If $$I=\int { \dfrac { 1 }{ { x }^{ 4 }\sqrt { { a }^{ 2 }+{ x }^{ 2 } } } dx, } $$ then $$I$$ equals

$$\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } \sqrt { { a }^{ 2 }{ +x }^{ 2 } } \right\} +C$$

$$\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } ({ a }^{ 2 }{ x }^{ 2 })^{3/2} \right\} +C$$

$$\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 2 }\sqrt { x } } ({ a }^{ 2 }{ x }^{ 2 })^{3/2} \right\} +C$$

MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 14

$\displaystyle \int { { e }^{ \tan ^{ -1 }{ x } } } \left[ \frac { 1+x+x^{ 2 } }{ 1+x^{ 2 } } \right] dx=$

0%
$\displaystyle x^{ 2 }{ e }^{ \tan ^{ -1 }{ x } }+c$
0%
$\displaystyle x{ e }^{ \tan ^{ -1 }{ x } }+c$
0%
$\displaystyle \dfrac{{ e }^{ \tan ^{ -1 }{ x } }}{x}+c$
0%
$\displaystyle \dfrac{{ e }^{ \tan ^{ -1 }{ x } }}{x^2}+c$

$\int {x\left( {\frac{{\ln {a^{{a^{\frac{x}{2}}}}}}}{{3{a^{\frac{{5x}}{2}}}{b^{3x}}}} + \frac{{\ln {b^{{b^{\frac{x}{2}}}}}}}{{2{a^{2x}}{b^{4x}}}}} \right)} dx\left( {where\,a,b \in R} \right)$ is equal to

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

$\displaystyle \int { { e }^{ \tan ^{ -1 }{ x } } } \left[ \dfrac { 1+x+x^{ 2 } }{ 1+x^{ 2 } } \right] dx=$

0%
$x^{ 2 }{ e }^{ \tan ^{ -1 }{ x } }+c$
0%
$x{ e }^{ \tan ^{ -1 }{ x } }+c$
0%
${ e }^{ \tan ^{ -1 }{ x } }+c$
0%
$\frac { 1 }{ 2 } { e }^{ \tan ^{ -1 }{ x } }+c$

The value of the integral

$\int\int xy(x+y)dx {\,}dy$ over the area between

$y=x^2$ and

$y=x$ is

0%
$\dfrac{3}{56}$
0%
$\dfrac{47}{56}$
0%
$\dfrac{33}{56}$
0%
$\dfrac{23}{56}$

$I=\int { \left\{ \log _{ e }( \log _{ e }{ x }) +\cfrac { 1 }{ { \left( \log _{ e }{ x } \right) }^{ 2 } } \right\} } dx$ is equal to

0%
$x\log _{ e }( \log _{ e }{ x } )+c$
0%
$x\log _{ e }( \log _{ e }{ x })-\cfrac { x }{ { \left( \log _{ e }{ x } \right) }^{ } } +c$
0%
$x\log _{ e }{ x } \log _{ e }{ x } +\cfrac { x }{ { \left( \log _{ e }{ x } \right) }^{ } } +c$
0%
none of these

$\left[ \int \log ( 1 + \cos x ) - x \tan \frac { x } { 2 } \right] d x$ is equal to ?

0%
$x \tan \frac { x } { 2 }$
0%
$\log ( 1 + \cos x )$
0%
$x \log ( 1 + \cos x )$
0%
None of these

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

0%
$(x+1)e^{x+x^{-1}}+c$
0%
$(x-1)e^{x+x^{-1}}+c$
0%
$-xe^{x+x^{-1}}+c$
0%
$xe^{x+x^{-1}}+c$

Solve:

$\int\dfrac{sin^32x}{cos^52x}dx$

0%
$\dfrac{{{{\tan }^4}2x}}{8} + C$
0%
$\dfrac{{{{\cos}^4}2x}}{8} + C$
0%
$\dfrac{{{{\sin}^4}2x}}{8} + C$
0%
$\dfrac{{{{\sec}^4}2x}}{8} + C$

$f\left (\dfrac {x - 4}{x + 2}\right ) = 2x + 1, (x\epsilon R - \left \{1, -2\right \})$ m then

$\int f(x) dx$ is equal to
(where

$C$ is a constant of integration).

0%
$12\log_{e}|1 - x| + 3x + C$
0%
$-12\log_{e}|1 - x| - 3x + C$
0%
$12\log_{e}|1 - x| - 3x + C$
0%
$-12\log_{e}|1 - x| + 3x + C$

Evaluate :

$\int { \cfrac { dx }{ x\cos { x } } }$

0%
$\ln { x } +\cfrac { { x }^{ 2 } }{ 2 } +\cfrac { { 3x }^{ 3 } }{ 3 } +...$
0%
$\ln { x } -\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { 4x }^{ 4 } }{ 16 } +...$
0%
$\ln { x } +\cfrac { { x }^{ 2 } }{ 4 } +\cfrac { { 5x }^{ 4 } }{ 96 } +...$
0%
$\ln { x } -\cfrac { { x }^{ 2 } }{ 3 } +\cfrac { { x }^{ 4 } }{ 9 } +...$

$\int_{}^{} {\frac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}dx}$ would be equal to

0%
$\frac{{\sin x + x\cos x}}{{x\sin x + \cos x}} + c$
0%
$\frac{{\sin x - x\cos x}}{{x\sin x + \cos x}} + c$
0%
$\frac{{\sin x - x\cos x}}{{x\sin x - \cos x}} + c$
0%
none of these

$\displaystyle \int_{}^{} {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx}$ is equal to :

0%
$- {e^x}\tan \frac{x}{2} + c$
0%
$- {e^x}\cot \frac{x}{2} + c$
0%
$- \frac{1}{2}{e^x}\tan \frac{x}{2} + c$
0%
$- \frac{1}{2}{e^x}\cot \frac{x}{2} + c$

Explanation

Let

$I=\displaystyle\int{{e}^{x}\dfrac{1-\sin{x}}{1-\cos{x}}dx}$

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

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

$=\displaystyle\int{{e}^{x}\left(\dfrac{{\csc}^{2}{\dfrac{x}{2}}}{2}-\cot{\dfrac{x}{2}}\right)dx}$

Let

$f\left(x\right)=-\cot{\dfrac{x}{2}}$

$\Rightarrow\,{f}^{\prime}{\left(x\right)}=\dfrac{1}{2}{\csc}^{2}{\dfrac{x}{2}}$

$\Rightarrow\,I=\displaystyle\int{{e}^{x}\left(\dfrac{{\csc}^{2}{\dfrac{x}{2}}}{2}-\cot{\dfrac{x}{2}}\right)dx}$

Using

$\displaystyle\int{{e}^{x}\left(f\left(x\right)+{f}^{\prime}{\left(x\right)}\right)dx}={e}^{x}f\left(x\right)+C$

$\Rightarrow\,I=-{e}^{x}\cot{\dfrac{x}{2}}+c$

If the primitive of

$\frac{1}{{f\left( x \right)}}\,$ is equal to

$\,{\left\{ {f\left( x \right)} \right\}^2} + c$ ,then f(x) is

0%
$x + D$
0%
$\frac{x}{2} + d$
0%
$\frac{{{x^2}}}{2} + d$
0%
${x^2} + d$

$\int \sqrt{\dfrac{x-5}{x-7}}dx=a\sqrt{x^{2}-12x+35+log}|x-6+\sqrt{x^{2}-12x+35}|+C$ then

$A$ =

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

$\dfrac{d}{dx}\ f(x)=g(x)$ for

$a\leq x\leq b$ then

$\int_{b}^{a}f(x)g(x)dx$ equals to:

0%
$f(2)-f(1)$
0%
$g(2)-g(1)$
0%
$\dfrac{[f(b)]^{2}-[f(a)]^{2}}{2}$
0%
$\dfrac{[g(b)]^{2}-[g(a)]^{2}}{2}$

The value of

$\displaystyle \int \left ( 3x^{2}\tan \dfrac{1}{x}-x\sec^{2}\dfrac{1}{x} \right )dx$ is

0%
$x^{3}\tan \dfrac{1}{x}+c$
0%
$x^{2}\tan \dfrac{1}{x}+c$
0%
$x\tan \dfrac{1}{x}+c$
0%
$\tan \dfrac{1}{x}+c$

Explanation

$I=\int \left(3x^{2} \tan \dfrac{1}{x}- x \sec^{2} \dfrac{1}{x} \right) dx$

$I= \int 3x^{2} \tan \dfrac{1}{x} dx- \int \sec^{2} \dfrac{1}{x}. x \ dx$

$I=I_1- I_2$

$I_1= \int 3x^{2} \tan \dfrac{1}{x} dx$

$=\tan \dfrac{1}{x}. \int 3x^{2} dx- \int \left( \dfrac{d}{dx} \tan \dfrac{1}{x}. \int 3x^{2} dx \right) dx+c$

$=\tan \dfrac{1}{x}. x^{3}- \int \sec^{2} \dfrac{1}{x}. \left( \dfrac{-1}{x^{2}} \right). x^{3} dx+c$

$=\tan \dfrac{1}{x}. x^{3}+ \int x \sec^{2} \dfrac{1}{x} dx+c$

Now

$I=I_{1}-I_{2}$

$=\tan \dfrac{1}{x}. x^{3}+ \int x. \sec^{2} \dfrac{1}{x} dx- \int x \sec^{2} \dfrac{1}{x} dx+ c$

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

$g(x)$ is a differentiable function satisfying

$\dfrac{d}{dx}{g(x)}=g(x)$ and

$g(0)=1,$ then

$\int { g\left( x \right) } \left( \dfrac { 2-sin2x }{ 1-cos2x } \right) dx$ is equal to

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

Let

$f:R\longrightarrow R,g : R\longrightarrow R$ be continuous functions. then the value of integeral.

$\int _{ \ell n\lambda }^{ \ell n/\lambda }{ \frac { f\left( \dfrac { { x }^{ 2 } }{ 4 } \right) \left[ f\left( x \right) -f\left( -x \right) \right] }{ g\left( \dfrac { { x }^{ 2 } }{ 4 } \right) \left[ g\left( x \right) +g\left( -x \right) \right] } } dx$ is:

0%
depend on $\lambda$
0%
a non-zero constant
0%
zero
0%
1

$I=\int { \dfrac { 1 }{ { x }^{ 4 }\sqrt { { a }^{ 2 }+{ x }^{ 2 } } } dx, }$ then

$I$ equals

0%
$\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } \sqrt { { a }^{ 2 }{ +x }^{ 2 } } \right\} +C$
0%
$\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 3x }^{ 3 } } ({ a }^{ 2 }{ x }^{ 2 })^{3/2} \right\} +C$
0%
$\dfrac { 1 }{ { a }^{ 4 } } =\left\{ \dfrac { 1 }{ x } \sqrt { { a }^{ 2 }+{ x }^{ 2 } } -\dfrac { 1 }{ { 2 }\sqrt { x } } ({ a }^{ 2 }{ x }^{ 2 })^{3/2} \right\} +C$
0%
None of these

Evaluate:

$\displaystyle \int \dfrac { 1 } { x ^ { 2 } \left( x ^ { 4 } + 1 \right) ^ { \frac { 3 } { 4 } } } d x ; x = 0$

0%
$\dfrac { \left( x ^ { 4 } - 1 \right) ^ { \frac { 1 } { 4 } } } { x } + c$
0%
$-\dfrac { \left( x ^ { 4 } + 1 \right) ^ { \frac { 1 } { 4 } } } { x } + c$
0%
$\dfrac { \sqrt { x ^ { 4 } + 1 } } { x } + c$
0%
None of these

Explanation

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

$=\displaystyle\int \frac{dx}{x^{2}\dfrac{(1+x^{4})^{3/4}.x^3}{x^{3}}}$

$I=\displaystyle\int \frac{dx}{x^{5}(\dfrac{(1+x^{4})^{1/4}}{x})^{3}}$

$u=\dfrac{(1+x^{4})^{1/4}}{x}$

$(ux)^{4}= x^{4}+1$

$u^{4}= 1+\dfrac{1}{x^{4}}$

$4u^{3}du = -\dfrac{4\, dx}{x^{5}}$

$u^{3}du=-\dfrac{dx}{x^{5}}$

$I=\displaystyle\int \dfrac{-u^{3}du}{u^{3}}=-\int du$

$= -u+c$

$=-\dfrac{(1+x^{4})^{1/4}}{x}+c$

$\int\dfrac{x^3-6x^2+11x-6}{\sqrt{x^2+4x+3}}dx=(Ax^2+bx+c)\sqrt{x^2+4x+3}+\lambda \int\dfrac{dx}{\sqrt{x^2+4x+3}}$ , then value of 'A' is

0%
$\dfrac{1}{3}$
0%
$1$
0%
$3$
0%
$-1/3$

$x\in \left(0, \dfrac{\pi}{2}\right)$ then

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

0%
$e^{-\pi/2}\sec\dfrac{x}{2}+c$
0%
$-e^{\dfrac{-x}{2}}\sec\dfrac{x}{2}+c$
0%
$e^{\dfrac{x}{2}}\sec\dfrac{x}{2}+c$
0%
$-e^{\dfrac{x}{2}}\sec\dfrac{x}{2}+c$

$\int _ { 0 } ^ { \pi / 4 } \tan ^ { 2 } x d x$ equals -

0%
$\pi / 4$
0%
$1 + ( \pi / 4 )$
0%
$1 - ( \pi / 4 )$
0%
$1 - ( \pi / 2 )$

Evaluate

$\int \dfrac{dx}{\sqrt{1-x}}$

0%
$\sin^{-1} \sqrt{x}$
0%
$-\sin^{-1} \sqrt{x}+c$
0%
$2\sqrt{1-x}+c$
0%
$-2\sqrt{1-x}+c$

Let

$[\cdot]$ denote the greatest integer function then the value of

$\displaystyle\int^{1.5}_0x[x^2]dx$ is?

0%
$0$
0%
$\dfrac{3}{2}$
0%
$\dfrac{7}{4}$
0%
$\dfrac{5}{4}$

The integral

$\int \dfrac { \sin ^{ { 2 } } x\cos ^{ { 2 } } x }{ \left( \sin ^{ { 5 } } x+\cos ^{ { 3 } } x\sin ^{ { 2 } } x+\sin ^{ { 3 } } x\cos ^{ { 2 } } x+\cos ^{ { 5 } } x \right) ^{ { 2 } } } dx$ is equal to :

0%
$\dfrac { - 1 } { 1 + \cot ^ { 3 } x } + C$
0%
$\dfrac { 1 } { 3 \left( 1 + \tan ^ { 3 } x \right) } + C$
0%
$\dfrac { - 1 } { 3 \left( 1 + \tan ^ { 3 } x \right) } + C$
0%
$\dfrac { 1 } { 1 + \cot ^ { 3 } x } + C$

The integral

$\int \dfrac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4}dx$ is equal to: (where C is a constant of integration)

0%
$\dfrac{x^4}{(2x^4 + 3x^2 + 1)^3} + C$
0%
$\dfrac{x^{12}}{6(2x^4 + 3x^2 + 1)^3} + C$
0%
$\dfrac{x^4}{6(2x^4 + 3x^2 + 1)^3} + C$
0%
$\dfrac{x^{12}}{(2x^4 + 3x^2 + 1)^3} + C$

Explanation

$\int \dfrac{3x^{13} + 2x^{11}}{(2x^4 + 3x^2 + 1)^4}dx$

$\int \dfrac{\lgroup \dfrac{3}{x^3} + \dfrac{2}{x^5} \rgroup dx}{\lgroup 2 + \dfrac{3}{x^2} + \dfrac{1}{x^4} \rgroup^4}$

Let

$\lgroup 2 + \dfrac{3}{x^2} + \dfrac{1}{x^4} \rgroup = t$

$-\dfrac{1}{2}\int\dfrac{dt}{t^4} = \dfrac{1}{6t^3} + C \Rightarrow \dfrac{x^{12}}{6(2x^4 + 3x^2 + 1)^4}+ C$

The value of

$\displaystyle \int^{\pi/4}_{-\pi /4} \dfrac{dx}{sec^2x(1-sinx)}$ is

0%
$\pi / 4$
0%
$\pi$
0%
$\pi / 2$
0%
$2 \pi$

Evaluate:

$\int { \sqrt { \dfrac { x }{ 4-{ x }^{ 3 } } } } dx$

0%
$\dfrac { 2 }{ 3 } \sin ^{ -1 }{ \left( \dfrac { { x }^{ \dfrac { 3 }{ 2 } } }{ 2 } \right) +c }$
0%
$\dfrac { 2 }{ 3 } \sin ^{ -1 }{ \left( { x }^{ \dfrac { 3 }{ 2 } } \right) +c }$
0%
$2\sin ^{ -1 }{ \left( \dfrac { { x }^{ \dfrac { 3 }{ 2 } } }{ 2 } \right) +c }$
0%
$\dfrac { 1 }{ 3 } \sin ^{ -1 }{ \left( \dfrac { { x }^{ \dfrac { 3 }{ 2 } } }{ 2 } \right) +c }$

$\displaystyle\int \dfrac{\cos x \, dx}{\sin^3x(1+\sin^6x)^{2/3}} = f(x)(1 + \sin^6x)^{1/\alpha} + c$
Where

$c$ is a constant of integration, then

$\lambda f\left(\dfrac{\pi}{3}\right)$ is equal to:

0%
$\dfrac{9}{8}$
0%
$-2$
0%
$2$
0%
$-\dfrac{9}{8}$

Explanation

$\displaystyle\int \dfrac{\cos x \, dx}{\sin^3x(1+\sin^6x)^{2/3}} = f(x)(1 + \sin^6x)^{1/\alpha} + c.....given$

Let

$\sin x = t$

$\cot x dx= dt$

$\displaystyle I= \int \dfrac{dt}{t^3(1+t^6)^{2/3}}$

$\displaystyle I=\int \dfrac{dt}{t^3.t^4\left(1 + \dfrac{1}{t^6}\right)^{2/3}}$

$\displaystyle I=\int \dfrac{dt}{t^7\left(1 + \dfrac{1}{t^6}\right)^{2/3}}$

Put

$1+\dfrac{1}{t^6}=r^3$

$\Rightarrow \dfrac{dt}{t^7} = \dfrac{-1}{2}r^2dr$

$= \dfrac{-1}{2}\displaystyle\int \dfrac{r^2dr}{r^2}$

$=\dfrac{-1}{2}r +c$

$= \dfrac{-1}{2}\left(\dfrac{\sin^6x + 1}{\sin^6x}\right)^{1/3}+c .......$ As

$r=\left (1+\dfrac{1}{t^6}\right )^{1/3}$ and

$t=\sin x$

$=\dfrac{-1}{2\sin^2x} (1 + \sin^6x)^{1/3}+c$

$f(x) = -\dfrac{1}{2} cosec^{2}x$ and

$\lambda = 3$

Now for

$x=\dfrac{\pi}3$

$cosec \dfrac{\pi}3=\dfrac{2}{\sqrt3}$

$\lambda f \left(\dfrac{\pi}{3}\right) = \dfrac{-1}2\times \left (\dfrac{2}{\sqrt3}\right )^2\times 3$

$\boxed{\lambda f \left(\dfrac{\pi}{3}\right) = -2}.....Answer$

Hence option

$'B'$ is the answer.

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 14 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 14 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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