If $$\int { f(x)dx=g(x) } $$, then $$\int { { f }^{ -1 }(x)dx } $$ is

$$x{ f }^{ -1 }(x)-g({ f }^{ -1 }(x))+C$$

MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 13

The value of

$\displaystyle \int e^{x} \left [\dfrac {1 + \sin x}{1 + \cos x}\right ] dx$ is

0%
$\dfrac {1}{2} e^{x} \sec \dfrac {x}{2} + C$
0%
$e^{x} \sec \dfrac {x}{2} + C$
0%
$\dfrac {1}{2} e^{x} \tan \dfrac {x}{2} + C$
0%
$e^{x} \tan \dfrac {x}{2} + C$

Explanation

Let

$\displaystyle I = \int e^{x} \left [\dfrac {1 + \sin x}{1 + \cos x}\right ] dx$

$\displaystyle = \int \left \{\dfrac {e^{x}}{(1 + \cos x)} + \dfrac {e^{x} \sin x}{(1 + \cos x)}\right \} dx$

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

$\displaystyle = \dfrac {1}{2} \int e^{x}\cdot \sec^{2} \dfrac {x}{2} \cdot dx + \int \underset {II}{e^{x}} \underset {I}{\tan} \dfrac {x}{2} \cdot dx$

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

(using integral by parts)

$\displaystyle= \dfrac {1}{2}\int e^{x} \cdot \sec^{2} \dfrac {x}{2} dx + e^{x} \cdot \tan \dfrac {x}{2} - \dfrac {1}{2} \int e^{x} \cdot \sec^{2} \dfrac {x}{2} \cdot dx$

$\displaystyle = e^{x} \cdot \tan \dfrac {x}{2} + C$

$\displaystyle\int { \cfrac { { x }^{ 2 }-1 }{ { x }^{ 4 }+{ x }^{ 2 }+1 } } dx$ is equal to

0%
$\log { \left( { x }^{ 4 }+{ x }^{ 2 }+1 \right) } +c$
0%
$\log { \cfrac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } +c }$
0%
$\cfrac { 1 }{ 2 } \log { \cfrac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } +c }$
0%
$\cfrac { 1 }{ 2 } \log { \cfrac { { x }^{ 2 }+x+1 }{ { x }^{ 2 }-x+1 } +c }$

Explanation

Let

$I=\int { \cfrac { { x }^{ 2 }-1 }{ { x }^{ 4 }+{ x }^{ 2 }+1 } } dx$

$\Rightarrow I=\int { \cfrac { 1-\cfrac { 1 }{ { x }^{ 2 } } }{ { x }^{ 2 }+1+\cfrac { 1 }{ { x }^{ 2 } } +1-1 } } dx$

$=\int { \cfrac { 1-\cfrac { 1 }{ { x }^{ 2 } } }{ { \left( x+\cfrac { 1 }{ x } \right) }^{ 2 }-1 } dx }$
Put

$x+\cfrac { 1 }{ x } =t\quad \Rightarrow 1-\cfrac { 1 }{ { x }^{ 2 } } dx=dt$

$\therefore I=\int { \cfrac { 1 }{ { t }^{ 2 }-1 } } dt$

$=\cfrac { 1 }{ 2\times 1 } \log { \cfrac { t-1 }{ t+1 } +c }$

$=\cfrac { 1 }{ 2 } \log { \cfrac { x+\cfrac { 1 }{ x } -1 }{ x+\cfrac { 1 }{ x } +1 } } +c$

$=\cfrac { 1 }{ 2 } \log { \cfrac { { x }^{ 2 }-x+1 }{ { x }^{ 2 }+x+1 } } +c$

$\displaystyle \int\frac{dx}{x(x^7+1)}$ is equal to

0%
$log\left(\dfrac{x^7}{x^7+1}\right)+c$
0%
$\dfrac{1}{7}log\left(\dfrac{x^7}{x^7+1}\right)+c$
0%
$log\left(\dfrac{x^7+1}{x^7}\right)+c$
0%
$\dfrac{1}{7}log\left(\dfrac{x^7+1}{x^7}\right)+c$

Explanation

$\displaystyle I= \int\dfrac{dx}{x(x^7+1)}$

multiplying and dividing by

$x^6$ we get

$\displaystyle I= \int\dfrac{x^6dx}{x^7(x^7+1)}$

put

$x^7 = t\rightarrow7x^6dx =dt$

$\displaystyle I= \dfrac17\int\dfrac{dt}{t(t+1)}$

$\displaystyle I=\dfrac17 \int\dfrac{dt}{t^2+t}\Rightarrow\int\dfrac{1}{ \begin{pmatrix}t^2+t+\dfrac14-\dfrac14\end{pmatrix}}dt$

$I=\dfrac17\displaystyle \int\dfrac{1}{(t+\dfrac12)^2-(\dfrac12)^2}dt$

$\displaystyle \Rightarrow I= \dfrac17\times\dfrac{1}{2\times\dfrac12}log\begin{pmatrix}\dfrac{t+\dfrac12-\dfrac12}{t+\dfrac12+\dfrac12}\end{pmatrix} + C$

$\begin{bmatrix}\because\displaystyle \int\dfrac{1}{x^2-a^2}dx =\dfrac{1}{2a}log\dfrac{x-a}{x+a}+ C\end{bmatrix}$

$\Rightarrow I= \dfrac17log\begin{pmatrix}\dfrac{t}{t+1}\end{pmatrix} + C$

$\Rightarrow I= \dfrac17log\begin{pmatrix}\dfrac{x^7}{x^7+1}\end{pmatrix} + C$

$\therefore \text {option B is correct}$

$\displaystyle \int {\frac{x^2}{(x\,\sin\,x+\cos\,x)^2}} dx$ is equal to

0%
$\displaystyle \frac{\sin\,x+\cos\,x}{x\,\sin\,x+\cos\,x}+C$
0%
$\displaystyle \frac{x\,\sin\,x-\cos\,x}{x\,\sin\,x+\cos\,x}+C$
0%
$\displaystyle \frac{\sin\,x-x\,\cos\,x}{x\,\sin\,x+\cos\,x}+C$
0%
None of these

Explanation

Since,

$\dfrac{d}{dx}(x\sin\, x+\cos\,x)=x\cos x+\sin x - \sin x=x\cos\,x$

$\therefore \displaystyle I=\int\frac{x^2dx}{(x\,\sin\,x+\cos\,x)^2}$

$=\displaystyle \int\frac{x}{\cos\,x}\cdot \frac{x\,\cos\,x}{(x\,\sin\,x+\cos\,x)^2}dx$

Solving by integration by parts,

$I=\displaystyle \frac{x}{\cos\,x}\cdot\left(\frac{-1}{x\,\sin\,x+\cos\,x}\right) - \int \dfrac{x\cos x + x\sin x}{\cos^2x}.\dfrac{-1}{(x\sin x+\cos x)}dx$

$=\displaystyle \frac{-x}{\cos\,x(x\,\sin\,x+\cos\,x)}+\int \sec^2 xdx$

$=\displaystyle \frac{-x} {\cos\,x(x\,\sin\,x+\cos\,x)}+\tan\,x+C$

$=\displaystyle \frac{-x}{\cos x(x\cos x + \sin x)}+\frac{\sin x}{\cos x}+C$

$= \displaystyle \frac{-x+\sin\,x(x\,\sin\,x+\cos\,x)}{\cos\,x(x\,\sin\,x+\cos\,x)}+C$

$=\displaystyle \frac{-x\,\cos^2\,x+\sin\,x\cdot \cos\,x}{\cos\,x(x\,\sin\,x+\cos\,x)}+C$

$=\displaystyle \left(\frac{\sin\,x-x\,\cos\,x}{\cos\,x+x\,\sin\,x}\right)+C$

$\displaystyle \int \displaystyle \frac{x\, -\, \sin x}{1\, +\, \cos x}\,dx\, =\, x \tan \left ( \displaystyle \frac{x}{2} \right )\, +\, p\, \log \left | \sec \left ( \displaystyle \frac{x}{2} \right ) \right |\, +\, c\, \Rightarrow\, p\, =$

0%
-4
0%
4
0%
2
0%
-2

Explanation

$\int \dfrac{x-sinx}{1+cosx}dx$

$=\int \dfrac{x}{1+cosx}dx+\int\dfrac{-sinx}{1+cosx}dx$

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

$=\dfrac{1}{2}\int xsec^{2}\dfrac{x}{2}dx-\int tan(\dfrac{x}{2})dx$

$=\dfrac{1}{2}[2xtan\dfrac{x}{2}-2\int tan\dfrac{x}{2}dx]-\int tan(\dfrac{x}{2})dx$

$=xtan\dfrac{x}{2}-2\int tan(\dfrac{x}{2})dx$

$=xtan\dfrac{x}{2}-4ln|sec\dfrac{x}{2}|+c$

Hence

$p=-4$

Let

$f(x)\, =\, 3x^{2}.\, \sin \,\displaystyle \frac{1}{x}\, -\, x\cos\, \displaystyle \frac{1}{x},\, x\, \neq\, 0, f(0)\, =\, 0\, f \left (\, \displaystyle \frac{1}{\pi} \right )\, =\, 0$ , then which of the following is/are not correct.

0%
$f(x)$ is continuous at $x = 0$
0%
$f(x)$ is non-differentiable at $x = 0$
0%
$f(x)$ is discontinuous at $x = 0$
0%
$f(x)$ is differentiable at $x = 0$

Explanation

$f(x)\, =\, 3x^{2}. \sin \displaystyle \frac{1}{x}\, -\, x.\cos\, \displaystyle \frac{1}{x}$

$\Rightarrow\, f(x)\, =\,\displaystyle \int \left ( 3x^{2}.\sin \displaystyle \frac{1}{x}\, -\, \cos\, \displaystyle \frac{1}{x} \right ) dx$

$=\, x^{3}\, \sin\, \displaystyle \frac{1}{x}.\, -\, \int \cos\, \frac{1}{x}\, \left (\, -\, \displaystyle \frac{1}{x^{2}} \right )\, x^{3}\, dx\, -\, \int x \cos \displaystyle \frac{1}{x}dx$

$=\, x^{3}\, \sin \, \displaystyle \frac{1}{x}\, +\, C$
Since

$f \left ( \displaystyle \frac{1}{\pi} \right )\, =\, 0\, +\, C \, \Rightarrow\, C\, =\, 0$

$\Rightarrow\, f(x)\left \{\begin{matrix} x^{3}\sin\, \displaystyle \frac{a}{x} & , & x \, \neq\, 0 \\ 0 & , & x = 0 \end{matrix} \right \}$

$F(x)$ is clearly continuous and differentiable at $x = 0$ zero with $f`(0) = 0$ .

$\displaystyle f\, (0)\, =\, \lim_{h \rightarrow 0} \frac{3h^2\, \sin \frac{1}{h}\, -\, h\, \cos \frac{1}{h}}{h}$ $\displaystyle =\, 3h\sin\frac{1}{h}\, -\, \cos \frac{1}{h}$

This limit doesn't exist, hence $f(x)$ is non-differentiable at $x = 0$ .

Also $\displaystyle \lim_{x \rightarrow 0}\, f`(x)\, =\, 0$ .

Thus $f`(x)$ is continuous at $x = 0$ .

$\displaystyle\int { \cfrac { \sqrt { x } }{ \sqrt { x } -\sqrt [ 3 ]{ x } } } dx$ is equal to

0%
$6\left\{ \cfrac { x }{ 6 }+ \cfrac { { x }^{ 6/5 } }{ 5 } +\cfrac { { x }^{ 1/2 } }{ 2 } +\cfrac { { x }^{ 1/3 } }{ 3 } +\log { \left( { x }^{ 1/6 }-1 \right) } \right\} +c$
0%
$6\left\{ \cfrac { x }{ 6 }+ \cfrac { { x }^{ 6/5 } }{ 5 } +\cfrac { { x }^{ 1/2 } }{ 3 } +\cfrac { { x }^{ 1/3 } }{ 2 } +\log { \left( { x }^{ 1/6 }-1 \right) } \right\} +c$
0%
$6\left\{ \cfrac { x }{ 6 }+ \cfrac { { x }^{ 6/5 } }{ 5 } +\cfrac { { x }^{ 1/2 } }{ 2 } +\cfrac { { x }^{ 1/3 } }{ 3 } +{ x }^{ 1/6 }+\log { \left( { x }^{ 1/6 }-1 \right) } \right\} +c$
0%
None of the above

Explanation

Let

$I=\displaystyle\int { \cfrac { \sqrt { x } }{ \sqrt { x } -\sqrt [ 3 ]{ x } } } dx$

$\displaystyle\int { \cfrac { { x }^{ 1/2 } }{ { x }^{ 1/2 }-{ x }^{ 1/3 } } } dx$

Let

$x={t}^{6}$

$\Rightarrow dx=6{ t }^{ 5 }dt$

$\therefore I=\displaystyle\int { \cfrac { { t }^{ 3 } }{ { t }^{ 3 }-{ t }^{ 2 } } } 6{ t }^{ 5 }dt$

$=6\displaystyle\int { \cfrac { { t }^{ 6 } }{ t-1 } dt } =6\int { \cfrac { { t }^{ 6 }-1+1 }{ t-1 } } dt$

After dividing we get,

$I=6\displaystyle\int { \left( { t }^{ 5 }+{ t }^{ 4 }+{ t }^{ 3 }+{ t }^{ 2 }+t+1+\cfrac { 1 }{ t-1 } \right) dt }$

$=6\left[ \cfrac { { t }^{ 6 } }{ 6 } +\cfrac { { t }^{ 5 } }{ 5 } +\cfrac { { t }^{ 4 } }{ 4 } +\cfrac { { t }^{ 3 } }{ 3 } +\cfrac { { t }^{ 2 } }{ 2 } +t+\log { \left( t-1 \right) } \right] +c$

$=6\left[ \cfrac { x }{ 6 } +\cfrac { { x }^{ 5/6 } }{ 5 } +\cfrac { { x }^{ 2/3 } }{ 4 } +\cfrac { { x }^{ 1/2 } }{ 3 } +\cfrac { { x }^{ 1/3 } }{ 2 } +{ x }^{ 1/6 }+\log { \left( { x }^{ 1/6 }-1 \right) } \right] +c\quad \left[ \because \quad x={ t }^{ 6 }\Rightarrow t={ x }^{ 1/6 } \right]$

Evaluate

$\displaystyle {\int \sin^{-1}\, \sqrt {\frac {x}{a\, +\, x}} dx}$

0%
$(a + x) \tan^{-1} \displaystyle \sqrt \frac {x}{a}\, -\, \sqrt {ax}\, +\, c$
0%
$\displaystyle \left(\frac { ax+1 }{ 2a } \right)\sin^{ -1 }\, \sqrt { \frac { x }{ a\, +\, x } } -\frac { 1 }{ 2 } \sqrt { \frac { x }{ a } } +c$
0%
$a \tan^{-1} \displaystyle \sqrt \frac {x}{a}\, -\, \sqrt {ax}\, +\, c$
0%
$(a + x) \tan^{-1} \displaystyle \sqrt {x}\, -\, \sqrt {ax}\, +\, c$

$\displaystyle \int x \frac{\ln (x + \sqrt{1 + x^{2}})}{\sqrt{1 + x^{2}}}\, dx$ equals to

0%
$\sqrt{1 + x^{2}} \ln (x + \sqrt{1 + x^{2}}) - x + c$
0%
$\displaystyle \frac{x}{2} \ln^{2} (x + \sqrt{1 + x^{2}}) - \frac{x}{\sqrt{1 + x^{2}}} + c$
0%
$\displaystyle \frac{x}{2} \ln^{2} (x + \sqrt{1 + x^{2}}) + \frac{x}{\sqrt{1 + x^{2}}} + c$
0%
$\sqrt{1 + x^{2}} \ln (x + \sqrt{1 + x^{2}}) + x + c$

Explanation

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

$\quad = \displaystyle \int \frac{2x}{2\sqrt{1 + x^{2}}}.\ln (x + \sqrt{1 + x^{2}}) dx$

Let

$I_1 = \displaystyle \int \frac{2x}{2\sqrt{1 + x^{2}}}.$

Put

$1+x^2=t\implies2x dx=dt$

$I_1=\displaystyle \int \dfrac{1}{2}t^{\tfrac{-1}{2}}dt=t^{\tfrac{1}{2}}= \sqrt{1 + x^{2}}$

Using integral by parts,

$\displaystyle \int u.v \ dx=u\int v\ dx$

$\displaystyle -\int \left ( \dfrac{du}{dx}\int v\ dx \right )dx$

$I = \displaystyle \ln (x + \sqrt{1 + x^{2}}) \int \frac{2x}{2\sqrt{1 + x^{2}}} dx -\int \left[ \frac{d\{\ln (x + \sqrt{1 + x^{2}})\}}{dx}. \int \frac{2x}{2\sqrt{1 + x^{2}}} dx\right ] dx+c$

$\quad = \displaystyle \ln (x + \sqrt{1 + x^{2}}). \sqrt{1 + x^{2}} -\int \left[ \frac{1}{x + \sqrt{1 + x^{2}}}. \left(1+\frac{x}{\sqrt{1+x^2}} \right).\sqrt{1 + x^{2}} \right ] dx+c$

$\quad = \displaystyle \ln (x + \sqrt{1 + x^{2}}). \sqrt{1 + x^{2}} -\int dx+c=\ln (x + \sqrt{1 + x^{2}}). \sqrt{1 + x^{2}} -x+c$

$\quad =\sqrt{1 + x^{2}} \ln (x + \sqrt{1 + x^{2}}) - x + c$

$\displaystyle \int { \cfrac { { x }^{ 3 } }{ \sqrt { 1+x^2 } } } dx$

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

Explanation

$I = \displaystyle \int \cfrac{x^3}{\sqrt{1 + x^2}} dx$

Put

$1+x^2=t\\2xdx=dt$

$I=\displaystyle \int \dfrac{t-1}{\sqrt t}\dfrac{dt}{2}$

$I=\displaystyle \int \sqrt t-\dfrac{1}{\sqrt t}\dfrac{dt}{2}$

$I=\cfrac{1}{2}\left[\cfrac{2}{3}t^{\tfrac{3}{2}}-2t^{\tfrac{1}{2}}\right]+c$

$I=\cfrac{(x^2+1)^{\tfrac{3}{2}}}{3}-(x^2+1)^{\tfrac{1}{2}}+c$

$I=\cfrac{(x^2+1)^{\tfrac{1}{2}}(1+x^2)}{3}-(x^2+1)^{\frac{1}{2}}+c$

$I=\dfrac{x^2\sqrt{1+x^2}}{3}-\dfrac{2\sqrt{1+x^t2}}{3}+c$

$\displaystyle \int { \sqrt { x } { e }^{ \sqrt { x } } } dx$ is equal to:

0%
$\displaystyle 2\sqrt { x } -{ e }^{ \sqrt { x } }-4\sqrt { { xe }^{ \sqrt { x } } } +c$
0%
$\displaystyle \left( 2x-4\sqrt { x } +4 \right) { e }^{ \sqrt { x } }+c$
0%
$\displaystyle \left( 2x+4\sqrt { x } +4 \right) { e }^{ \sqrt { x } }+c$
0%
$\displaystyle \left( 1-4\sqrt { x } \right) { e }^{ \sqrt { x } }+c$

Explanation

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

Let

$\sqrt{x} = t$

So,

$x = t^2$ and

$dx = 2tdt$

$I = \displaystyle \int 2t^2e^t dt$

$I = 2t^2e^t -\displaystyle \int \dfrac{d(2t^2)}{dt} e^t dt$

$\therefore I = 2t^2e^t -\displaystyle \int 4te^t dt$

$\therefore I = 2t^2e^t - 4te^t + 4\displaystyle \int e^t dt$

$\therefore I = 2t^2e^t - 4te^t + 4e^t + c$

$\therefore I = e^t(2t^2 - 4t + 4) + c$

$\therefore I = e^{\sqrt{x}} (2x - 4\sqrt{x} + 4) + c$

$\displaystyle\int \dfrac{cos^{n-1}}{sin^{n+1}}dx, n\neq 0$ is ____

0%
$\dfrac{cot^nx}{n}$
0%
$\cfrac{-cot^{n-1}x}{n-1}$
0%
$\cfrac{-cot^nx}{n}$
0%
$\cfrac{cot^{n-1}x}{n-1}$

Explanation

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

$=\int (\dfrac{cosx}{sinx})^{n}.\dfrac{1}{cosx.sinx}.dx$

$=\int cot^{n}x.\dfrac{sinx}{cosx}.\dfrac{1}{sin^{2}x}dx$

$=\int cot^{n}x.tanx.cosec^2x dx$

$=\int cot^{n-1}x.cosec^{2} x dx$

Let

$cot(x)=t$ , then

$cosec^{2}xdx=-dt$
Hence

$I=-\int t^{n-1}.dt$

$=\dfrac{-t^{n}}{n}$

$=-\dfrac{cot^{n}x}{n}$

$\int { { ({ x }^{ 2 }+5) }^{ 3 } } dx$

0%
$\cfrac { { x }^{ 7 } }{ 7 } +3{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$
0%
$\cfrac { { x }^{ 6 } }{ 7 } +3{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$
0%
$\cfrac { { x }^{ 7 } }{ 7 } +3{ x }^{ 5 }-25{ x }^{ 3 }+125x+C$
0%
$\cfrac { { x }^{ 7 } }{ 7 } +4{ x }^{ 5 }+25{ x }^{ 3 }+125x+C$

Explanation

$\int (x^2 + 5)^3 dx$

$= \int (x^6 + 3(x^2)^2 \times 5 + 3x^2 \times 5^2 + 5^3) dx$

$= \int (x^6 + 15x^4 + 75x^2 + 125) dx$

$= \cfrac{x^7}{7} + 15 \times \cfrac{x^5}{5} + 75 \times \cfrac{x^3}{3} + 125x + c$

$= \cfrac{x^7}{7} + 3x^5 + 25x^3 + 125x +c$

Evaluate:

$\displaystyle \int \dfrac{(x-1)e^x}{(x+1)^3}dx=$

0%
$\dfrac{e^x}{x+1}$
0%
$\dfrac{e^x}{(x+1)^2}$
0%
$\dfrac{e^x}{(x+1)^3}$
0%
$\dfrac{x\cdot e^x}{(x+1)}$

Explanation

Consider,

$\displaystyle I=\int \dfrac{(x-1)e^{x}}{(x+1)^{3}}dx$

$\displaystyle I =\int \dfrac{xe^{x}-e^{x}-2e^x+2e^x}{(x+1)^{3}}dx$

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

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

Applying ILATE rule to the first integral, gives us

$\displaystyle I =\dfrac{e^{x}}{(x+1)^{2}}-\int \left(\dfrac{-2e^{x}}{(x+1)^{3}}\right)dx-\int\dfrac{2e^{x}}{(x+1)^{3}}dx$

$\displaystyle I =\dfrac{e^{x}}{(x+1)^{2}}+\int \left(\dfrac{2e^{x}}{(x+1)^{3}}\right)dx-\int\dfrac{2e^{x}}{(x+1)^{3}}dx$

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

$\int { { x }^{ 4 }{ e }^{ 2x } } dx=$

0%
$\cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$
0%
$\cfrac { { e }^{ 2x } }{ 2 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$
0%
$\cfrac { { e }^{ 2x } }{ 8 } \left( 2{ x }^{ 4 }+4{ x }^{ 3 }+6{ x }^{ 2 }+6x+3 \right) +C$
0%
$\cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }+4{ x }^{ 3 }+6{ x }^{ 2 }+6x+3 \right) +C$

Explanation

$\int x^4.e^{2x}dx$

Applying by parts rule of integration,

$\dfrac{x^4}{2} e^{2x}-\dfrac{4}{2}\int x^3.e^{2x}dx$

Applying by-parts again, multiple times till the time powers in

$x$ in the integral vanishes,

$\dfrac{x^4}{2} e^{2x}-2[\dfrac{x^3}{2} e^{2x}-\dfrac{3}{2}\int x^2.e^{2x}dx]$

$\dfrac{x^4}{2} e^{2x}-x^3.e^{2x}+3[\dfrac{x^2}{2}.e^{2x}-\int x.e^{2x} dx]$

$\dfrac{x^4}{2} e^{2x}-x^3.e^{2x}+3\dfrac{x^2}{2}.e^{2x}-3[\dfrac{x}{2}.e^{2x}-\int\dfrac{e^{2x}}{2}dx]$

$\dfrac{x^4}{2} e^{2x}-x^3.e^{2x}+3\dfrac{x^2}{2}.e^{2x}-3\dfrac{x}{2}.e^{2x}+3\dfrac{e^{2x}}{4}+c$

$\rightarrow \cfrac { { e }^{ 2x } }{ 4 } \left( 2{ x }^{ 4 }-4{ x }^{ 3 }+6{ x }^{ 2 }-6x+3 \right) +C$

$\displaystyle \int \dfrac{\text{cosec}^2x - 2010}{\cos^{2010x}}dx=-\dfrac{f(x)}{(g(x))^{2010}}+C$ ; then the number of solutions where equation

$\dfrac{f(x)}{g(x)}=\left \{ x \right \}$ in

$[0, 2\pi]$ is/are:

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

Given :

$\displaystyle \int { \cfrac { (\text{cosec}^{ 2 }{ x } -2010)dx }{ { \cos ^{ 2010 }{ x } } } }$

Let

$\cfrac { \cot { x } }{ \cos ^{ 2010 }{ x } } =t\\ dt=\cfrac { \cos ^{ 2010 }{ x } (-\text{cosec}^{ 2 }{ x } +2010)dx }{ ({ \cos ^{ 2010 }{ x)^{ 2 } } } } \\ -dt=\cfrac { (\text{cosec}^{ 2 }{ x } -2010)dx }{ \cos ^{ 2010 }{ x } } \\ \displaystyle \int { -dt } =-t+C\\ =-\cfrac { \cot { x } }{ \cos ^{ 2010 }{ x } } \\ f(x)=\cot { x } \\ g(x)=\cos { x } \\ \cfrac { f(x) }{ g(x) } =\text{cosec }{ x } =\{ x\}$

Hence there is no solution.

Let

$f(x)$ be a positive function. Let

$I_{1} = \int_{1 - k}^{k} xf\left \{x(1 - x)\right \} dx$ ,

$I_{2} = \int_{1 - k}^{k} f\left \{x(1 - x) \right \} dx$ ,
where

$2k - 1 > 0$ , then

$\dfrac {I_{1}}{I_{2}}$ is

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

Explanation

$I_{1} = \displaystyle \int_{1 - k}^{k} xf \left \{x (1 - x)\right \}dx$

$= \displaystyle \int_{1 - k}^{k} (1 - x)f\left \{(1 - x)(1 -(1- x))\right \}dx$

$(\because \int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx)$

$=\displaystyle \int_{1 - k}^{k} (1 - x)f\left \{(1 - x)x\right \}dx$

$= \displaystyle \int_{1 - k}^{k} f\left \{x(1 - x)\right \} dx$

$- \displaystyle \int_{1 - k}^{k} xf \left \{x(1 - x)\right \}dx$
Therefore,

$I_{1} = I_{2} - I_{1}\Rightarrow 2I_{1} = I_{2}$

$\Rightarrow \dfrac {I_{1}}{I_{2}} = \dfrac {1}{2}$

$\displaystyle \int \dfrac{x + \sqrt[3]{x^2} + \sqrt[6]{x}}{x(1+\sqrt[3]{x})} dx = ax^{\frac{2}{3}} + b\ \tan^{-1}(\sqrt[6]{x}) +c$ then:

0%
$a= \frac{3}{2}$
0%
$a= \frac{3}{4}$
0%
$a= \frac{-3}{2}$
0%
$a= \frac{-3}{4}$

Explanation

$\int { \dfrac { x+{ x }^{ 2/3 }+{ x }^{ 1/6 } }{ x\left( { x }^{ 1/3 }+1 \right) } } dx$

$u={ x }^{ 1/6 }$

$du=\dfrac { 1 }{ 6 } \dfrac { dx }{ { x }^{ 5/6 } }$

${ x }^{ 1/3 }={ u }^{ 2 }$

${ x }^{ 2/3 }={ u }^{ 4 }$

$x={ u }^{ 6 }$

$=6\int { \dfrac { { u }^{ 5 }+{ u }^{ 3 }+1 }{ { u }^{ 2 }+1 } } du$

Polynomial division

$\Rightarrow 6\int { \left( \dfrac { 1 }{ { u }^{ 2 }+1 } +{ u }^{ 3 } \right) } du$

$\Rightarrow 6{ \tan }^{ -1 }\left( u \right) +\dfrac { { 6u }^{ 4 } }{ 4 } +c$

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

Let

$y = y(x), y(1)=1\ and\ y(e) ={e^2}$ . Consider

$J = \int {{{x + y} \over {xy}}} dy$ ,

$I = \int {{{x + y} \over {{x^2}}}} dx$ ,

$J - I = g\left( x \right)$ and g(1) = 1, then the value of g(e) is

0%
$2e+1$
0%
$e+1$
0%
${e^2}$ -e+1
0%
${e^2}$ +e-1

Explanation

$I=\int { \dfrac { x+y }{ xy } } dy\quad ;\quad I=\int { \dfrac { x+y }{ { x }^{ 2 } } } dx$

$I=\int { \left( \dfrac { 1 }{ y } +\dfrac { 1 }{ x } \right) } dy\quad ;\quad I=\int { \left( \dfrac { 1 }{ x } +\dfrac { y }{ { x }^{ 2 } } \right) } dx$

$I\Rightarrow Iny+\dfrac { y }{ x } \quad ;\quad I=Inx-\dfrac { y }{ x }$

$y\left( x \right) =J-I$

$\Rightarrow Iny-Inx+\dfrac { 2y }{ x }$

$g\left( x \right) \Rightarrow In\left( y/x \right) +\dfrac { 2y }{ x } =In\left( \dfrac { y\left( x \right) }{ 2 } \right) +\dfrac { 2y\left( x \right) }{ x }$

$g\left( 1 \right) =Iny+2y+1$

$g\left( 2 \right) =In\left| \dfrac { { e }^{ 2 } }{ e } \right| +\dfrac { 2{ e }^{ 2 } }{ e }$

$\Rightarrow 2e+1$

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

$\int { { f }^{ -1 }(x)dx }$ is

0%
$x{ f }^{ -1 }(x)+C$
0%
$f({ g }^{ -1 }(x))+C$
0%
$x{ f }^{ -1 }(x)-g({ f }^{ -1 }(x))+C$
0%
${ g }^{ -1 }(x)+C$

The value of

$\displaystyle \int{ \sqrt{\dfrac{e^x - 1}{e^x + 1}}dx}$ is equal to

0%
$ln (e^x + \sqrt{e^{2x} - 1} - sec^{-1} (e^x) + C$
0%
$ln (e^x + \sqrt{e^{2x} - 1} + sin^{-1} (e^x) + C$
0%
$ln ( e^x - \sqrt{e^{2x} - 1} - sec^{-1} (e^x) + C$
0%
none of these

Evaluate :

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

0%
$2\ell n[x \sin x+\cos x]-\dfrac {X^{2}}{x \tan x+1}+C$
0%
$x\ell n[x \sin x+\cos x]-\dfrac {X^{2}}{x \tan x+1}+C$
0%
$2\ell n[x \sin x+\cos x]-\dfrac {X^{2}}{x \tan x-1}+C$
0%
none

$\displaystyle\int { { x }^{ 13/2 }.{ \left( 1+{ x }^{ 5/2 } \right) }^{ 1/2 }dx } =P{ \left( 1+{ x }^{ 5/2 } \right) }^{ 7/2 }+Q{ \left( 1+{ x }^{ 5/2 } \right) }^{ 5/2 }+R{ \left( 1+{ x }^{ 5/2 } \right) }^{ 3/2 }+C$ , then

$P,\ Q$ and

$R$ are

0%
$P=\frac { 4 }{ 35 } ,\ Q=-\frac { 8 }{ 25 } ,\ R=\frac { 4 }{ 15 }$
0%
$P=\frac { 4 }{ 35 } ,\ Q=\frac { 8 }{ 25 } ,\ R=\frac { 4 }{ 15 }$
0%
$P=-\frac { 4 }{ 35 } ,\ Q=-\frac { 8 }{ 25 } ,\ R=\frac { 4 }{ 15 }$
0%
$P=\frac { 4 }{ 35 } ,\ Q=-\frac { 8 }{ 25 } ,\ R=-\frac { 4 }{ 15 }$

Explanation

Take

$1+x^{5/2}=t^2$

$\Rightarrow \dfrac{5}{2}x^{3/2}dx=2t \space dt$

Rewrite the given integral as

$\Rightarrow \int{x^5(1+x^{5/2})^{1/2}}x^{3/2}dx$

Substitutng

$t$ we get

$\Rightarrow \dfrac{4}{5}\int(t^2-1)^2.t^2 dt$

$\Rightarrow \dfrac{4}{5}\int(t^4-2t^2+1).t^2 dt$

$\Rightarrow \dfrac{4}{5}\int(t^6-2t^4+t^2) dt$

$\Rightarrow \dfrac{4}{5} (\dfrac{t^7}{7}-\dfrac{2t^5}{5}+\dfrac{t^3}{3})+C$

$\Rightarrow \dfrac{4t^7}{35}-\dfrac{8t^5}{25}+\dfrac{4t^3}{15}+C$

$\Rightarrow \dfrac{4}{35}(1+x^{5/2)^{7/2}}-\dfrac{8}{25}(1+x^{5/2})^{5/2}+\dfrac{4}{15}(1+x^{5/2})^{3/2}+C$

Comparing this with given expression we get

$P=\dfrac{4}{35}, Q=\dfrac{-8}{25}, R=\dfrac{4}{15}$

The value of

$\int (x\ e^{\ell n \sin x}-\cos x)dx$ is equal to:

0%
$x \cos x+C$
0%
$\sin x-x\ \cos +C$
0%
$-e^{\ell n x}\cos x+C$
0%
$\sin x+x \cos x+C$

Explanation

$\int(xe^{ln \space sinx}-cosx)dx$

We know that

$e^{ln \space a}=a$

This property can be substituted in the above integral, we get,

$\Rightarrow \int(xe^{ln \space sinx}-cosx)dx=\int{(x\space sinx-cosx)dx}$

$\Rightarrow \int{x \space sinx\space dx}- \int cosx \space dx$

Use integration by parts for the first integral,

$\int{udv}=uv-\int {vdu}$ we get

$\Rightarrow \int{x \space sinx\space dx}- \int cosx \space dx=(-x \space cosx+sinx)-sinx +C$

$\Rightarrow \int{x \space sinx\space dx}- \int cosx \space dx=-x \space cosx+ C$

$\Rightarrow \int(xe^{ln \space sinx}-cosx)dx=-x\space cosx+ C$

Since

$e^{ln\space x}=x$ , we can write the above integral as

$\int(xe^{ln \space sinx}-cosx)dx=-e^{ln\space x} \space cosx+ C$

One of the roots of the equation

$2000x^6+100x^5+10x^3+x-2=0$ is of the form

$\dfrac{m+\sqrt{n}}{r}$ . When 'm' is non zero integer and n and r relatively prime natural numbers. Then

$\dfrac{m+n+r}{100}=?$

0%
$100$
0%
$2$
0%
$3$
0%
$0$

$\int {x{{\sin }^{ - 1}}xdx}$ =?

0%
$\frac{1}{4}{\sin ^{ - 1}}(x)(2x + 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
0%
$\frac{1}{4}{\sin ^{ - 1}}(x)(2x^2 - 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
0%
$\frac{1}{4}{\cos ^{ - 1}}(x)(2x - 1) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c$
0%
$\frac{1}{4}{\sin ^{ - 1}}(x)(2x - 1) - \frac{{x\sqrt {1 - {x^2}} }}{4} + c$

Solve :

$\int x . \sin^2 x \, dx$

0%
$\dfrac {x^2}{4}+x\dfrac {\sin 2x}{6}+\dfrac {\cos 2x}{8}$
0%
$\dfrac {x^2}{4}-x\dfrac {\sin 2x}{4}-\dfrac {\cos 2x}{8}$
0%
$\dfrac {x^2}{6}+x\dfrac {\sin 2x}{4}-\dfrac {\cos 2x}{8}$
0%
$\dfrac {x^2}{4}-x\dfrac {\cos 2x}{4}-\dfrac {\sin 2x}{8}$

The value of

$\displaystyle\int\left(\dfrac{x^2+\cos^2x}{1+x^2}\right)cosec^2xdx$ equals?

0%
$\cot x+\tan^{-1}x+c$
0%
$\cot x-\tan^{-1}x+c$
0%
$-(\cot x+\tan^{-1}x)+c$
0%
$-\cot x-\cot^{-1}x+c$

The value of the integer

$I=\int_{1}^{\infty} \dfrac{(x^{2}-x)}{x^{3}\sqrt{(x^{2}-1)}}dx$ is

0%
$0$
0%
$2/3$
0%
$4/3$
0%
$none\ of\ these$

Explanation

Let,

$x = \sec t \implies dx = \sec t \tan t dt$

$x : 1 \rightarrow \infty, \sec t : 0 \rightarrow \pi/2$

$I = \int_{1}^{\infty} \dfrac{(x^2 - x)}{x^3 \sqrt{x^2 - 1}} dx$

$I = \int_{1}^{\infty} \dfrac{(x - 1)}{x^2 \sqrt{x^2 - 1}} dx$

$I = \int_{0}^{\pi/2} \dfrac{(\sec t - 1)}{\sec^2 t \sqrt{\sec^2 t - 1}} \sec t \tan t dt$

$I = \int_{0}^{\pi/2} \dfrac{(\sec t - 1)}{\sec^2 t . \tan t} \sec t \tan t dt$

$I = \int_{0}^{\pi/2} \dfrac{(\sec t - 1)}{\sec t } dt$

$I = \int_{0}^{\pi/2} (1 - \cos t ) dt$

$I = [t - \sin t]_{0}^{\pi/2}$

$I = \dfrac{\pi }{2} - 1$ (Ans)

Option D is correct.

The integral

$\int _{ \tfrac { \pi }{ 12 } }^{ \tfrac { \pi }{ 4 } }{ \dfrac { 8\cos { 2x } }{ { \left( \tan { x } +\cot { x } \right) }^{ 3 } } dx }$ equals:

0%
$\dfrac {15}{128}$
0%
$\dfrac {5}{64}$
0%
$\dfrac {13}{32}$
0%
$\dfrac {13}{256}$

Explanation

$\displaystyle\int^{{\tfrac{\pi}{4}}}_{\tfrac{\pi}{12}} \dfrac{8\ cos2x}{{(tan x + cot x)^3}}dx$

$\displaystyle \int^{\tfrac{\pi}{4}}_{{\tfrac{\pi}{12}}} \frac{8\ cos2x}{(\dfrac{sinx}{cosx}+\dfrac{cosx}{sinx})^3}dx$

$\displaystyle \int^{\tfrac{\pi}{4}}_{\tfrac{\pi}{12}} \dfrac{8\ cos2x\ sin^3x \ cos^3x}{(sin^2x +cos^2x )^3}dx$

$\displaystyle \int^{\tfrac{\pi}{4}}_{\tfrac{\pi}{12}} \dfrac{8\ cos2x \ (8\ sin^3x\ cos^3x)}{(sin^2x +cos^2x )^3}\dfrac{dx}{8}$

$\displaystyle \int^{\tfrac{\pi}{4}}_{\tfrac{\pi}{12}} \dfrac{8\ cos2x (sin2x)^3 }{(sin^2x +cos^2x )^3}\frac{dx}{8}$

$\int^{\tfrac{\pi}{4}}_{\tfrac{\pi}{12}} {cos2x (sin2x)^3 }{}dx$

put

$\displaystyle sin 2x=t\rightarrow2\ cos2x.dx=dt$

when

$x=\frac{\pi}{12}\rightarrow t= \frac12$

and

$x=\frac{\pi}{4}\rightarrow t= 1$

$\displaystyle\int^{1}_{\frac12} {cos2x.t^3 }{}\frac{dx}{2cos2x}$

$\displaystyle \int^{1}_{\frac12} {\frac{t^3 }{2}}{}dx$

$\displaystyle \frac12[\frac{t^4}{4}]^{1}_{\frac12}$

$\displaystyle \frac18[1-\frac{1}{16}]$

$\displaystyle \frac{15}{128}$

Hence option A is correct

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.