If $$f(3-x)=f(x)$$, then $$\int_{1}^{2}x\ f(x)\ dx$$ is equal to

$$\dfrac{3}{2}\int_{1}^{2}f(x)\ dx$$

$$\dfrac{3}{2}\int_{1}^{2}f(2-x)\ dx$$

$$\dfrac{1}{2}\int_{1}^{2}f(x)\ dx$$

If $$\int \log \left( a ^ { 2 } + x ^ { 2 } \right) d x = h ( x ) ,$$ then $$h ( x ) = $$

$$x \log \left( a ^ { 2 } + x ^ { 2 } \right) - 2 x + 2 a \tan ^ { - 1 } \left( \dfrac { x } { a } \right) + c$$

$$x \log \left( a ^ { 2 } + x ^ { 2 } \right) + 2 \tan ^ { - 1 } \left( \dfrac { x } { a } \right) + c$$

$$x ^ { 2 } \log \left( a ^ { 2 } + x ^ { 2 } \right) + x + a \tan ^ { - 1 } \left( \dfrac { x } { a } \right) + c$$

$$x ^ { 2 } \log \left( a ^ { 2 } + x ^ { 2 } \right) + 2 x - a ^ { 2 } \tan ^ { - 1 } \left( \dfrac { x } { a } \right) + c$$

MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 8

The value of

$\int \dfrac {1}{\sqrt {\sin^{3}x \cos^{5}x}} dx$ is

0%
$\dfrac {-2}{\sqrt {\tan x}} + \dfrac {2}{3} (\tan x)^{3/2} + C$
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$\dfrac {2}{\sqrt {\tan x}} - \dfrac {2}{3} (\tan x)^{3/2} + C$
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$\dfrac {-2}{\sqrt {\tan x}} + \dfrac {2}{3} (\tan x)^{1/2} + C$
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None of these

Explanation

Let

$I=\int { \cfrac { 1 }{ \sqrt { \sin ^{ 3 }{ x } \cos ^{ 5 }{ x } } } dx }$

$I=\int { \cfrac { 1 }{ \sqrt { \cfrac { \sin ^{ 3 }{ x } }{ \cos ^{ 3 }{ x } } \cos ^{ 8 }{ x } } } dx } =\cfrac { 1 }{ \cos ^{ 4 }{ x } \sqrt { \tan ^{ 3 }{ x } } } dx$

$I=\int { \cfrac { \sec ^{ 4 }{ x } }{ \sqrt { \tan ^{ 3 }{ x } } } dx } =\int { \cfrac { \sec ^{ 3 }{ x } .\sec ^{ 2 }{ x } }{ \sqrt { \tan ^{ 3 }{ x } } } dx }$

$I=\int { \cfrac { \left( 1+\tan ^{ 2 }{ x } \right) }{ \sqrt { \tan ^{ 3 }{ x } } } \sec ^{ 2 }{ x } dx }$

Let

$\tan { x } =t$

$\sec ^{ 2 }{ x } dx=dt$

So,

$I=\int { \cfrac { 1+{ t }^{ 2 } }{ \sqrt { { t }^{ 3 } } } dt } =\int { \cfrac { 1+{ t }^{ 2 } }{ { t }^{ 3/2 } } dt }$

$I=\int { { t }^{ -3/2 }dt } +\int { { t }^{ 1/2 }dt }$

$I=\cfrac { { t }^{ -1/2 } }{ -1/2 } +\cfrac { { t }^{ 3/2 } }{ 3/2 } +c$

So,

$I=\cfrac { -2 }{ \sqrt { \tan { x } } } +\cfrac { 2 }{ 3 } { \left( \tan { x } \right) }^{ 3/2 }+c$

where C is arbitary constant

Evaluate :

$\int x \sin 3x \ dx$

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$-\dfrac{1}{3}x \cos 3x+\dfrac{1}{9} \sin 3x +C$
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$-\dfrac{1}{3} x \cos 3x +\dfrac{1}{3} \sin 3x +C$
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$\dfrac{1}{3} x \sin 3x -\dfrac{1}{3} \cos 3x+C$
0%
None of these

Explanation

$I=\int x \sin 3x \ dx$

$I=x[\int \sin 3x \ dx]-\int [\dfrac{d}{dx}(x) \times \int \sin 3x \ dx]dx$

$I=x \times -\dfrac{1}{3} \cos 3x -\int [-\dfrac{1}{3} \cos 3x] dx$

$I=-\dfrac{1}{3}x \cos 3x +\dfrac{1}{3} \int \cos 3x \ dx =-\dfrac{1}{3}x\cos 3x +\dfrac{1}{9} \sin 3x +C$

$y=(x+\sqrt{x^{2}-a^{2}})^{n}$ then

$(x^{2}-a^{2})(\dfrac{dy}{dx})^{2}$ =

0%
$n^{2}y$
0%
$-n^{2}y$
0%
$ny^{2}$
0%
$n^{2}y^{2}$

Explanation

$y = (x+\sqrt{x^2-a^2})^n$

${y}' = n(x+\sqrt{x^2-a^2})^{n-1} (1+\dfrac{1(2x)}{2\sqrt{x^2-a^2}})$

$= n \dfrac{y}{x+\sqrt{x^2-a^2}}(1+\dfrac{x}{\sqrt{x^2-a^2}})$

$= n \dfrac{y}{(x+\sqrt{x^{2}-a^2})}+\dfrac{(\sqrt{x^2+a^2+x})}{\sqrt{x^2-a^2}}$

$({y}')^2 = (\dfrac{ny}{x^2-a^2})^2$

$({y}')^2 = \dfrac{n^{2}y^2}{x^2-a^2}$

$\therefore (x^{2}-a^2) ({y}') = (x^2-a^2)\dfrac{x^2y^2}{(x^2-a^2)}$

$= n^{2}y^2$ option is D is correct.

1151063_1144362_ans_7dd05037f1194946a155a55fbb0dcae0.jpg

$\int e^{x/2}\sin \left( \frac { \pi }{ 4 } +\frac { x }{ 2 } \right) dx=$

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$\sqrt{2}\ e^{x/2}\sin \frac { x }{ 2 }+c$
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$\sqrt{2}\ e^{x/2}\cos\frac { x }{ 2 }+c$
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$-\sqrt{2}\ e^{x/2}\sin \frac { x }{ 2 }+c$
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$-\sqrt{2}\ e^{x/2}\cos\frac { x }{ 2 }+c$

Explanation

$\displaystyle \int e^{x/2}\sin \left (\dfrac {\pi}{4}+\dfrac {x}{z}\right)dx$

$=\dfrac {1}{D}e^{x/2} \sin \left (\dfrac {\pi}{4}+\dfrac {x}{z}\right)$

$=e^{x/2}\dfrac {1}{2+D}\sin \left (\dfrac {\pi}{4}+\dfrac {x}{z}\right)=e^{x/2}\left (\dfrac {2-D}{4-D^2}\right)\sin \left (\dfrac {\pi}{4}+\dfrac {x}{z}\right)dx$

$\dfrac { { e }^{ x/2 }2\sin { \left( \dfrac { \pi }{ 4 } +\dfrac { x }{ 2 } \right) } -\cos { \left( \dfrac { \pi }{ 4 } +\dfrac { x }{ 2 } \right) } }{ 4-\left( -1 \right) } +C$

$\dfrac { { e }^{ x/2 }2\sin { \left( \dfrac { \pi }{ 4 } +\dfrac { x }{ 2 } \right) } -\cos { \left( \dfrac { \pi }{ 4 } +\dfrac { x }{ 2 } \right) } }{ 5 } +C$

after using

$\sin (A+B)$ and

$\cos (A-B)$

$=\sqrt 2 e^{x/2}\sin \left (\dfrac {x}{2}\right)+C$

$\displaystyle \int \dfrac{x \,\ell nx}{(x^2 - 1)^{3/2}}dx$ equals

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$arc \,\sec x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$
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$\sec^{-1} x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$
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$\cos^{-1} x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$
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$\sec x - \dfrac{\ell n\,x}{\sqrt{x^2 - 1}} + C$

Explanation

$\displaystyle \displaystyle \int \dfrac {x ln x}{(x^2-1)^{3/2}}dx$

$lnx\displaystyle \int { \dfrac { x }{ { \left( { x }^{ 2 }-1 \right) }^{ 3/2 } } dx } -\displaystyle \int { \left[ \left( \dfrac { d }{ dx } lnx \right) \displaystyle \int { \dfrac { x }{ { \left( { x }^{ 2 }-1 \right) }^{ 3/2 } } } \right] dx }$

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

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

$\therefore \displaystyle \int { \dfrac { xlnx }{ { \left( { x }^{ 2 }-1 \right) }^{ 3/2 } } dx } =arc\sec { x-\dfrac { lnx }{ \sqrt { { x }^{ 2 }-1 } } +C }$

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

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

Explanation

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

$\sin^{-1}x+\cos^{-1}x=\pi /2$

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

$\Rightarrow \dfrac{2}{\pi}\left[x\sin^{-1}x+\sqrt{1-x^2}-x\cos^{-1}x+\sqrt{1-x^2}\right]+c$

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

$\int { { x }^{ 2 } } \sqrt { { x }^{ 6 }-1 } dx=$

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$\frac { 1 }{ 6 } \left\{ { x }^{ 3 }\sqrt { { x }^{ 6 }-1 } -\log { \left( { x }^{ 3 }+\sqrt { { x }^{ 6 }-1 } \right) } \right\} +c$
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$\frac { 1 }{ 6 } \left\{ { x }^{ 3 }\sqrt { { x }^{ 6 }-1 } +\log { \left( { x }^{ 3 }+\sqrt { { x }^{ 6 }-1 } \right) } \right\} +c$
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$\frac { 1 }{ 6 } \left\{ { x }^{ 3 }\sqrt { { x }^{ 6 }-1 } -\sin ^{ -1 }{ \left( { x }^{ 3 } \right) } \right\} +c$
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$\frac { 1 }{ 6 } \left\{ { x }^{ 3 }\sqrt { { x }^{ 6 }-1 } +\sin ^{ -1 }{ \left( { x }^{ 3 } \right) } \right\} +c$

Explanation

$\int \:x^2\sqrt{x^6-1}dx$

$\mathrm{Apply\:u-substitution:}\:u=x^6$

$=\int \frac{\sqrt{u-1}}{6\sqrt{u}}du$

$=\frac{1}{6}\cdot \int \frac{\sqrt{u-1}}{\sqrt{u}}du$

$\mathrm{Apply\:Integration\:By\:Parts:}\:u=\sqrt{u-1},\:v'=\frac{1}{\sqrt{u}}$

$=\frac{1}{6}\left(2u^{\frac{1}{2}}\sqrt{u-1}-\int \sqrt{\frac{u}{u-1}}du\right)$

$=\frac{1}{6}\left(2u^{\frac{1}{2}}\sqrt{u-1}-\left(\sqrt{\frac{u}{u-1}}\left(u-1\right)+\ln \left|\sqrt{u}+\sqrt{u-1}\right|\right)\right)$

$=\frac{1}{6}\left(2\sqrt{x^6}\sqrt{x^6-1}-\left(\sqrt{\frac{x^6}{x^6-1}}\left(x^6-1\right)+\ln \left|\sqrt{x^6}+\sqrt{x^6-1}\right|\right)\right)$

$=\frac{1}{6}\left(x^3\sqrt{x^6-1}-\ln \left|x^3+\sqrt{x^6-1}\right|\right)+C$

Let

$A = \int _ { 0 } ^ { 1 } \frac { e ^ { t } } { t + 1 }$ dt, then the value of

$\int _ { 0 } ^ { 1 } \frac { t e ^ { t ^ { 2 } } } { t ^ { 2 } + 1 } d t$ is:-

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$A ^ { 2 }$
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$\frac { 1 } { 2 } A$
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2 $A$
0%
$\frac { 1 } { 2 } A ^ { 2 }$

The value of the integral

$\int _{ -\pi /2 }^{ \pi /2 }{ \left[ { x }^{ 2 }+log\frac { \pi -x }{ \pi +x } \right] }$ cos x dx is

0%
0
0%
$\frac { { \pi }^{ 2 } }{ 2 } -4$
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$\frac { { \pi }^{ 2 } }{ 2 } +4$
0%
$\frac { { \pi }^{ 2 } }{ 2 }$

$\int _ { 0 } ^ { 1 } \frac { d x } { \sqrt { x + 1 } + \sqrt { x } } d x =$

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$\frac { 4 } { 3 } ( \sqrt { 2 } + 1 )$
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$\frac { 4 } { 3 } ( \sqrt { 2 } - 1 )$
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$\frac { 3 } { 4 } ( \sqrt { 2 } - 1 )$
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$\frac { 3 } { 4 } ( \sqrt { 2 } - 2 )$

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

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$x\ \tan \dfrac {x}{2}+c$
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$x\ \cot \dfrac {x}{2}$
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$x\ \sin\dfrac {x}{2}+c$
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$x\ \cos \dfrac {x}{2}$

Explanation

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

$\displaystyle \int \dfrac {x+2\sin \left (\dfrac {x}{2}\right) \cos \dfrac {x}{2}}{2\cos^2x/2}dx =\displaystyle \int \dfrac {xdx}{2\cos^2 (x/2)}+\displaystyle \int tan \dfrac {x}{2}dx$

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

integrate first integral by parts

$U=x$

$du=dx$

$du=\dfrac {1}{2}\sec^2 x/2 dx$

$u=\tan x/2$

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

$=x\tan (x/2)+c$

Solve

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

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$\dfrac{(x-1)}{2}(x-\dfrac{cos2x}{2})+C$
0%
$\dfrac{(x-1)}{2}(x-\dfrac{sin2x}{2})+C$
0%
$\dfrac{(x+1)}{2}(x-\dfrac{sin2x}{2})+C$
0%
$None\ of\ these$

Explanation

Solving By parts, We get

$\int xsin^2xdx$

$\implies \dfrac{x}{2}\int(1-cos2x)dx-\dfrac{1}{2}\int1.(1-cos2x)dx$

$\implies \dfrac{(x-1)}{2}\int(1-cos2x)dx$

$\implies \dfrac{(x-1)}{2}(x-\dfrac{sin2x}{2})+C$

$\displaystyle\int {\dfrac{1}{{1 + \sin x}}dx = \tan \left( {\frac{x}{2} + a} \right) + b}$ , then

0%
$a = - \dfrac{\pi }{4},b \in R$
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$a = \dfrac{\pi }{4},b \in R$
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$a = \dfrac{{5\pi }}{4},b \in R$
0%
None of these

Explanation

$\displaystyle \int \dfrac {dx}{1+\sin x}=\tan \left (\dfrac {x}{2}+a \right)+b$

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

$=\displaystyle \int (\sec^2 x-\tan x \sec x)dx$

$=\tan x-\sec x-c=$

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

$=\dfrac { 2\cos { \dfrac { x }{ 2 } } \sin { \dfrac { x }{ 2 } } -\left( \sin ^{ 2 }{ \dfrac { x }{ 2 } } +\cos ^{ 2 }{ \dfrac { x }{ 2 } } \right) }{ \left( \cos { \dfrac { x }{ 2 } } -\sin { \dfrac { x }{ 2 } } \right) \left( \cos { \dfrac { x }{ 2 } } +\sin { \dfrac { x }{ 2 } } \right) } =-\dfrac { \left( \cos { \dfrac { x }{ 2 } } -\sin { \dfrac { x }{ 2 } } \right) }{ \left( \cos { \dfrac { x }{ 2 } } +\sin { \dfrac { x }{ 2 } } \right) }$

$=\dfrac { -\left( \dfrac { 1 }{ \sqrt { 2 } } \cos { \dfrac { x }{ 2 } } +\dfrac { 1 }{ \sqrt { 2 } } \sin { \dfrac { x }{ 2 } } \right) =\sin { \left( \dfrac { x }{ 2 } -\dfrac { \pi }{ 4 } \right) } }{ -\left( \dfrac { 1 }{ \sqrt { 2 } } \cos { \dfrac { x }{ 2 } } -\dfrac { 1 }{ \sqrt { 2 } } \sin { \dfrac { x }{ 2 } } \right) =\sin { \left( \dfrac { x }{ 2 } -\dfrac { \pi }{ 4 } \right) } } =\tan { \left( \dfrac { x }{ 2 } -\dfrac { \pi }{ 4 } \right) }$

$a=\dfrac { \pi }{ 4 } ,b\in R$

Solve

$\int {{x^3}\log xdx }$

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$\dfrac{{{x^4}\log x}}{4} + C$
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$I=\dfrac{{{x}^{4}}}{4}\log x-\dfrac{{{x}^{4}}}{16}+C$
0%
$\dfrac{1}{{8}}\left[ {{x^4}\log x - 4{x^2}} \right] + C$
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$\dfrac{1}{{16}}\left[ {4{x^4}\log x + {x^4}} \right] + C$

Explanation

Consider the given integral.

$I=\int{{{x}^{3}}\log xdx}$

$I=\log x\left( \dfrac{{{x}^{4}}}{4} \right)-\int{\dfrac{1}{x}}\left( \dfrac{{{x}^{4}}}{4} \right)dx$

$I=\dfrac{{{x}^{4}}}{4}\log x-\dfrac{1}{4}\int{{{x}^{3}}}dx$

$I=\dfrac{{{x}^{4}}}{4}\log x-\dfrac{1}{4}\left( \dfrac{{{x}^{4}}}{4} \right)+C$

$I=\dfrac{{{x}^{4}}}{4}\log x-\dfrac{{{x}^{4}}}{16}+C$

Hence, this is the answer.

Evaluate :

$\displaystyle\int {\dfrac{9cosx-sinx}{4sinx+5cosx}dx}$

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$x+\ln(4\sin x+\cos x)+c$
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$x+\ln(\sin x-5\cos x)+c$
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$x+\ln(4\sin x+5\cos x)+c$
0%
None of these

Explanation

$\to \displaystyle \int \dfrac{9\cos x - \sin x}{4\sin x + 5 \cos x} dx$

It is in the form

$\dfrac{f(x)}{g(x)}$ , so we will write

$f(x)$ as

$A(g(x)) + B(g'(x))$ .

$\to f(x) = 9\cos x - \sin x$

$\to g(x) = 4\sin x + 5 \cos x$

$g'(x) = 4 \cos x - 5 \sin x$

So,

$f(x) = g(x) + g'(x)$

$\to \displaystyle \int 1+ \dfrac{g'(x)}{g(x)} dx \Rightarrow x + ln (g(x))+C$

Answer

$= x + ln (4\sin x + 5\cos x) + C$

1048864_1177850_ans_edd56aa53b5f407cbaafdd20171be0dc.png

$\int { \cfrac { \sin { x } }{ \sin { \left( x-\alpha \right) } } } dx=Ax+B\log { \sin { \left( x-\alpha \right) } } +c$ , then the value of

$(A,B)$ is-

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$\left( \sin { \alpha } ,\cos { \alpha } \right)$
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$\left( \cos { \alpha } ,\sin { \alpha } \right)$
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$\left( -\sin { \alpha } ,\cos { \alpha } \right)$
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$\left(- \cos { \alpha } ,\sin { \alpha } \right)$

Explanation

$\displaystyle \int \frac{sin(x-\alpha +\alpha )}{sin(x-\alpha )}dx$

$\displaystyle \int \frac{sin(x-\alpha )cos\alpha +cos(x-\alpha )sin\alpha }{sin(x-\alpha )}$

$\displaystyle \int (cos\alpha +sin\alpha \,cot(x-\alpha ))dx$

$xcos\alpha +sin\alpha ln sin(x-\alpha )+c$

$A = cos\alpha$

$B = sin\alpha$

1084285_1173392_ans_6ec6706d0f664d52868d64482df94d43.jpg

Solve

$\displaystyle \int_{1}^{2}\dfrac{\sqrt{x}}{\sqrt{3-x}+\sqrt{x}}\ dx$

0%
$1$
0%
$\dfrac{1}{2}$
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$\dfrac{3}{2}$
0%
0

$f(3-x)=f(x)$ , then

$\int_{1}^{2}x\ f(x)\ dx$ is equal to

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$\dfrac{3}{2}\int_{1}^{2}f(2-x)\ dx$
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$\dfrac{3}{2}\int_{1}^{2}f(x)\ dx$
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$\dfrac{1}{2}\int_{1}^{2}f(x)\ dx$
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$\int_{1}^{2}f(x)\ dx$

$f(x)=\int { \left( \dfrac { x^2+\sin^2x }{ 1+x^2 } \right) } \sec^2 x dx$ and f(0)=0,then f (1) equals:

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$1-\dfrac { \pi }{ 4 }$
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$-\dfrac { \pi }{ 4 }$
0%
$\tan 1-\dfrac { \pi }{ 4 }$
0%
$\tan 1+1$

Explanation

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

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

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

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

$f(x)= tanx - tan^{-1}x+ \alpha$

$f(0)=0\Rightarrow \alpha=0$

$f(1)=tan 1 -\dfrac { \pi }{ 4 }$

The value of

$\int$

$(x e^{ln sinx} - cos x)$ dx is equal to :

0%
-x cos x + C
0%
sin x - x cos x + C
0%
$- e^{lnx} cos x + C$
0%
sin x + x cos x + C

Explanation

$\begin{array}{l} I=\int { \left( { x\, { e^{ \log \, \, \sin x } }-\cos x } \right) } dx \\ =\int { \left( { x\, \sin x-\cos x } \right) } dx \\ =-x\cos x+\int { \cos x\, dx } -\int { \cos x\, dx } +C \\ =-x\cos x+C \end{array}$

Hence,

Option

$A$ is correct answer.

The integrating factor of the differential equation

$\dfrac{dy}{dx}\left(x\log _e\:x\right)+y=2\log _e\:x$ is given by

0%
$x$
0%
$e^x$
0%
$\log _e\:x$
0%
$\log _e\left(\log _e\:x\right)$

Explanation

$\dfrac{dy}{dx}\left(x\log_{e}{x}\right)+y=2\log_{e}{x}$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{1}{\left(x\log_{e}{x}\right)}y=\dfrac{2\log_{e}{x}}{\left(x\log_{e}{x}\right)}$

$\Rightarrow \dfrac{dy}{dx}+\dfrac{1}{\left(x\log_{e}{x}\right)}y=\dfrac 2x$ which is of the form

$\dfrac{dy}{dx}+py=q$

Integrating factor

$={e}^{\int{p}dx}$ where

$p=\dfrac{1}{\left(x\log_{e}{x}\right)}$

Integrating factor

$={e}^{\int{\dfrac{1}{\left(x\log_{e}{x}\right)}}dx}$

Take

$t=\log_{e}{x}\Rightarrow dt=\dfrac{1}{x}dx$

$\int{\dfrac{1}{\left(x\log_{e}{x}\right)}}dx=\int{\dfrac{dt}{t}}=\ln{\left|t\right|}\ where\ t=\log_{e}{x}$

Integrating factor

$={e}^{\int{p}dx}={e}^{\ln t}=t=\log_{e}{x}$

$\displaystyle \int \dfrac{\{f(x) \cdot \phi' (x) - f'(x) \cdot \phi(x) \}}{f(x) \cdot \phi(x)} \{log \,\phi(x) - log \,f(x) \}dx$ is equal to

0%
$log \dfrac{\phi(x)}{f(x)} + k$
0%
$\dfrac{1}{2} \left \{log \dfrac{\phi(x)}{f(x)} \right \}^2 + k$
0%
$\dfrac{\phi(x)}{f(x)} log \dfrac{\phi(x)}{f(x)} + k$
0%
None of these

Explanation

$I=\displaystyle\int\left\{\dfrac{f(x)\phi'(x)-f'(x).\phi(x)}{f(x).\phi(x)}\right\}\left\{\log\phi(x)-\log f(x)\right\}dx$

$=\displaystyle\int \left\{\dfrac{f(x)\phi'(x)-f'(x)\phi(x)}{f(x)\phi(x)}\right\}f(x)\log\dfrac{\phi(x)}{f(x)}dx$

Let

$\log\dfrac{\phi (x)}{f(x)}=t$

$\dfrac{f(x)}{\phi(x)}\left(\dfrac{f(x)\phi'(x)-f'(x)\phi(x)}{f^2(x)}\right)dx=dt$

$=\displaystyle\int t dt$

$=\dfrac{t^2}{2}+k$

$=\dfrac{1}{2}\left[\log \dfrac{(\phi(x)}{f(x)}\right]^2+k$

$\therefore I=\dfrac{1}{2}\left[\log \dfrac{\phi(x)}{f(x)}\right]^2+k$ .

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

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$log (\sec x + \tan x) + c$
0%
$log (\sec x + \tan x)^{\dfrac{1}{2}} + c$
0%
$log \,\sec x (\sec x + \tan x) + c$
0%
None of these

Explanation

$\displaystyle\int \sqrt{1+2\tan x(\tan x+\sec x)}dx$

$=\displaystyle\int \sqrt{1+2\tan^2x+2\tan x\sec x}dx$

$=\displaystyle\int \sqrt{\sec^2x-\tan^2x+2\tan^2x+2\tan x\sec x}dx$

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

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

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

$=\displaystyle\int \sec xdx+\displaystyle\int \tan xdx$

$=log(\sec x+\tan x)+log(\sec x)+c$

$=log(\sec x(\sec x+\tan x)+c$

$\therefore \displaystyle\int \sqrt{1+2\tan x(\tan x+\sec x)}dx=log(\sec x(\sec x+\tan x))+c$ .

$\int(tanx -cotx)^{2}dx=$

0%
$tanx+x+c$
0%
$tanx-x+c$
0%
$tanx-cotx+c$
0%
$tanx-cotx-4x+cc$

Explanation

Now,

$\displaystyle\int (\tan x-\cot x)^2 dx$

$=\displaystyle\int (\tan^2x-2+\cot^2x)dx$

$=\displaystyle\int (\sec^2x-4+\cos ec^2x)dx$

$=\tan x-4x-\cot x+c$ . [ Where

$c$ is integrating constant]

$\int \log \left( a ^ { 2 } + x ^ { 2 } \right) d x = h ( x ) ,$ then

$h ( x ) =$

0%
$x \log \left( a ^ { 2 } + x ^ { 2 } \right) + 2 \tan ^ { - 1 } \left( \dfrac { x } { a } \right) + c$
0%
$x ^ { 2 } \log \left( a ^ { 2 } + x ^ { 2 } \right) + x + a \tan ^ { - 1 } \left( \dfrac { x } { a } \right) + c$
0%
$x \log \left( a ^ { 2 } + x ^ { 2 } \right) - 2 x + 2 a \tan ^ { - 1 } \left( \dfrac { x } { a } \right) + c$
0%
$x ^ { 2 } \log \left( a ^ { 2 } + x ^ { 2 } \right) + 2 x - a ^ { 2 } \tan ^ { - 1 } \left( \dfrac { x } { a } \right) + c$

Explanation

$I=\displaystyle\int{\log{\left({a}^{2}+{x}^{2}\right)}dx}$

Integrating by parts, we get

Let

$u=\log{\left({a}^{2}+{x}^{2}\right)}\Rightarrow\,du=\dfrac{2x\,dx}{{a}^{2}+{x}^{2}}$

$dv=dx\Rightarrow\,v=x$

$I=x\log{\left({a}^{2}+{x}^{2}\right)}-\displaystyle\int{x\times\dfrac{2x\,dx}{{a}^{2}+{x}^{2}}}$

$=x\log{\left({a}^{2}+{x}^{2}\right)}-2\displaystyle\int{\dfrac{{x}^{2}\,dx}{{a}^{2}+{x}^{2}}}$

$=x\log{\left({a}^{2}+{x}^{2}\right)}-2\displaystyle\int{\dfrac{\left({x}^{2}+{a}^{2}-{a}^{2}\right)\,dx}{{a}^{2}+{x}^{2}}}$

$=x\log{\left({a}^{2}+{x}^{2}\right)}-2\displaystyle\int{\dfrac{\left({x}^{2}+{a}^{2}\right)\,dx}{{a}^{2}+{x}^{2}}}+2\displaystyle\int{\dfrac{{a}^{2}\,dx}{{a}^{2}+{x}^{2}}}$

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

$=x\log{\left({a}^{2}+{x}^{2}\right)}-2x+2{a}^{2}\times\dfrac{1}{a}{\tan}^{-1}{\dfrac{x}{a}}+c$

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

Evaluate:

$\displaystyle\int \frac { 1 } { ( x + 2 ) \sqrt { x + 1 } } d x$

0%
$2 \tan ^ { - 1 } ( \sqrt { x + 1 } ) + c$
0%
$\tan ^ { - 1 } ( \sqrt { x - 1 } ) + c$
0%
$2 \tanh ^ { - 1 } ( \sqrt { x + 1 } ) - c$
0%
$\tan ^ { - 1 } ( \sqrt { x + 1 } ) + c$

Explanation

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

$Substitute\, u=\sqrt { x+1 }$

$du=\dfrac 1{2\sqrt{x+1}}dx$

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

$=2{ \tan ^{ -1 } }u+C$

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

Hence the correct option is

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

Evaluate:

$\int x ^ { x } \ln ( e ^x ) d x$

0%
$x ^ { x } + C$
0%
$x \cdot \ln x + C$
0%
$( \ln x ) ^ { x } + C$
0%
$x ^ { \ln x } + C$

Explanation

Let

$x^x=z$

So,

$x\log x=\log z$

Now on differentiating

$(1+\log x)=\cfrac{1}{z}\cfrac{dz}{dx}$

$\implies dz=x^x(1+\log x)dx$

Let

$I=\int x^x(\log e^x)dx$

$\implies I=\int x^x(\log e+\log e^x)dx$

$\implies I=\int x^x(1+\log e^x)dx$

Let

$x^x=z$

We get

$x^x(1+\log x)dx=dz$

So,

$I=\int dz=z+c=x^x+C$

$\displaystyle \int e^{x}[f(x)+f'(x)]dx=e^{x}.f(x)+C$

0%
True
0%
False

$\int \log _ { 10 } x d x =$

0%
$x \log _ { 10 } x + c$
0%
$x \left( \log _ { 10 } x + \log _ { 10 } e \right) + c$
0%
$\log _ { 10 } x + c$
0%
$x \left( \log _ { 10 } x - \log _ { 10 } e \right) + c$

Explanation

$\int log_{10}xdx\\=\int(\dfrac{log_ex}{log_e10})dx\\=(\dfrac{1}{log_e10})\int logx\cdot1\cdot dx\\(\dfrac{1}{log_e10})[(logx)\cdot x-\int(\dfrac{1}{x}\cdot x dx)]\\=(\dfrac{d1}{log_e10})(xlogx-x)+c\\=x((\dfrac{log_ex}{log_e10})-(\dfrac{1}{log_e10}))+c\\=x(log_{10}x-log_{10}e)+C$

$\int { f\left( x \right)dx=f\left( x \right) } ,$ then

$\int { \left\{ f\left( x \right) \right\}^2 }$ dx is equal to :

0%
$\frac { 1 }{ 2 } \left\{ f\left( x \right) \right\}^2$
0%
${ \left\{ f\left( x \right) \right\}^3 }$
0%
$\dfrac { { \left\{ f\left( x \right) \right\} ^{ 3 } } }{ 3 }$
0%
${ \left\{ f\left( x \right) \right\}^2 }$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 8 - 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 8 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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