int x^{3}e^{x^{2}}dx=$$
  • $$\displaystyle \frac{1}{2}e^{x^{2}}(x^{2}-1)+c$$
  • $$e^{x^{2}}(x^{2}-1)+c$$
  • $$\dfrac{1}{2}e^{x^{2}}(x^{2}+1)+c$$
  • $$e^{x^{2}}(x^{2}+1)+c$$
$$\displaystyle \int\dfrac{e^{(x^{2}+4\ln x)}-x^{3}e^{x^{2}}}{x-1}dx$$ is equal to
  • $$(\displaystyle \dfrac{e^{3\ln x}-e^{\ln x}}{2x})e^{x^{2}}+c$$
  • $$\displaystyle \dfrac{(x-1)xe^{x^{2}}}{2}+c$$
  • $$\displaystyle \dfrac{(x-1)xe^{x^{2}}}{2x}+c$$
  • $$\dfrac{1}{2}(x^{2}-1)e^{x^{2}}+c$$
lf $$f(x)$$ is a polynomial of nth degree then $$\displaystyle \int e^{x}f(x)dx=$$
  • $$e^{x}[f(x)-f^{'}(x)+f^{''}(x)-f^{'''}(x)+\ldots\ldots+(-1)^{n}f^{n}(x)]$$ Where $$f^{n}(x)$$ denotes nth order derivative of w.r.t. $$x$$
  • $$e^{x}[f(x)+f(x)+f^{'}(x)+f^{''}(x)+\ldots\ldots+(-1)^{n}J^{n}(x)]$$
  • $$e^{x}[f(x)+f^{'}(x)+f^{'}(x)+f^{'''}(x)+\ldots\ldots+(-1)^{n}f^{2n}(x)]$$
  • $$d[f(x)+f(x)+f^{'}(x)+f^{'''}(x)+\ldots\ldots+\{-1)^{n}f^{3n}(x)]$$
$$ \displaystyle \int_{0}^{\pi }\left \{ f(x) +f^{''}(x)\right \}\sin\, x\, dx=5,f(\pi )=$$ Then $$f(0)$$ equals 
  • $$2$$
  • $$3$$
  • $$5$$
  • $$4$$
$$\displaystyle \int \frac{\mathrm{cosec} ^{2}x-2005}{\cos ^{2005}x}dx$$ is equal to
  • $$\displaystyle \frac{\cot x}{(\cos x)^{2005}}+c$$
  • $$\displaystyle \frac{\tan x}{(\cos x)^{2005}}+c$$
  • $$\displaystyle \frac{(-\tan x)}{(\cos x)^{2005}}+c$$
  • none of these
Evaluate: $$\displaystyle \int x\left ( \frac{\ln a^{a^{x/2}}}{3a^{5x/2}b^{3x}}+\frac{\ln b^{b^{x}}}{2a^{2x}b^{4x}} \right )$$
  • $$\displaystyle \frac{1}{6\ln a^{2}b^{3}}a^{2x}b^{3x}\ln \frac{a^{2x}b^{3x}}{e}+k$$
  • $$-\displaystyle \frac{1}{6\ln a^{2}b^{3}}\frac{1}{a^{2x}b^{3x}}\ln \frac{1}{ea^{2x}b^{3x}}+k$$
  • $$\displaystyle \frac{1}{6\ln a^{2}b^{3}}\frac{1}{a^{2x}b^{3x}}\ln (a^{2x}b^{3x})+k$$
  • $$\displaystyle \frac{1}{6\ln a^{2}b^{3}}\frac{1}{a^{2x}b^{3x}}\ln (a^{2x}b^{4x})+k$$
If $$\displaystyle I_{n}=\int (\ln x)^{n} dx,$$ then $$I_{n}+nI_{n-1}=$$
  • $$\displaystyle \frac{\left ( \ln x \right )^{n}}{x}+C$$
  • $$x(\ln x)^{n-1}+C$$
  • $$x(\ln x)^{n}+C$$
  • none of these
What is the value of a + b ?
  • 2
  • 1
  • $$\frac{3}{2}$$
  • $$\frac{5}{2}$$
$$\displaystyle \int {\dfrac{dx}{\sin^2x \cos^2x}}$$
  • $$\tan x-\cot x+c$$
  • $$\tan x-x+1$$
  • $$\tan x-x$$
  • $$\tan x+x$$
If $$\displaystyle \int {\displaystyle  \frac { \sec^{ 2 }x-2010 }{ \sin^{ 2010 }x } dx=\displaystyle \frac { P(x) }{ (\sin x)^{ 2010 } } +c}$$ , where $$c$$ is arbitrary constant then value of $$P\left( \displaystyle \frac { \pi  }{ 3 }  \right) $$
  • $$0$$
  • $$\displaystyle \frac { 1 }{ \sqrt { 3 } } $$
  • $$\sqrt { 3 } $$
  • $$\displaystyle \frac { 3\sqrt { 3 } }{ 2 } $$
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