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

$$\int {{{(\sin x)}^{99}}{{(\cos x)}^{ - 101}}dx = \_\_\_\_\_\_\_ + C.} $$
  • $$\dfrac{{{{(\tan x)}^{100}}}}{{100}}$$
  • $$\dfrac{{{{(\tan x)}^2}}}{2}$$
  • $$\dfrac{{{{(\tan x)}^{98}}}}{{98}}$$
  • $$\dfrac{{{{(\tan x)}^{97}}}}{{97}}$$
Evaluate $$\int {{x^2}\log x\,dx\, }$$.
  • $$\cfrac{{{x^2}}}{2}\log x - \cfrac{1}{{9}}{x^2}+c$$
  • $$\cfrac{{{x^3}}}{3}\log x - \cfrac{1}{{9}}{x^2}+c$$
  • $$\cfrac{{{x^3}}}{3}\log x - \cfrac{1}{9}{x^3}+c$$
  • $$\cfrac{{{x^3}}}{3}\log x + \cfrac{1}{9}{x^3}+c$$
The value of $$\int { { e }^{ x } } .\dfrac { { x }^{ 2 }+1 }{ { \left( x+1 \right)  }^{ 2 } } dx$$ is 
  • $${ e }^{ x }\left( \dfrac { x-1 }{ x+1 } \right) +C$$
  • $${ e }^{ x }\left( \dfrac { x+1 }{ x-1 } \right) +C$$
  • $${ e }^{ x }.x+C$$
  • None of these
$$\int {{e^x}(\log \sin x + \cot x)\,dx = } \_\_\_\_\_\_ + C.$$
  • $${e^x}\cot x$$
  • $${e^x}\log \sin x$$
  • $${e^x}\tan x$$
  • None of these
$$\int {\sqrt {1 - \cos x} \,dx = \_\_\_\_\_\_\_ + C;\,2\pi < x < 3\pi } $$
  • $$ - 2\sqrt 2 \cos \dfrac{x}{2}$$
  • $$ - \sqrt 2 \cos \dfrac{x}{2}$$
  • $$ 2\sqrt 2 \cos \dfrac{x}{2}$$
  • $$\dfrac{{ - 1}}{2}\sqrt 2 \cos \dfrac{x}{2}$$
The integral  $$\int \cos \left( \log _ { e } x \right) d x$$  is equal to :
(where  $$C$$  is a constant of integration)
  • $$\dfrac { { x } }{ 2 } \left[ \sin \left( \log _{ { { e } } }{ x } \right) -\cos \left( \log _{ { { e } } }{ x } \right) \right] +{ C }$$
  • $$\dfrac { { x } }{ 2 } \left[ \cos \left( \log _{ { { e } } }{ x } \right) +\sin \left( \log _{ { { e } } }{ x } \right) \right] +{ C }$$
  • $${ x }\left[ \cos \left( \log _{ { { e } } }{ x } \right) +\sin \left( \log _{ { { e } } }{ x } \right) \right] +{ C }$$
  • $$x \left[ \cos \left( \log _ { e } x \right) - \sin \left( \log _ { c } x \right) \right] + C$$
Evaluate: $$\displaystyle\int { \dfrac { { x }^{ 4 }+1 }{ 1+{ x }^{ 6 } }  } dx$$
  • $${ tan }^{ -1 }(x)-{ tan }^{ -1 }({ x }^{ 3 })+c$$
  • $${ tan }^{ -1 }(x)-\frac { 1 }{ 3 } { tan }^{ -1 }({ x }^{ 3 })+c$$
  • $${ tan }^{ -1 }(x)+{ tan }^{ -1 }({ x }^{ 3 })+c$$
  • $${ tan }^{ -1 }(x)+\frac { 1 }{ 3 } { tan }^{ -1 }({ x }^{ 3 })+c$$
$$\int { \dfrac { { x }^{ 2 }+1 }{ { x }^{ 4 }+1 }  } dx=$$
  • $$\dfrac { 1 }{ \sqrt { 2 } } { \tan }^{ -1 }(\dfrac { { x }^{ 2 }+1 }{ \sqrt { 2x } } )+c$$
  • $${ \tan }^{ -1 }(\dfrac { { x }^{ 2 }+1 }{ \sqrt { 2x } } )+c$$
  • $$\dfrac { 1 }{ \sqrt { 2 } } {\tan }^{ -1 }(\dfrac { { x }^{ 2 }-1 }{ \sqrt { 2x } } )+c$$
  • $${ \tan }^{ -1 }(\dfrac { { x }^{ 2 }-1 }{ \sqrt { 2x } } )+c$$
$$\int { \frac { dx }{ \sqrt [ 4 ]{ { (x+1) }^{ 5 }{ (x+2) }^{ 3 } }  }  } $$ is equal to :
  • $$4{ (\frac { x+1 }{ x+2 } ) }^{ 1/4 }+c$$
  • $$-4{ (\frac { x+1 }{ x+2 } ) }^{ -1/4 }+c$$
  • $$-4{ (\frac { x+2 }{ x+1 } ) }^{ 1/4 }+c$$
  • None of these
$$\displaystyle\int{{e}^{{\tan}^{-1}{x}}\left(\dfrac{1+x+{x}^{2}}{1+{x}^{2}}\right)dx}$$ is equal to
  • $${ -e }^{ { tan }^{ -1 }x }+c$$
  • $${ e }^{ { tan }^{ -1 }x }+c$$
  • $${ -xe }^{ { tan }^{ -1 }x }+c$$
  • $${ xe }^{ { tan }^{ -1 }x }+c$$
$$\int { \cfrac { \cos { x } -1 }{ \sin { x } +1 }  } { e }^{ x }dx$$ is equal to:
  • $$\cfrac { { e }^{ x }\cos { x } }{ 1+\sin { x } } +c$$
  • $$c-\cfrac { { e }^{ x }\sin { x } }{ 1+\sin { x } } $$
  • $$c-\cfrac { { e }^{ x }}{ 1+\sin { x } } $$
  • $$c-\cfrac { { e }^{ x }\cos { x } }{ 1+\sin { x } } $$
If $$\int { \cfrac { 1 }{ \left( 1+x \right) \sqrt { x }  }  } dx=f\left( x \right) +A$$, where A is any arbitary constant, then the function f(x) is
  • $$2\tan^{-1}{x}$$
  • $$2\tan^{-1}{\sqrt{x}}$$
  • $$2\cot^{-1}{\sqrt{x}}$$
  • $$log_{e}{(1+x)}$$
$$\int { \dfrac { dx }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+4) }  } =$$
  • $$\frac { 1 }{ 3 } \tan ^{ -1 }{ x } -\frac { 1 }{ 3 } \tan ^{ -1 }{ \frac { x }{ 2 } } +c$$
  • $$\frac { 1 }{ 3 } \tan ^{ -1 }{ x } +\frac { 1 }{ 3 } \tan ^{ -1 }{ \frac { x }{ 2 } } +c$$
  • $$\frac { 1 }{ 3 } \tan ^{ -1 }{ x } -\frac { 1 }{ 6 } \tan ^{ -1 }{ \frac { x }{ 2 } } +c$$
  • $$\tan ^{ -1 }{ x } -2\tan ^{ -1 }{ \frac { x }{ 2 } } +c$$
Integrate: $$\displaystyle \int \dfrac{x}{\sqrt{x+4}}dx$$
  • $$\dfrac 23(x+4)^{\tfrac 32}-8\sqrt{x+4}$$
  • $$\dfrac 23(x+4)^{\tfrac 32}+8\sqrt{x+4}$$
  • $$\dfrac 23(x+4)^{\tfrac 32}+4\sqrt{x+4}$$
  • None of these
The angle made by the tangent line at (1, 3) on the curve $$y=4x-{ x }^{ 2 }$$ with $$\overline{ OX } $$ is 
  • $$\tan ^{-1}2$$
  • $$\tan ^{-1}(1/2)$$
  • $$\tan ^{-1}-2$$
  • None of these
If $$\displaystyle \int { \left( u\cfrac { dv }{ dx }  \right)  } dx=uv-\int { wdx } $$, then $$w=$$
  • $$\cfrac { du }{ dx } \cfrac { dv }{ dx } $$
  • $$v\cfrac { du }{ dx } $$
  • $$\cfrac { d }{ dx } (uv) $$
  • $$u\cfrac { dv }{ dx } $$
$$\displaystyle \int { \cfrac { x\tan ^{ -1 }{ x }  }{ { \left( 1+x^{ 2 } \right)  }^{ 3/2 } }  } dx=$$
  • $$\cfrac { x+\tan ^{ -1 }{ x } }{ { \sqrt { \left( 1+x^{ 2 } \right) } }^{ } } +c$$
  • $$\cfrac { x-\tan ^{ -1 }{ x } }{ { \sqrt { \left( 1+x^{ 2 } \right) } }^{ } } +c$$
  • $$\cfrac { \tan ^{ -1 }{ x }-x }{ { \sqrt { \left( 1+x^{ 2 } \right) } }^{ } } +c$$
  • None of these
$$\displaystyle \int \dfrac{dx}{9x^2 + 1} =$$_____.
  • $$\dfrac{1}{3} \tan^{-1} (2x) + c$$
  • $$\dfrac{1}{3} \tan^{-1} x + c$$
  • $$\dfrac{1}{3} \tan^{-1} (3x) + c$$
  • $$\dfrac{1}{3} \tan^{-1} (6x) + c$$
Evaluate $$\displaystyle \int \:e^x\left(\log \left(x\right)+\dfrac{1}{x^2}\right)dx$$
  • $$e^x(\log{x}+\cfrac{1}{x^2})$$
  • $$e^x(\log{x}+\cfrac{1}{x})$$
  • $$e^x(\log{x}-\cfrac{1}{x^2})$$
  • $$e^x(\log{x}-\cfrac{1}{x})$$
$$\int { \left( 2+\log { x }  \right) { \left( ex \right)  }^{ x }dx } =.....+C;x>1$$
  • $${ \left( ex \right) }^{ x }$$
  • $${ x }^{ x }$$
  • $${ \left( ex \right) }^{ -x }$$
  • $${ e }^{ { x }^{ x } }$$
If $$\displaystyle \int { e^ { \sec}  } (\sec x\, \tan x \, f(x)+ \sec x\, \tan x+ \tan^2x)dx=e^{\sec x}f(x)+c. Then f(x) is$$
  • $$\sec x+x\ tan x+\dfrac{1}{2}$$
  • $$x \sec x+x \tan x+\dfrac{1}{2}$$
  • $$x \sec x+x^2 \tan x+\dfrac{1}{2}$$
  • $$\sec x+ \tan x+\dfrac{1}{2}$$
$$\int _{ a }^{ b }{ \dfrac{dx}{2 + 3x} } =$$
  • $$\dfrac{1}{3} log_e (2 + 3b)$$
  • $$\dfrac{1}{3} log_e (2 + 2a)$$
  • $$\dfrac{1}{3} log_e \left(\dfrac{2 + 3b}{2 + 3a}\right)$$
  • $$\dfrac{1}{3} log_e \left(\dfrac{2 + 3a}{2 + 3b}\right)$$
$$\displaystyle\int \dfrac{2x^3-1}{x^4+x}dx$$ is equal to?
  • $$ln\left|\dfrac{x^3+1}{x}\right|+c$$
  • $$ln\left|\dfrac{x^3+1}{x^2}\right|+c$$
  • $$\dfrac{1}{2}\ln\left|\dfrac{x^3+1}{x^2}\right|+c$$
  • $$\dfrac{1}{2}ln\left|\dfrac{x^3+1}{x}\right|+c$$
If $$\displaystyle\int 2^{2x}\cdot 2^xdx=A\cdot 2^{2^x}+c$$, then $$A=?$$
  • $$\dfrac{1}{log 2}$$
  • $$log 2$$
  • $$(log 2)^2$$
  • $$\dfrac{1}{(log 2)^2}$$
If $$\displaystyle \int \dfrac{dx}{x^3(1+x^6)^{2/3}} = xf(x) (1 + x^6)^{\frac{1}{3}} + C$$
where $$C$$ is a constant of integration, then the function $$f(x)$$ is equal to - 
  • $$-\dfrac{1}{6x^3}$$
  • $$\dfrac{3}{x^2}$$
  • $$-\dfrac{1}{2x^2}$$
  • $$-\dfrac{1}{2x^3}$$
Evaluate : $$ \int \dfrac { sec x}{( sec x + tan x ) } dx $$
  • $$ tan x+ sec x+ C $$
  • $$ tan x -sec x + C $$
  • $$ -tan x + sec x + C $$
  • $$ -tan x - sec x + C $$
What is $$\displaystyle \int \dfrac{dx}{2x^2 - 2x + 1}$$ equal to ?
  • $$\dfrac{\tan^{-1} (2x - 1)}{2} + c$$
  • $$2 \tan^{-1} (2x - 1) += c$$
  • $$\dfrac{\tan^{-1} (2x + 1)}{2} + c$$
  • $$\tan^{-1} (2x - 1) + c$$
$$\displaystyle \int\frac{x\cos x\log x-\sin x}{x(\log x)^{2}}dx=$$
  • $$\displaystyle \frac { \sin { x }  }{ \log { x }  } +C$$
  • $$\displaystyle \frac { \cos { x }  }{ \log { x }  } +C$$
  • $$\displaystyle \frac { \log { x }  }{ \sin { x }  } +C$$
  • $$\displaystyle \frac { \log { x }  }{ \cos { x }  } +C$$
If $$\int { x\sin { x }  } dx=-x\cos { x } +\alpha $$, then $$\alpha$$ is equal to
  • $$\sin x+C$$
  • $$\cos x+C$$
  • $$C$$
  • None of these
Evaluate : $$\displaystyle \int \dfrac { cos 2x}{cos^2 x sin^2 x }dx  $$
  • $$ - cot x -tan x +C $$
  • $$ -cot x + tan x + C$$
  • $$ cot x -tan x +C $$
  • $$ cot x + tan x + C$$
0:0:1


Answered Not Answered Not Visited Correct : 0 Incorrect : 0

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers