If $${ I }_{ n }=\int { \sin ^{ n }{ x } dx },$$, then $${ nI }_{ n }-\left( n-1 \right) { I }_{ n-2 }=$$

$$\cos ^{ n-1 }{ x\sin{ x } }$$

$$-\cos ^{ n-1 }{ x\sin{ x } }$$

MCQ Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 6

$\int e^x (\sin x + 2 \cos x) \sin x\, dx$ is equal to

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$e^x \cos x + C$
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$e^x \sin x + C$
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$e^x \sin^2 x + C$
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$e^x \sin 2x + C$
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$e^x (\cos x + \sin x) + C$

Explanation

Let

$I = \int e^x (\sin x + 2 \cos x) \sin x dx$

$= \int e_{II}^x \sin_I^2 x dx + \int 2e^x \sin x \cos x dx$
Applying integration by parts in first integral, we get

$I = e^x \sin^2 x - \int 2 \sin x \cos x e^x dx + 2 e^x \sin x \cos x dx + C$

$= e^x \sin^2 x + C$

$\displaystyle\int { \cfrac { 1 }{ 7 } \sin { \left( \cfrac { x }{ 7 } +10 \right) } dx }$ is equal to

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$\cfrac { 1 }{ 7 } \cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$
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$-\cfrac { 1 }{ 7 } \cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$
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$-\cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$
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$-7\cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$
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$\cos { \left( x+70 \right) } +C$

Explanation

Let

$I=\int { \cfrac { 1 }{ 7 } \sin { \left( \cfrac { x }{ 7 } +10 \right) } dx }$

$=\cfrac { 1 }{ 7 } \int { \sin { \left( \cfrac { x }{ 7 } +10 \right) } } =\cfrac { 1 }{ 7 } \cfrac { -\cos { \left( \cfrac { x }{ 7 } +10 \right) } }{ \cfrac { 1 }{ 7 } }$

$=-\cos { \left( \cfrac { x }{ 7 } +10 \right) } +C$

$\displaystyle \int \dfrac{xe^x}{(1 + x)^2} dx$ is equal to

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$\dfrac{-e^x}{x + 1} + C$
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$\dfrac{e^x}{x + 1} + C$
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$\dfrac{xe^x}{x + 1} + C$
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$\dfrac{-xe^x}{x + 1} + C$
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$\dfrac{e^x}{(x + 1)^2} + C$

Explanation

Let

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

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

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

Applying integration by parts in first integral, we get

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

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

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

$\displaystyle\int { \sqrt { 1+\sin { x } } \cdot f\left( x \right) dx } =\dfrac { 2 }{ 3 } { \left( 1+\sin { x } \right) }^{ { 3 }/{ 2 } }+C$ , then

$f\left( x \right)$ is equal to

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$\cos { x }$
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$\sin { x }$
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$\tan { x }$
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$1$

Explanation

$\int { \sqrt { 1+\sin { x } } } .f\left( x \right) dx=\dfrac { 2 }{ 3 } { (1+\sin { x } ) }^{ { 3 }/{ 2 } }+c$

On differentiating both sides, we get

$\sqrt { 1+\sin { x } } .f\left( x \right) =\dfrac { 2 }{ 3 } .\dfrac { 3 }{ 2 } { (1+\sin { x } ) }^{ { 1 }/{ 2 } }.\cos { x } +0\\ \Rightarrow f\left( x \right) =\cos { x }$

$\displaystyle\int { { x }^{ x }\log { \left( ex \right) } dx }$ is equal to

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${ x }^{ x }+c$
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$x\cdot \log { x } +c$
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${ \left( \log { x } \right) }^{ x }+c$
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${ x }^{ \log { x } }+c$

Explanation

Let

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

$=\displaystyle\int { { x }^{ x }\left( 1+\log { x } \right) dx }$
Let

$t={ x }^{ x }={ e }^{ x\log { x } }$

$\Rightarrow \dfrac { dt }{ dx } ={ x }^{ x }\left\{ x\cdot \dfrac { 1 }{ x } +\log { x } \right\}$

$\Rightarrow dt={ x }^{ x }\left( 1+\log { x } \right) dx$
Therefore,

$I=\displaystyle\int { t+c } ={ x }^{ x }+c$

$\int \dfrac {2x + 5}{\sqrt {7 - 6x - x^{2}}} dx = A\sqrt {7 - 6x - x^{2}} + B\sin^{-1} \left (\dfrac {x + 3}{4}\right ) + C$
(where

$C$ is a constant of integration), then the ordered pair

$(A, B)$ is equal to

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$(-2, -1)$
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$(2, -1)$
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$(-2, 1)$
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$(2, 1)$

Explanation

Note that

$7-6x-x^2 = 16 - (x+3)^2$ and

$\dfrac{d}{dx} (7-6x-x^2) = -2x-6$

So, we have

$\int \dfrac{2x+5}{\sqrt{7-6x-x^2}} dx = \int \dfrac{2x+6}{\sqrt{7-6x-x^2}} dx - \int \dfrac{1}{\sqrt{16-(x+3)^2}} dx\\ = -2\sqrt{7-6x-x^2} - \sin^{-1} (\dfrac{x+3}{4}) + C$

So, we have

$A = -2, \ B = -1$ .

Thus option A is the correct answer.

$f\left( x \right)$ satisfies the relation

$f\left( \dfrac { 5x-3y }{ 2 } \right) =\dfrac { 5f\left( x \right) -3f\left( y \right) }{ 2 } \forall \ x,y\in R$ and

$\ f\left( 0 \right) =3$ and

$f^{ ' }\left( 0 \right) =2$ , then the period of

$\sin { \left( f\left( x \right) \right) }$ is

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$2\pi$
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$\pi$
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$3\pi$
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$4\pi$

$\displaystyle\int\displaystyle\frac{e^{\cot^{-1}x}}{1+x^2}(x^2-x+1)dx$

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$\displaystyle\frac{e^{\cot^{-1}x}}{1+x^2}$
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$x\cdot e^{\cot^{-1}x}$
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$e^{\cot^{-1}x}$
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$-e^{\cot^{-1}x}$

Explanation

Consider,

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

$\Rightarrow$

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

$\Rightarrow$

$\displaystyle \int e^{cot^{-1}x}=xe^{cot^{-1}x}+\int \dfrac{xe^{cot^{-1}x}}{1+x^2}+C$

$\Rightarrow$

$\displaystyle \int e^{cot^{-1}x}-\int \dfrac{xe^{cot^{-1}x}}{1+x^2}dx = xe^{cot^{-1}x}+C$

$\displaystyle \Rightarrow I = xe^{cot^{-1}x}+C$

Hence, the answer is option (B).

Let

$I=\displaystyle \int _{ \pi /4 }^{ \pi /3 }{ \cfrac { \sin { x } }{ x } } dx$ . Then?

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$\cfrac { 1 }{ 2 } \le I\le 1\quad$
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$4\le I\le 2\sqrt { 30 }$
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$\cfrac { \sqrt { 3 } }{ 8 } \le I\le \cfrac { \sqrt { 2 } }{ 6 }$
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$1\le I\le \cfrac { 2\sqrt { 3 } }{ \sqrt { 2 } }$

Explanation

$I=\displaystyle \int^{\tfrac{\pi}{3}}_{\tfrac{\pi}{4}}\dfrac{sinx}{x}dx$

$\dfrac{sinx}{x}$ is a decreasing function in given interval

difference of limits

$=\dfrac{\pi}{3}-\dfrac{\pi}{4}=\dfrac{\pi}{12}$

so,

$\dfrac{\pi}{12}\cdot\dfrac{sin\dfrac{\pi}{3}}{\dfrac{\pi}{3}} \le I \le \dfrac{\pi}{12} \cdot\dfrac{sin\dfrac{\pi}{4}}{\dfrac{\pi}{4}}$

$\dfrac{\sqrt3}{8} \leq I \leq \dfrac{\sqrt2}{6}$

$\int { { \left( ex \right) }^{ x }\left( 2+\log { x } \right) } dx=....+c,x\in { R }^{ + }-\left\{ 1 \right\}$

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${ x }^{ x }$
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${ \left( ex \right) }^{ x }\quad$
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${ e }^{ x }$
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$\left( 1+\log { x } \right) { \left( ex \right) }^{ x }$

Explanation

$\int (ex)^x(2+\log x)dx=\int (ex)^xdx+\int e^x.x^x(1+\log x)dx$

Integrating by parts ,

$\Rightarrow \int(ex)^x +e^x(\int x^x(1+\log x)) -\int(\dfrac{d}{dx}(e^x).(\int x^x(1+\log x)dx))dx$

$\Rightarrow \int (ex)^x+ e^xx^x-\int (ex)^x$

$\Rightarrow (ex)^x$

$\displaystyle \int \tan^{-1} x dx=$ __________

$+c$ .

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$x\tan^{-1}x+\dfrac{1}{2}log\left|\dfrac{\tan^{-1}x}{x^2+1}\right|$
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$x\tan^{-1}x-\dfrac{1}{2}log\left|x^2+1\right|$
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$x\tan^{-1}x+\dfrac{1}{2}log\left|x^2+1\right|$
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$\dfrac{1}{1+x^2}$

Explanation

$I=\displaystyle\int \tan^{-1}x$ dx
Here taking u

$=\tan^{-1}x$ and v

$=1$

$\therefore \dfrac{du}{dx}=\dfrac{1}{1+x^2}$ and

$\displaystyle \int v dx=\displaystyle \int 1 dx=x$

$I=u\displaystyle\int v dx-\displaystyle\int \left(u'\displaystyle\int v dx\right) dx$
(Rule of integration by parts)

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

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

$I=x \tan^{-1}x-\dfrac{1}{2}\displaystyle\int \dfrac{f'(x)}{f(x)}dx$ ,
Where

$f(x)=1+x^2$

$=x\tan^{-1}x-\dfrac{1}{2}log |f(x)|+c$

$\therefore I=x\tan^{-1}x-\dfrac{1}{2}log |1+x^2|+c$ .

Solve

$\int { { e }^{ 7x+\log { x } } } dx=.....+c$

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${ e }^{ 7x }\left[ 7x-1 \right]$
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$\cfrac { { e }^{ x } }{ 98 } \left[ 7x-1 \right] \quad$
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$\cfrac { { e }^{ x } }{ 49 } \left[ 7x-1 \right] \quad$
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${ e }^{ 7x }\quad$

Explanation

$I=\int { \left( { e }^{ 7x+\log { x } } \right) } dx$

$=\int { { e }^{ 7x }.{ e }^{ \log { x } } } dx=\int { { e }^{ 7x }.x } dx$
Take

$u=x;v={ e }^{ 7x }$

$\therefore \cfrac { du }{ dx } =1;\int { v } dx=\cfrac { { e }^{ 7x } }{ 7 }$

$I=u\int { v } dx-\int { \left( u'\int { v } dx \right) } dx$
(rule of integration by parts)

$\therefore I=\cfrac { x{ e }^{ 7x } }{ 7 } -\int { \left( 1.\cfrac { { e }^{ 7x } }{ 7 } \right) } dx=\cfrac { { e }^{ 7x }(x) }{ 7 } -\cfrac { { e }^{ 7x } }{ 49 } +c$

$\therefore I=\cfrac { { e }^{ 7x } }{ 49 } \left( 7x-1 \right)$

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

$\displaystyle\int f^{-1}(x)dx=$ _____________

$+$ c.

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$x\cdot f(x)-g(f^{-1}(x))$
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$xf^{-1}(x)-g(f^{-1}(x))$
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$xf^{-1}(x)-g(f(x))$
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$x\cdot f^{-1}(x)$

Explanation

Now lets use integration by parts:

$\int { f^{ -1 }\left( x \right) dx } =xf^{ -1 }\left( x \right) -\int { x } \frac { d }{ dx } \left[ f^{ -1 }\left( x \right) \right] dx$

Now in the second integral, lets use substitution

$x=f(u)$ ;

$\\ \int { f^{ -1 }\left( x \right) dx } =xf^{ -1 }\left( x \right) -\int { f\left( u \right) } d\left[ f^{ -1 }\left( f\left( u \right) \right) \right] \\ \int { f^{ -1 }\left( x \right) dx } =xf^{ -1 }\left( x \right) -g\left( u \right) +C$

But

$u=f^{ -1 }\left( x \right)$ ,

$\therefore \int { f^{ -1 }\left( x \right) dx } =xf^{ -1 }\left( x \right) -g\left( f^{ -1 }\left( x \right) \right) +C$

The value of

$\int \dfrac{d^2}{dx^2} \, (tan^{-1}x) \, dx$ is equal to

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$\dfrac{1}{1 \, + \, x^2}$ + C
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$tan^{-1}x$ + C
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x tan - $\dfrac{1}{2} \, log (1 \, + \, x^2) \, + \, C$
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$\dfrac{1 \, + \, x^2}{2}$ + C

Explanation

$\displaystyle Q \int \dfrac{d^2}{dx^2} (\tan^{-1 }x) dx$

$\displaystyle = \int \dfrac{d}{dx} \dfrac{d}{dx} (\tan^{-1 }x)dx$

$\displaystyle = \int \dfrac{d}{dx }\dfrac{1}{1+x^2} dx$

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

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

put

$\displaystyle \Rightarrow 1+ x^2 = t \Rightarrow 2x \, dx = dt$

$\displaystyle = \int \dfrac{-1}{t^2} dt = \dfrac{-t^{-2+1}}{-2 +1} +c =\dfrac{1}{t} +c$

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

1070052_993509_ans_9e7d8a7435844177bed9c8b6f311bf45.jpeg

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

0%
${ x }^{ 3 }\tan ^{ -1 }{ x }$
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$\cfrac { { x }^{ 3 } }{ 3 } \tan ^{ -1 }{ x }$
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${ x }^{ 2 }\tan ^{ -1 }{ x }$
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$\cfrac { { x }^{ 2 } }{ 2 } \tan ^{ -1 }{ x }$

Explanation

Given

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

$\int (3x^2.\tan^{-1}x.dx)+\int(\dfrac{x^3}{1+x^2})dx$

Integrating by parts

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

$\dfrac{d}{dx}x^3=3x^2;\int \dfrac{1}{1+x^2}=tan^{-1}x$

$\Rightarrow\int 3x^2tan^{-1}x.dx + x^3 \tan^{-1}x-\int 3x^2tan^{-1}x.dx$

$\Rightarrow x^3\tan^{-1}x+c$

$I= \int \frac{x+2}{(x+1)^2}dx;$ then I is equal to

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$\log (x+1)+\dfrac{1}{x+1}+c$
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$\log (x+2)-\dfrac{1}{x+1}+c$
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$\log (1+x)-\dfrac{1}{x+1}+c$
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$\log (x+2)+\dfrac{1}{x+1}+c$

Explanation

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

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

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

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

$\left[ \because \int { { x }^{ n }dx=\dfrac { { x }^{ n+1 } }{ n+1 } +c } \right]$

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

$\left[ \because \int { \dfrac { 1 }{ x } dx=\log { x+c } } \right]$

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

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

Hence, the answer is

$\log\left(x+1\right)-\dfrac{1}{x+1}+c.$

$g(1)=g(2)$ then the value of

$\int _{ 1 }^{ 2 }{ { \left[ f\{ g(x)\} \right] }^{ -1 } } f'\{ g(x)\} g'(x)dx\quad is$

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$1$
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$2$
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$0$
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none of these

Explanation

$\because g(1)=g(2)\\ \int _{ 1 }^{ 2 }{ { [f\{ g(x)\} ] }^{ -1 } } f^{ ' }\{ g(x)\} g^{ ' }\left( x \right) dx\\ \int _{ 1 }^{ 2 }{ \cfrac { 1 }{ f\{ g(x)\} } } f^{ ' }\{ g(x)\} g^{ ' }\left( x \right) dx\quad [\because \int { \cfrac { 1 }{ x } dx } =\log { x } ]\\ \therefore { =[\log { f(g(x)) } ] }_{ 1 }^{ 2 }\\ \Rightarrow \log { f(g(2)) } -\log { f(g(1)) } \\ \Rightarrow \log { f(t) } -\log { f(t) } \quad [let\quad g(1)=g(2)=t]\\ =0$

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

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${{{2^x}.{e^x}} \over {1 - \ell n2}} + C$
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${{{2^x}.{e^x}} \over {1 + \ell n2}} + C$
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${{{2^x}.{e^x}} \over { - 1 + \ell n2}} + C$
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${{{{\left( {2e} \right)}^x}} \over {\ell n\left( {2e} \right)}} + C$

Explanation

$\displaystyle\int 2^{x}.e^{x}d{x}=\int e^{(1+\ln 2)x}d{x}=\dfrac{e^{(1+\ln 2)x}}{1+\ln 2}+C=\dfrac{2^{x}.e^{x}}{1+\ln 2}+c$

$y= \int \sqrt{1+\sin2x} dx; y$ is equal to-

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$\sin x - \cos x+c$
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$\sin x+ \cos x +c$
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$\cos x - \sin x +c$
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None of these

Explanation

$\int { \sqrt { 1+\sin { 2x } } dx }$

$=\int { \sqrt { \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } +2\sin { x } \cos { x } } dx }$

$=\int { \sqrt { { \left( \cos { x } +\sin { x } \right) }^{ 2 } } dx }$

$=\int { \left( \cos { x } +\sin { x } \right) dx }$

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

$=\sin x-\cos x+c.$

Hence, the answer is

$\sin x-\cos x+c.$

$I^n \, = \, {I_{0}}^{\pi/4} \, tan ^n \, X sec ^2 X dx$ then

$I_1 , \, I _2 \, , \, I_3 \,$ are in

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A. P
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G . P
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H . M
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none

Explanation

$I_n \, = \,\left [ \dfrac {tan^{n + 1}x}{n \ + \, 1} \right ]_{0}^{\pi4} \, = \, \dfrac {1}{n \, + \, 1}$

$I_! , \, I _2 \, , \, I_3 \,$ are

$\dfrac{1}{2} \, , \, \dfrac{1}{3} \, , \, \dfrac{1}{4} \,$ Which are in H . P and there reciprocals are in A . P .

$\displaystyle \int _0^{\pi/2} \sin x \cos x dx$ is equal to:

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$\dfrac 1 2$
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$\dfrac 14$
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$2$
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$1$

Explanation

$\displaystyle \int _0^{\pi/2} \sin x \cos x dx$

$\sin x=t\implies \cos x dx=dt$

$x\to 0\to \dfrac \pi 2$

$t\to 0\to 1$

$\Rightarrow \displaystyle \int _0^{\pi/2} t dt$

$\Rightarrow\left.\dfrac {t^2}2\right|^1_0$

$\Rightarrow\dfrac 12-0=\dfrac 12$

The value of

$\displaystyle \int \dfrac{dx}{4x^2 +9}$ is:

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

Explanation

Use Integral formula:

$\int \frac{dx}{a^{2}+x^{2}}=\frac{1}{a}tan^{-1}\left (\frac{x}{a} \right )$

now,

$\int \frac{dx}{9+4x^{2}}=\frac{1}{4}\int \frac{dx}{\frac{9}{4}+x^{2}}$

$=\frac{1}{4}\int \frac{dx}{\left ( \frac{3}{2} \right )^{2}+x^{2}}$

$=\left (\frac{1}{4} \right )\left ( \frac{2}{3} \right )tan^{-1}(\frac{2x}{3})+c$

$=\frac{1}{6}tan^{-1}(\frac{2x}{3})+c$

Evaluate :

$\displaystyle \int \sqrt{\dfrac{a - x}{x}} dx$

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

Find the derivative of

$\dfrac{e^{x}}{\sin x}$ .

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$e^x\text{cosec }x[\cot x+1]$
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$e^x\text{cosec }x[1-\cot x]$
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$e^x\sec x[\cot x+1]$
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$e^x\sec x[\cot x-1]$

Explanation

On Differentiation we get

$\dfrac d{dx} e^xcosec x=-e^x.cosec x\cot x+e^x.cosec x=e^x.cosec x[1-\cot x]$

${ I }_{ n }=\int { \sin ^{ n }{ x } dx },$ , then

${ nI }_{ n }-\left( n-1 \right) { I }_{ n-2 }=$

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$\sin ^{ n-1 }{ x\cos { x } }$
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$\cos ^{ n-1 }{ x\sin{ x } }$
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$-\sin ^{ n-1 }{ x\cos { x } }$
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$-\cos ^{ n-1 }{ x\sin{ x } }$

Explanation

${I_n} = \int_{}^{} {{{\sin }^n}xdx}$

$= \int_{}^{} {{{\sin }^{n - 1}}x\sin xdx}$

$= - {\sin ^{n - 1}}x\cos x - \int_{}^{} {\left( {n - 1} \right){{\sin }^{n - 2}}x\cos x \times - \cos xdx}$

$= - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int_{}^{} {{{\sin }^{n - 2}}x\left( {1 - {{\sin }^2}x} \right)dx}$

$\ = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int_{}^{} {\left( {{{\sin }^{n - 2}}x - {{\sin }^n}x} \right)dx}$

$= - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int_{}^{} {{{\sin }^{n - 2}}xdx - \left( {n - 1} \right)\int_{}^{} {{{\sin }^n}xdx} }$

$\Rightarrow {I_n} = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right){I_{n - 2}} - \left( {n - 1} \right){I_n}$

$\Rightarrow n{I_n} = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right){I_{n - 2}}$

$\Rightarrow n{I_n} - \left( {n - 1} \right){I_{n - 2}} = - {\sin ^{n - 1}}x\cos x$

Solve

$\displaystyle \int {\frac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}dx}$

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

Explanation

$\smallint \,\,\dfrac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}dx$

$\smallint x{e^x}.\,\dfrac{1}{{{{\left( {x + 1} \right)}^2}}}dx$

$x{e^x}\,\smallint \,\,\dfrac{{dx}}{{{{\left( {x + 1} \right)}^2}}}\, - \,\smallint \left[ {\left( {x{e^2} + {e^x}} \right) \times \,\,\,\smallint \,\dfrac{{dx}}{{{{\left( {1 + x} \right)}^2}}}} \right]dx$

$x{e^x}\left( {\dfrac{{ - 1}}{{x + 1}}} \right) - \,\smallint \dfrac{{ - \left( {x{e^x} + {e^x}} \right)}}{{x + 1}}dx$

$\dfrac{{ - x{e^x}}}{{x + 1}} + \,\,\smallint \,{e^x}\,dx$

$\dfrac{{ - x{e^x}}}{{x + 1}} + {e^x} + c$

$\dfrac{{ - x{e^x} + {e^x}\left( {x + 1} \right)}}{{x + 1}} + c$

$\dfrac{{{e^x}}}{{x + 1}} + c$

The value of

$\int { { e }^{ { \tan ^{ -1 } { x } } }} \dfrac { \left( 1+x{ +x }^{ 2 } \right) }{ 1{ +x }^{ 2 } } dx$ is ?

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$\tan ^{ -1 }{ x } +C$
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$e^{\tan^{-2}{x}}+2x+C$
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$e^{\tan^{-1}{x}}+C$
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${ xe }^{ \tan ^{ -1 }{ x+C } }$

Explanation

$\displaystyle \tan^{-1} x = t \quad \Rightarrow \quad x = \tan t$

$\displaystyle dx. \frac{1}{1+x^2} = dt.$

$\displaystyle I = \int e^t (1+\tan t + \tan^2 t) dt$

$\displaystyle = \int e^t (\tan t+ \sec^2 t ) dt \quad \left[\because \int e^x (f(x) + f'(x). dx = e^x f(x) +c\right]$

$\displaystyle = e^t \tan t +c$

$\displaystyle = e^{\tan^{-1} x} \tan (\tan^{-1}x) + c$

$\displaystyle I = x \, e^{\tan^{-1}x} + c$

$\displaystyle \int {\dfrac{{{2^x}}}{{\sqrt {1 - {4^x}} }}dx = K{{\sin }^{ - 1}}\left( {{2^x}} \right) + C} ,$ then K is equal to

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$\ell n2$
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$\dfrac{1}{2}\ell n2$
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$\dfrac{1}{2}$
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$\dfrac{1}{{\ell n2}}$

Explanation

Consider

$I = \displaystyle \int \dfrac {2^x dx }{\sqrt {1-4^x}} = \displaystyle \int \dfrac {2^x dx}{\sqrt{1-(2^x)^2}}$

put

$2^x=t \Rightarrow 2^x \ln 2 \ dx = dt$

$I=\dfrac{1}{\ln2}\displaystyle \int \dfrac {dt}{\sqrt {1-t^2}} = \dfrac{1}{\ln 2} \sin^{-1}t + c$

Compare with

$K \sin^{-1}2^x+c$

We get

$K=\dfrac{1}{\ln2}$

$\displaystyle \int \sqrt{\dfrac{a+x}{a-x}}dx$ is equal to-

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$a\sin^{-1} (x/a)-\sqrt{a^{2}-x^{2}}+c$
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$a\cos^{-1} (x/a)-\sqrt{a^{2}-x^{2}}+c$
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$a\sin^{-1} (x/a)-\sqrt{a^{2}+x^{2}}+c$
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$a\cos^{-1} (x/a)-\sqrt{a^{2}+x^{2}}+c$

Explanation

Consider the given integral.

$I=\int{\sqrt{\dfrac{a+x}{a-x}}dx}$

$I=\int{\sqrt{\dfrac{a+x}{a-x}}\times \sqrt{\dfrac{a+x}{a+x}}dx}$

$I=\int{\dfrac{a+x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}$

$I=a\int{\dfrac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}-\dfrac{1}{2}\int{\dfrac{-2x}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx}$

$I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)-\dfrac{1}{2}\int{\left\{ \dfrac{d}{dx}\left( {{a}^{2}}-{{x}^{2}} \right){{\left( {{a}^{2}}-{{x}^{2}} \right)}^{-1/2}} \right\}dx}$

$I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)-\dfrac{1}{2}\dfrac{{{\left( {{a}^{2}}-{{x}^{2}} \right)}^{1/2}}}{\left( 1/2 \right)}+C$

$I=a{{\sin }^{-1}}\left( \dfrac{x}{a} \right)-\sqrt{{{a}^{2}}-{{x}^{2}}}+C$

Hence, this is the correct answer.

Evaluate:

$\displaystyle\int \dfrac{\sin^3x}{( \cos^4x+ 3 \cos^2x+ 1)\tan^{-1} ( \sec x+ \cos x)}dx$

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$\tan^{-1} ( \sec x + \cos x) +c$
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$\log_e | \tan^{-1} ( \sec x + \cos x) |+c$
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$\dfrac{1}{(\sec x+ \cos x) ^2}+c$
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$\tan^{-1} ( \cot x+ \text{cosec} x)+c$

Explanation

we have to evaluate

$I= \int\dfrac{sin^3\ x}{(cos^4x+3\ cos^2+1)tan^{-1}(sec\ x+ cos\ x)}dx$

Let us assume

$tan^{-1}(sec\ x+ cos\ x)=t$

then,

$\Rightarrow \dfrac{1}{1+ (sec\ x+cos\ x)^2}.(sec\ x.tan\ x- sin\ x)dx=dt$

$\Rightarrow \dfrac{1}{1+ (\dfrac{1}{cos\ x}+cos\ x)^2}.(\dfrac{sin\ x}{cos^2\ x}- sin\ x)dx=dt$

$\Rightarrow \dfrac{cos^2\ x}{cos^2\ x+ (1+cos^2\ x)^2}.\dfrac{(sin\ x- sin\ xcos^2\ x)}{cos^2\ x}dx=dt$

$\Rightarrow \dfrac{sin\ x(1- cos^2x)}{cos^2x+1+ cos^4x+2cos^2x}dx=dt$

$\Rightarrow \dfrac{sin^3x}{ cos^4x+3cos^2x+ 1}dx=dt$

$I= \int\dfrac{sin^3x}{(cos^4x+3 cos^2+1)tan^{-1}(sec\ x+ cos\ x)}dx\Rightarrow\int\dfrac{dt}{t}$

$I = log_e\begin{vmatrix}t\end{vmatrix}+ c$

$I = log_e\begin{vmatrix}tan^{-1}(sec\ x +cos\ x)\end{vmatrix}+ C$

$\therefore \text{ Correct answer is option B.}$

CBSE Questions for Class 12 Commerce Applied Mathematics Indefinite Integrals Quiz 6 - 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 6 - MCQExams.com

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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