If $$ \displaystyle \int_{0}^{1} \frac{\tan^{-1} x}{x}dx$$ is equal to

$$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin x}{x}dx$$

$$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x}dx$$

$$\displaystyle \frac{1}{2} \int_{0}^{\frac{\pi}{2}}\frac{\sin x}{x}dx$$

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

Given,

$f(x) = \begin{vmatrix} 0 & {x^2 - \sin x} & {\cos x - 2} \\ {\sin x -x^2} & 0 & {1 - 2x} \\ {2 - \cos x} & {2x - 1} & 0 \end{vmatrix}$ , then

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

0%
$\frac{x^3}{3} - x^2 \sin x + \sin 2x + C$
0%
$\frac{x^3}{3} - x^2 \sin x + \cos 2x + C$
0%
$\frac{x^3}{3} - x^2 \cos x + \cos 2x + C$
0%
None of the above

Explanation

We have,

$f(x) = \begin{vmatrix} 0 & {x^2 - \sin x} & {\cos x - 2} \\ {\sin x -x^2} & 0 & {1 - 2x} \\ {2 - \cos x} & {2x - 1} & 0 \end{vmatrix}$

$\Rightarrow$

$f(x) = \begin{vmatrix} 0 & {\sin x - x^2} & {2 - \cos x} \\ {x^2 - \sin} & 0 & {2x - 1} \\ {\cos x - 2} & {1 - 2x} & 0 \end{vmatrix}$
[Interchanging rows and columns]

$\Rightarrow$

$f(x) = (-1)^3 \begin{vmatrix} 0 & {x^2 - \sin x} & {\cos x - 2} \\ {\sin x -x^2} & 0 & {1 - 2x} \\ {2 - \cos x} & {2x - 1} & 0 \end{vmatrix}$
[Taking (-1) common from each column]

$\Rightarrow$

$f(x) = - f(x)$

$\Rightarrow$

$f(x) = 0$

$\Rightarrow$

$\displaystyle \int f(x) dx = 0$

Let

$f(x)=\displaystyle \int_{-1}^{x}e^{t^2}dt$ and

$h(x)=f(1+g(x))$ where

$g(x)$ is defined for all

$x, g'(x)$ exists for all

$x,$ and

$g(x) < 0$ for

$x>0.$ If

$h'(1)=e$ and

$g'(1)=1,$ then the possible values which

$g(1)$ can take

0%
$0$
0%
$-1$
0%
$-2$
0%
$-4$

Explanation

Given,

$f(x)=\displaystyle \int_{-1}^{x}e^{t^2}dt; h(x)=f(1+g(x)); g(x)<0$ for

$x>0$

Now,

$h(x)=\displaystyle \int_{-1}^{1+g(x)}e^{t^2}dt\quad (given)$

Differentiating, we get

$h'(x)=e^{(1+g(x))^2}.g'(x)$

Now,

$h'(1)=e$

$\therefore e^{(1+g(1))^2}.g'(1)=e$

$\Rightarrow (1+g(1))^2=1$

$\Rightarrow 1+g(1)=\pm1$

$\Rightarrow g(1)=0\,(not\, possible)$

$g(1)=-2$

$\displaystyle \int_{0}^{t} \dfrac{b x \cos 4 x-a \sin 4 x}{x^{2}} d x=\dfrac{a \sin 4 t}{t}-1,$ where

$0<t<\dfrac{\pi}{4}$ then the values of

$a, b$ are equal to

0%
$\dfrac{1}{4}, 1$
0%
$-1,4$
0%
$2,2$
0%
$2,4$

Explanation

$\displaystyle I =b \int_{0}^{t} \dfrac{1}{x} \cos 4 x d x-a \int_{0}^{t} \dfrac{1}{x^{2}} \sin 4 x dx$

$=b I_{1}-a I_{2}$

$\displaystyle I_{2} =\int_{0}^{t} \dfrac{1}{x^{2}} \sin 4 x dx$

$\displaystyle =\left\{\left[-\dfrac{1}{x} \sin 4 x\right]_{0}^{t}+4 \int_{0}^{t} \dfrac{\cos 4 x}{x} d x\right\}$

$=\left[-\dfrac{\sin 4 t}{t}+4+4 I_{1}\right],\left\{\lim _{x \rightarrow 0} \dfrac{\sin 4 x}{x}=4\right\}$

$\therefore I=b I_{1}-a\left\{-\dfrac{\sin 4 t}{t}+4+4 I_{1}\right\}$

$\displaystyle =(b-4 a) \int_{0}^{t} \dfrac{1}{x} \cos 4 x d x+\dfrac{a \sin 4 t}{t}-4 a$

$=\dfrac{a \sin 4 t}{t}-1 \\$

$\displaystyle \therefore (b-4 a) \int_{0}^{t} \dfrac{1}{x} \cos 4 x d x=4 a-1\\$
L.H.S. is a function of

$t,$ whereas R.H:S. is a constant. Hence, we must have

$b-4 a=0$ and

$4 a-1=0\\$

$\therefore a=\dfrac{1}{4}, b=1$

The value of

$\int_{a}^{b}(x-a)^{3}(b-x)^{4} d x$ is

0%
$\dfrac{(b-a)^{4}}{6^{4}}$
0%
$\dfrac{(b-a)^{8}}{280}$
0%
$\dfrac{(b-a)^{7}}{7^{3}}$
0%
None of these

Explanation

Let

$I=\int_{a}^{b}(x-a)^{3}(b-x)^{4} d x$

Put

$x=a \cos ^{2} \theta+b \sin ^{2} \theta, \Rightarrow d x=2(b-a) \sin \theta \cos \theta d \theta$

then

$\int_{a}^{b}(x-a)^{3}(b-x)^{4} d x$

$=2(b-a) \int_{0}^{\pi / 2}\left(a \cos ^{2} \theta+b \sin ^{2} \theta-a\right)^{3}\left(b-a \cos ^{2} \theta\right)\\$

$=2(b-a)^{8} \int_{0}^{\pi / 2} \sin ^{7} \theta \cos ^{9} \theta d \theta\\$

$=2(b-a)^{8} \int_{0}^{\pi / 2} \sin ^{7} \theta\left(1-\sin ^{2} \theta\right)^{4} \cos \theta d \theta\\$

$=2(b-a)^{8} \int_{0}^{1} x^{7}\left(1-x^{2}\right)^{4} d x\\$

$=2(b-a)^{8} \int_{0}^{1} x^{7}\left(1-4 x^{2}+6 x^{4}-4 x^{6}+x^{8}\right) d x\\$

$=2(b-a)^{8}\left[\dfrac{1}{8}-\dfrac{4}{10}+\dfrac{6}{12}-\dfrac{4}{14}+\dfrac{1}{16}\right]=\dfrac{(b-a)^{8}}{280}$

the value of

$I_{n+2}-I_n$ is equal to

0%
$n\pi$
0%
$\pi$
0%
$-\pi$
0%
$0$

Explanation

Given

$I_n = \displaystyle \int_{-\pi}^{\pi} \dfrac{sinnx}{(1+\pi^x)sinx}dx \qquad ...(i)$

Using

$\displaystyle \int_{a}^{b}f(x)dx=\displaystyle \int_{a}^{b}f(b+a-x)dx,$ we get

$I_n=\displaystyle \int_{-\pi}^{\pi} \dfrac{\pi^x sinnx}{(1+\pi^x) sinx}dx\qquad..(ii)$

On adding Eqs.(i) and (ii), we have

$2I_n=\displaystyle \int_{-\pi}^{\pi} \dfrac{sinnx}{sinx}dx=2\displaystyle \int_{0}^{\pi} \dfrac{sinnx}{sinx}dx [\because f(x)=\dfrac{sinnx}{sinx}$ is an even function]

$\Rightarrow I_n=\displaystyle \int_{0}^{\pi}\dfrac{sinnx}{sinx}dx$

Now,

$I_{n+2}-I_n=\displaystyle \int_{0}^{\pi} \dfrac{sin(n+2)x-sinnx}{sinx}dx$

$=\displaystyle \int_{0}^{\pi} \dfrac{2cos(n+1)x.sinx}{sinx}dx$

$=2\displaystyle \int_{0}^{\pi}cos(n+1)dx=2\left[\dfrac{sin(n+1)x}{(n+1)}\right]_{0}^{\pi}=0\qquad...(iii)$

$\therefore I_{n+2}-I_n=0$

The value of the definite integral

$\displaystyle \int_{0}^{\infty} \dfrac{dx}{(1+x^a)(1+x^2)}(a>0)$ is

0%
$\dfrac{\pi}{4}$
0%
$\dfrac{\pi}{2}$
0%
${\pi}$
0%
Some function of a

$\displaystyle I_{n}=\int_{0}^{1} \dfrac{d x}{\left(1+x^{2}\right)^{n}},$ where

$n \in N,$ which of the following statements hold good?

0%
$2 n I_{n+1}=2^{-n}+(2 n-1) I_{n}$
0%
$I_{2}=\dfrac{\pi}{8}+\dfrac{1}{4}$
0%
$I_{2}=\dfrac{\pi}{8}-\dfrac{1}{4}$
0%
$I_{3}=\dfrac{3 \pi}{32}+\dfrac{1}{4}$

Explanation

$\displaystyle I_{n}=\int_{0}^{1} \dfrac{d x}{\left(1+x^{2}\right)^{n}}=\int_{0}^{1}\left(1+x^{2}\right)^{-n} d x$

$\displaystyle =\left.\dfrac{x}{\left(1+x^{2}\right)^{n}}\right|_{0} ^{1}-\int_{0}^{1}(-n)\left(1+x^{2}\right)^{-n-1} 2 x \times x d x$

$\displaystyle =\dfrac{1}{2^{n}}+2 n \int_{0}^{1} \dfrac{x^{2} dx}{\left(1+x^{2}\right)^{n+1}}$

$\displaystyle =\dfrac{1}{2^{n}}+2 n \int_{0}^{1} \dfrac{1+x^{2}-1}{\left(1+x^{2}\right)^{n+1}} d x\\$

$\Rightarrow 2 n I_{n+1}=2^{-n}+(2 n-1) I_{n}\\$

$\Rightarrow 2 I_{2}=\dfrac{1}{2}+I_{1}=\dfrac{1}{2}+\left.\tan ^{-1} x\right|_{0} ^{1}\\$

$\Rightarrow I_{2}=\dfrac{1}{4}+\dfrac{\pi}{8}\\$

Also

$4 I_{3}=2^{-2}+3 I_{2}\\$

$=\dfrac{1}{4}+3\left(\dfrac{1}{4}+\dfrac{\pi}{8}\right)=\dfrac{1}{4}+\dfrac{3 \pi}{32}$

Abscissae of point of intersection are in

0%
AP
0%
GP
0%
HP
0%
None of these

Explanation

We have

$g(x)=g(0)+xg'(0)+\dfrac{x^2}{2}g''(0)=-bx^2+cx-6$

$h(x)=g(x)=4x^4-ax^3+bx^2-cs+6=0$ has 4 distinct real roots. Using Descarte's rule of signs.
Given biquadratic equation has 4 distinct positive roots.
Let the roots be

$x_1,x_2,x_3$ and

$x_4$
Now,

$\dfrac{\dfrac{1}{x_1}+\dfrac{2}{x_2}|+\dfrac{3}{x_3}+\dfrac{4}{x_4}}{4}\geq \sqrt{\dfrac{24}{x_1x_2x_3x_4}}$

$\Rightarrow 2\geq2 \Rightarrow \dfrac{1}{x_1}=\dfrac{2}{x_2}=\dfrac{3}{x_3}=\dfrac{4}{x_4}=k$

$\Rightarrow \dfrac{1}{x_1}.\dfrac{2}{x_2}.\dfrac{3}{x_3}.\dfrac{4}{x_4}=k^4$

$\Rightarrow \dfrac{24}{3/2}=k^4 \Rightarrow k=2$

$\therefore$ Roots are

$\dfrac{1}{2},1,\dfrac{3}{2}$ and

$2.$ Also,a=20 and c=25

The value of the definite integral

$\displaystyle \int_{0}^{\pi/2}sinx\space sin2x\space sin3xdx$ is equal to

0%
$\dfrac{1}{3}$
0%
$-\dfrac{2}{3}$
0%
$-\dfrac{1}{3}$
0%
$\dfrac{1}{6}$

Explanation

$=\displaystyle \int_{0}^{\pi/2}\sin x\sin2x\sin3xdx$

$=\displaystyle \int_{0}^{\pi/2}sin\left(\dfrac{\pi}{2}-x\right) sin2\left(\dfrac{\pi}{2}-x\right) sin3\left(\dfrac{\pi}{2}-x\right)dx$

$\Rightarrow I=\displaystyle \int_{0}^{\pi/2}-cosxsin2xcos3xdx$

$\therefore 2I=\displaystyle \int_{0}^{\pi/2}-sin2x(cosxcos3x-sin3xsinx)dx$

$=\displaystyle \int_{0}^{\pi/2}-sin2xcos(4x)dx$

$=-\displaystyle \int_{0}^{\pi/2}sin2x(2cos^22x-1)dx$
Put

$cos2x=t \Rightarrow -sin2x\times 2dx=dt$

$\therefore 2I=\displaystyle \int_{1}^{-1}\dfrac{1}{2}(2t^2-1)dt=\dfrac{1}{2}\left[ \dfrac{2t^3}{3}-t \right]^{-1}$

$=\dfrac{1}{2}\left[ \dfrac{2}{3}(-1)^3-(-1)-\left( \dfrac{2}{3}(1)^3 -1 \right)\right]$

$=\dfrac{1}{2}\left[ -\dfrac{2}{3}+1-\dfrac{2}{3}+1\right]=\dfrac{1}{2}\left[\dfrac{2}{3}\right]\qquad I\Rightarrow \dfrac{1}{6}$

The value of

$\int_{2}^{5} \dfrac {\sqrt x}{\sqrt {x} + \sqrt {7-x}} dx$
is :

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

Explanation

Let

$I = \int_{2}^{5} \dfrac {\sqrt x}{\sqrt {x} + \sqrt {7-x}} dx$ .................(1)

$\Rightarrow \int_{2}^{5} \dfrac {\sqrt 2+5-x}{\sqrt {2+5-x} + \sqrt {7-2-5+x}} dx$

$\Rightarrow \int_{2}^{5} \dfrac {\sqrt 7 -x}{\sqrt {7-x} + \sqrt {x}} dx$ ............(2)

Adding equation (1) and (2)

$2I = \int_{2}^{5} \dfrac {\sqrt x + \sqrt{7-x}}{\sqrt {7-x} + \sqrt {x}} dx$
=

$\int_{2}^{5} 1 dx = []_{2}^{5} = 5-2 = 3$

$I = \dfrac {3}{2}$

hence,

$\int_{2}^{5} \dfrac {\sqrt x}{\sqrt {x} + \sqrt {7-x}} dx=\dfrac{3}{2}$
option C is correct.

$A(x) = \int_{0}^{x} \theta ^2 d\theta$
then value of

$A(3)$ will be :

0%
9
0%
27
0%
3
0%
81

Explanation

$A(x) = \int_{0}^{x} \theta ^2 d\theta$
=

$[\dfrac {\theta ^3}{3}]_{0}^{x}$
=

$\dfrac {1}{3} x^3$

$\Rightarrow A(3) = \dfrac {1}{3} 3^3$
=

$\dfrac {1}{3} \times 3 \times 3 \times 3$
=

$3 \times 3$
=

$9$

Thus, option A is correct

$\int_{a-c}^{b-c} f(x+c) dx$ is:

0%
$\int_{a}^{b} f(x+c) dx$
0%
$\int_{a}^{b} f(x) dx$
0%
$\int_{a-2c}^{b-2c} f(x) dx$
0%
$\int_{a}^{b} f(x+2c) dx$

$\int_{0}^{log 5} \displaystyle \frac{e^{x}\sqrt{e^{x}-1}}{e^{x}+3} dx =$

0%
$3+ 2 \pi$
0%
$4 - \pi$
0%
$2 + \pi$
0%
$\pi -4$

Explanation

Let

$I=\int { \cfrac { { e }^{ x }\sqrt { { e }^{ x }-1 } }{ { e }^{ x }+3 } }$
Substitute

$t=e^{ x }\Rightarrow dt=e^{ x }dx$ , we get

$I=\int { \cfrac { \sqrt { t-1 } }{ t+3 } dt }$
Substitute

$u=\sqrt { t-1 } \Rightarrow du=\cfrac { 1 }{ 2\sqrt { t-1 } } dt$ , we get

$I=2\int { \cfrac { { u }^{ 2 } }{ { u }^{ 2 }+4 } du } =2\int { \left( 1-\cfrac { 4 }{ { u }^{ 2 }+4 } \right) du } =2\int { du } -2\int { \cfrac { 4 }{ { u }^{ 2 }+4 } } du\\ =2u-4{ tan }^{ -1 }\left( \cfrac { u }{ 2 } \right) =2\sqrt { t-1 } -4{ tan }^{ -1 }\left( \cfrac { \sqrt { t-1 } }{ 2 } \right) \\ =2\sqrt { e^{ x }-1 } -4{ tan }^{ -1 }\left( \cfrac { \sqrt { e^{ x }-1 } }{ 2 } \right)$
Therefore,

$\int _{ 0 }^{ \log { 5 } }{ Idx } ={ \left[ 2\sqrt { e^{ x }-1 } -4{ tan }^{ -1 }\left( \cfrac { \sqrt { e^{ x }-1 } }{ 2 } \right) \right] }_{ 0 }^{ \log { 5 } }=4-\pi$

$\displaystyle I_n=\int_{0}^{\tfrac{\pi}{4}} \tan^nx\sec^2xdx,$ then

$I_1, I_2, I_3..$ are in

0%
A.P.
0%
G.P.
0%
H.P.
0%
A.G.P.

Explanation

$\displaystyle I_n=\int_{0}^{\tfrac{\pi}{4}} \tan^nx\sec^2xdx,$

Substituting

$t=\tan x \Rightarrow dt =\sec^2xdx$

$I_n=\int_{0}^{1}t^ndt$ =

$\dfrac{1}{n\;+\;1}$

$\Rightarrow I_n\;=\;\dfrac{1}{n\;+\;1}$

Reciprocal of the terms

$I_1,I_2,I_3..$ are in A.P, therefore

$\dfrac{1}{n+1}, \dfrac{1}{n+2}, \dfrac{1}{n+3}$ are in H.P.

Hence, option 'C' is correct.

Evaluate the integral

$\displaystyle \int_{0}^{\pi/4}\frac{ {s}i {n}\theta+ {c} {o} {s}\theta}{9+16 {s}i {n}2\theta} \ {d}\theta$

0%
$\displaystyle \frac{1}{20} \log 2$
0%
$\displaystyle \frac{1}{20} \log 3$
0%
$\log 3$
0%
$\log 2$

Explanation

$\displaystyle \int_{0}^{\pi/ 4} \dfrac {\sin \theta + \cos \theta}{9 + 16\sin 2\theta} d\theta$

Let $\sin \theta - \cos \theta = t$

$\therefore (\cos \theta + \sin \theta) d\theta = dt$

$= \displaystyle \int_{-1}^{0} \dfrac {dt}{9 + 16(1 – t^{2})} (\because t^{2} = 1 – 2\cdot \sin \theta \cdot \cos \theta)$

$= \displaystyle \int_{-1}^{0} \dfrac {dt}{25 – 16t^{2}}$

$= \dfrac {1}{16} \displaystyle \int_{-1}^{0} \dfrac {dt}{\left (\dfrac {5}{4}\right )^{2} – t^{2}}$

$= \dfrac {1}{16}\times \dfrac {1}{2\times \dfrac {5}{4}}\left [ln \left |\dfrac {\dfrac {5}{4} +t}{\dfrac {5}{4} – t}\right |\right ]_{-1}^{0}$

$\left (\because \displaystyle \int \dfrac {1}{a^{2} – x^{2}} dx = \dfrac {1}{2a} ln \left |\dfrac {a + x}{a – x}\right | + c\right )$

$= \dfrac {1}{40} \left [ln |1| - ln \left |\dfrac {\dfrac {1}{4}}{\dfrac {9}{4}}\right |\right ]$

$= \dfrac {1}{40} ln 9 = \dfrac {ln 3}{20}$

$\therefore \displaystyle \int_{0}^{\pi/ 4} \dfrac {\sin \theta + \cos \theta}{9 + 16 \sin 2\theta} d\theta = \dfrac {ln 3}{20}$ .

The value of the definite integral

$\displaystyle \int_{ 0 }^{\sqrt{{\ln (\displaystyle \frac{\pi}{2})}}}\cos(e^{x^{2}})2xe^{x^{2}} \:dx$ is:

0%
$1$
0%
$1+ (\sin1)$
0%
$1- (\sin1)$
0%
$(\sin1)-1$

Explanation

Let,

$z=e^{x^2}\implies dz=2xe^{x^2}dx$

When

$x=0,z=1$ and when

$x=\sqrt{\ln{(\dfrac{\pi}{2})}},z=\dfrac{\pi}{2}$

Hence, integration becomes:-

$\displaystyle\int_{1}^{\pi/2} \cos{z}dz$

$=\left[\sin{z}\right]_{1}^{\pi/2}$

$=1-(\sin{1})$

Hence, the answer is option-(C).

Evaluate:

$\displaystyle \int_{0}^{\pi}\frac{dx}{5+4\cos x}$

0%
$\dfrac{\pi}{2}$
0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{3}$
0%
$-\pi$

Explanation

Putting

$\cos{x}=\dfrac{1-\tan^2{(x/2)}}{1+\tan^2{(x/2)}}$ ,

$\displaystyle\int_{0}^{\pi}\dfrac{dx}{5+4\dfrac{1-\tan^2{(x/2)}}{1+\tan^2{(x/2)}}}$

$=\displaystyle\int_{0}^{\pi} \dfrac{\{1+\tan^2{(x/2)\}}dx}{9+\tan^2{(x/2)}}$

$=\displaystyle\int_{0}^{\pi} \dfrac{\sec^2{(x/2)}dx}{\tan^2{(x/2)}+3^2}$

Let ,

$z=\tan{(x/2)}\implies dz=\dfrac{1}{2}\sec^2{(x/2)}dx\implies \sec^2{(x/2)}dx=2dz$

And when,

$x=0, z=0$ and when

$x=\pi, z=\infty$

Hence, integration becomes:-

$\displaystyle\int_{0}^{\infty} \dfrac{2dz}{z^2+3^2}$

$=\dfrac{2}{3}\left[\tan^{-1} {\dfrac{z}{3}}\right]_{0}^{\infty}$

$=\dfrac{2}{3}\left[\tan^{-1} {\infty}-\tan^{-1} {0}\right]$

$=\dfrac{2}{3}\left(\dfrac{\pi}{2}-0\right)$

$=\dfrac{\pi}{3}$

$\int ^{\pi /2}_{-\pi /2} cost . \sin(2t-\displaystyle \frac{\pi}{4})dt=$

0%
$\displaystyle \frac{\sqrt{2}}{3}$
0%
$-\displaystyle \frac{\sqrt{2}}{3}$
0%
$\displaystyle \frac{\sqrt{3}}{1}$
0%
$\displaystyle \frac{1}{\sqrt{3}}$

Explanation

$I=\int { \left( cost \right) \left( sin\left( 2t-\cfrac { \pi }{ 4 } \right) \right) } dt=-\int { sin\left( \cfrac { \pi }{ 4 } -2t \right) \left( cost \right) dt } \\ =-\cfrac { 1 }{ 2 } \int { \left( sin\left( \cfrac { 1 }{ 4 } \left( \pi -12t \right) \right) +sin\left( \cfrac { 1 }{ 4 } \left( \pi -4t \right) \right) \right) dt } \\ =-\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -12t \right) \right) dt } -\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -4t \right) \right) dt }$
Let

${ I }={ I }_{ 1 }+{ I }_{ 2 }$ , such that

${ I }_{ 1 }=-\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -12t \right) \right) dt }$
Substituting

$x=\pi -4t\Rightarrow dx=-4dt$

${ I }_{ 1 }=\cfrac { 1 }{ 8 } \int { sin\cfrac { x }{ 4 } dx } =-\cfrac { 1 }{ 2 } cos\cfrac { x }{ 4 } =-\cfrac { 1 }{ 2 } sin\left( 3t+\cfrac { \pi }{ 4 } \right)$ ...(1)
And

${ I }_{ 2 }=-\cfrac { 1 }{ 2 } \int { sin\left( \cfrac { 1 }{ 4 } \left( \pi -4t \right) \right) dt }$
Substituting

$y=\pi -12t\Rightarrow dy=-12dt$ , we get

${ I }_{ 2 }=\cfrac { 1 }{ 24 } \int { sin\cfrac { y }{ 4 } dy } =-\cfrac { 1 }{ 6 } cos\cfrac { y }{ 4 } \quad =-\cfrac { 1 }{ 6 } sin\left( t+\cfrac { \pi }{ 4 } \right)$ ...(2)
Therefore from (1) and (2)

$\int _{ -\cfrac { \pi }{ 2 } }^{ \cfrac { \pi }{ 2 } }{ \left( cott \right) sin\left( 2t-\cfrac { \pi }{ 4 } \right) dt } =\left[ -\cfrac { 1 }{ 2 } sin\left( 3t+\cfrac { \pi }{ 4 } \right) -\cfrac { 1 }{ 6 } sin\left( t+\cfrac { \pi }{ 4 } \right) \right] _{ -\cfrac { \pi }{ 2 } }^{ \cfrac { \pi }{ 2 } }=-\cfrac { \sqrt { 2 } }{ 3 }$

$\displaystyle \int_{0}^{a}\frac{dx}{x+\sqrt{a^{2}-x^{2}}}=$

0%
$\pi$
0%
$\dfrac{\pi}{3}$
0%
$-\pi$
0%
$\dfrac{\pi}{4}$

Explanation

$\displaystyle I=\int _{ 0 }^{ a }{ \cfrac { dx }{ x+\sqrt { { a }^{ 2 }-{ x }^{ 2 } } } }$

Substituting

$x=a\sin t\Rightarrow dx=a\cos tdt$

$\displaystyle I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { \cos tdt }{ \sin t+\cos t } }$ ...(1)

Using property

$\int _{ a }^{ b }{ f\left( x \right) dx } =\int _{ a }^{ b }{ f\left( a+b-x \right) dx }$

$\displaystyle I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ \cfrac { \sin tdt }{ \cos t+\sin t } }$ ...(2)

Adding (1) and (2), we get

$\displaystyle 2I=\int _{ 0 }^{ \cfrac { \pi }{ 2 } }{ dt } =\cfrac { \pi }{ 2 } \Rightarrow I=\cfrac { \pi }{ 4 }$

$\displaystyle \int_{0}^{\pi/2}\sqrt{\cos x}\sin^{5}xdx=$

0%
$\displaystyle \frac{34}{231}$
0%
$\displaystyle \frac{64}{231}$
0%
$\displaystyle \frac{30}{321}$
0%
$\displaystyle \frac{128}{231}$

Explanation

$\int_{0}^{\dfrac{\pi}{2}} \sqrt{cos x}sin^5 x   dx$
$\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x}(sin^4x)sin x   dx$
$-\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos   x}(1-cos^2 x)^2d cos x$
$-\left [ \int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x} dcos x+\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x} cos^4 x dx - 2\int_{0}^{\dfrac{\pi}{2}}\sqrt{cos x} cos^2 x   dx \right ]$
$-\left [ \dfrac{2}{3} cos^{\dfrac{3}{2}} x \int_{0}^{\dfrac{\pi}{2}}+\dfrac{2}{11} cos^{\dfrac{11}{2}}x\int_{0}^{\dfrac{\pi}{2}}-2(\dfrac{2}{7}) cos^{\dfrac{7}{2}}x \int_{0}^{\dfrac{\pi}{2}} \right ]$
$=-\left [\dfrac{2}{3} (-1)+\dfrac{2}{11}(-1)-2(\dfrac{2}{7})(-1)\right ]$
$=\dfrac{64}{231}$

$\displaystyle \int_{1}^{2}\mathrm{x}^{2\mathrm{x}}[1+\log \mathrm{x}]\mathrm{d}\mathrm{x}=$

0%
$\displaystyle \frac{9}{2}$
0%
$\displaystyle \frac{11}{2}$
0%
$\displaystyle \frac{13}{2}$
0%
$\displaystyle \frac{15}{2}$

Explanation

Let

$\displaystyle {x}^{x} = t$
Diiferentiating both sides,

$\displaystyle {x}^{x}(1 + \log x) dx = dt$
Substituting the above obtained values back into the expression,

$\displaystyle \int {x}^{2x}(1 + log x) dx = \int t dt$

$\displaystyle = \frac{{t}^{2}}{2} + c$
Putting the value of t back, we get the integral as

$\displaystyle \frac{{({x}^{x})}^{2}}{2} + c$
Substituting the limits from 1 to 2, we get

$= \displaystyle \left( \frac{{({2}^{2})}^{2}}{2} \right) - \left( \frac{{({1}^{1})}^{2}}{2}\right)$
.

$\displaystyle = 8 - \frac{1}{2}$
.

$\displaystyle = \frac{15}{2}$

$\mathrm{a}>\mathrm{b}$ then

$\displaystyle \int_{0}^{\pi}\frac{\mathrm{d}\mathrm{x}}{\mathrm{a}+\mathrm{b}\sin \mathrm{x}}=$

0%
$\displaystyle \frac{2}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\cot^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$
0%
$\displaystyle \frac{1}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\cot^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$
0%
$\displaystyle \frac{1}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\tan^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$
0%
$\displaystyle \frac{2}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}\tan^{-1}(\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}})$

$\displaystyle \int_{0}^{\infty}e^{-\mathrm{x}^{2}}\mathrm{d}\mathrm{x}=\frac{\sqrt{\pi}}{2}$ , then

$\displaystyle \int_{0}^{\infty}e^{-ax^{2}}dx,\ \mathrm{a}>0$ is

0%
$\displaystyle \frac{\sqrt{\pi}}{2}$
0%
$\displaystyle \frac{\sqrt{\pi}}{2a}$
0%
$2 \displaystyle \frac{\sqrt{\pi}}{a}$
0%
$\displaystyle \frac{1}{2}\sqrt{\frac{\pi}{a}}$

Explanation

$\int _{ 0 }^{ \infty }{ { e }^{ { -ax }^{ 2 } } } dx\\ =\int _{ 0 }^{ \infty }{ { e }^{ -(\sqrt { a } x)^{ 2 } } } dx$

Let,

$\sqrt { a } .{ x }=1$

$\sqrt { a } .dx=dt\\ dx=\cfrac { dt }{ \sqrt { a } } \\ I=\int _{ 0 }^{ \infty }{ e^{ -t^{ 2 } } } .\cfrac { dt }{ \sqrt { a } } \\ =\cfrac { 1 }{ \sqrt { a } } \int _{ 0 }^{ \infty }{ e^{ -t^{ 2 } } } dt\\ =\cfrac { 1 }{ \sqrt { a } } \times \cfrac { \sqrt { \pi } }{ 2 } \\ =\cfrac { 1 }{ 2 } \sqrt { \cfrac { \pi }{ a } }$

Hence the correct answer is

$\cfrac { 1 }{ 2 } \sqrt { \cfrac { \pi }{ a } }$

Let

$\displaystyle\frac{d}{dx}(F(x))=\frac{e^{\displaystyle\sin{x}}}{x}$ ,

$x>0$ . If

$\displaystyle\int_1^4{\frac{2e^{\displaystyle\sin{x^2}}}{x}dx}=F(k)-F(1)$ , then the possible value of

$k$ is

0%
10
0%
14
0%
16
0%
18

Explanation

Substitute

${ x }^{ 2 }=z$
Then

$\displaystyle \int _{ 1 }^{ 4 }{ \frac { 2{ e }^{ \sin { { x }^{ 2 } } } }{ x } dx } =\int _{ 1 }^{ 16 }{ \frac { { e }^{ \sin { z } } }{ z } dz } =\int _{ 1 }^{ 16 }{ \frac { d }{ dz } \left\{ F\left( z \right) \right\} } dz$

$={ \left( F\left( z \right) \right) }_{ 1 }^{ 16 }=F\left( 16 \right) -F\left( 1 \right)$

$\therefore F\left( 16 \right) -F\left( k \right) =F\left( 16 \right) -F\left( 1 \right)$
Gives

$k=16$

Observe the following Lists

List-I	List-II
A: $\displaystyle \int_{-2}^{2}\frac{1}{4+x^{2}}dx$	1) $\displaystyle \frac{\pi}{3}$
B: $\displaystyle \int_{1}^{2}\frac{1}{x\sqrt{x^{2}-1}}dx$	2) 0
C: $\displaystyle \int_{0}^{\pi}\cos 3x.\cos 2xdx$	3) $\displaystyle \frac{\pi}{4}$
	4) $\displaystyle \frac{\pi}{2}$

0%
A-3, B-1, C-4
0%
A-3, B-1, C-2
0%
A-1, B-3, C-2
0%
A-4, B-1, C-2

Explanation

$\displaystyle \int _{ -2 }^{ 2 }{ \frac { 1 }{ 4+{ x }^{ 2 } } } dx=\left[ \frac { 1 }{ 2 } \tan ^{ -1 }{ \frac { x }{ 2 } } \right] _{ -2 }^{ 2 }$

$\displaystyle =\left[ \frac { 1 }{ 2 } \tan ^{ -1 }{ 1 } -\frac { 1 }{ 2 } \tan ^{ -1 }{ \left( -1 \right) } \right] =\frac { \pi }{ 8 } +\frac { \pi }{ 8 } =\frac { \pi }{ 4 }$

B)

$\displaystyle \int _{ 1 }^{ 2 }{ \frac { 1 }{ x\sqrt { { x }^{ 2 }-1 } } } dx=\left[ \sec ^{ -1 }{ x } \right] _{ 1 }^{ 2 }$

$\displaystyle =\left[ \sec ^{ -1 }{ 2 } -\sec ^{ -1 }{ 1 } \right] =\left[ \frac { \pi }{ 3 } -0 \right] =\frac { \pi }{ 3 }$

C)

$\displaystyle \int _{ 0 }^{ \pi }{ \cos { 3x } \cos { 2x } } dx=\frac { 1 }{ 2 } \int _{ 0 }^{ \pi }{ \left( \cos { x } +\cos { 5x } \right) } dx$

$\displaystyle =\frac { 1 }{ 10 } \left[ 5\sin { x } +\sin { 5x } \right] _{ 0 }^{ \pi }=0$

$\displaystyle \int_{0}^{\pi/3}\frac{\cos x}{3+4\sin x}dx=k\log(\frac{3+2\sqrt{3}}{3})$ , then

$\mathrm{k}$ is equal to

0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{1}{8}$

Explanation

$\displaystyle\int_{0}^{\pi/3}\dfrac{\cos{x}}{3+4\sin{x}}dx$

Let

$z=3+4\sin{x}\implies dz=4\cos{x}dx\implies \cos{x}dx=\dfrac{dz}{4}$

When

$x=0,z=3$ and when

$x=\dfrac{\pi}{3},z=3+2\sqrt{3}$

Hence, integration becomes:-

$\dfrac{1}{4}\displaystyle\int_{3}^{3+2\sqrt{3}}\dfrac{1}{z} dz$

$=\dfrac{1}{4}\left[\log{z}\right]_{3}^{3+2\sqrt{3}}$

$=\dfrac{1}{4}\log{\left(\dfrac{3+2\sqrt{3}}{3}\right)}$

$\implies k=\dfrac{1}{4}$

Hence, answer is option-(C).

The value of the integral

$\displaystyle \int_{0}^{3\alpha}\text{cosec}(x -\alpha)\text{cosec}(x-2\alpha) dx$ is

0%
$2\displaystyle \sec\alpha\log(\frac{1}{2}\text{cosec}\alpha)$
0%
$2\displaystyle \sec\alpha\log(\frac{1}{2}\sec\alpha)$
0%
$2 \text{cosec}\alpha\log(\sec\alpha)$
0%
$2 \displaystyle \text{cosec}\alpha\log(\frac{1}{2}\sec\alpha)$

Explanation

$\displaystyle \int_{0}^{3\alpha}\text{cosec}(x -\alpha)\text{cosec}(x-2\alpha) dx$

$=\displaystyle \int_{0}^{3\alpha}\frac{1}{\sin (x-\alpha)\sin (x-2\alpha)}dx$

$\displaystyle \frac{1}{\sin \alpha}\int_{0}^{3\alpha} \displaystyle \frac{\sin [(x-\alpha)-(x-2\alpha)]}{\sin (x-\alpha)\sin (x-2\alpha)}dx$

$\Rightarrow \displaystyle \frac{1}{\sin \alpha}\int_{0}^{3\alpha}[\cot(x-2\alpha)-\cot(x-\alpha)]dx$

$=\displaystyle \frac{1}{\sin \alpha}(\log [\sin (x-2\alpha)]|_{0}^{3\alpha}-\log (\sin (x-\alpha))|_{0}^{3\alpha})$

$=\displaystyle \frac{1}{\sin \alpha}[\log (\sin \alpha)-\log (\sin (-2\alpha))-\log (\sin (2\alpha))+\log (\sin (-\alpha))]$

$=2\displaystyle \frac{1}{\sin \alpha}\log \left | \displaystyle \frac{\sin \alpha}{2\sin \alpha \cos \alpha} \right |$

$=2 \text{cosec}\alpha\ \log\left(\dfrac{1}{2} \sec\alpha\right)$

$I_{n}=\displaystyle \int_{0}^{\frac{\pi}{4}}\tan^{n} dx,$ then

$\dfrac{1}{I_{2}+I_{4}},\dfrac{1}{I_{3}+I_{5}},\dfrac{1}{I_{4}+I_{6}}\cdots$ form

0%
an A.P
0%
a G.P
0%
a H.P
0%
an AGP

Explanation

$I_n = \displaystyle \int_{0}^{\pi/4}\tan^{n}x\ dx$

$I_{n+2}=\displaystyle \int_{0}^{\pi/4}\tan^{n}x(\tan^{2}x)dx$

$=\displaystyle \int_{0}^{\pi/4}\tan^{n}x(\sec^{2}x-1)dx$

$=\displaystyle \int_{0}^{\pi/4}\tan^{n}x\ d(\tan x)-I_n$

$\Rightarrow I_{n+2} + I_n=\dfrac{1}{n+1}$

$I_{2}+I_{4}=\dfrac13$

$I_{3}+I_{5}=\dfrac14$

$I_{4}+I_{6}=\dfrac15$

$So,\dfrac{1}{I_{2}+I_{4}};\dfrac{1}{I_{3}+I_{5}};\dfrac{1}{I_{4}+I_{6}}$ forms an AP

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

$g(x)=max\left \{ f(t) ;0\leq t< x \right \}$ , then

$\overset {1}{\underset { 0 }{ \int } } g (x) dx = ?$

0%
$\dfrac{29}{24}$
0%
$\dfrac{7}{6}$
0%
$\dfrac{5}{4}$
0%
none of these

Explanation

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

$g(x)=max\left \{ f(t) ;0\leq t< x \right \}$

y = g(x)

So,

$\overset {1}{\underset { 0 }{ \int } }g (x) dx=\overset {1/2}{\underset { 0 }{ \int } } f(x)dx+\left [ 1-\dfrac{1}{2} \right ]\times\dfrac{5}{4}$

$=\left [ -\dfrac{x^3}{3}+\dfrac{x^2}{2}+x\right ]_0^{1/2}+\dfrac{5}{8}=\dfrac{29}{24}$

$\displaystyle \int_{0}^{1}\frac{3^{x+1}-4^{x-1}}{12^{x}}dx=$

0%
$\displaystyle \frac{9}{4}log_{4}\, e$
0%
$\displaystyle \frac{9}{4}log_{4}\, e-\frac{1}{6}log_{3}e$
0%
$\displaystyle log_{4}3$
0%
None of these

Explanation

$\int_{0}^{1}\dfrac{3^{x+1}-4^{x-1}}{12^{x}} dx$

$=\displaystyle \int_{0}^{1}(\dfrac{3}{4^{\mathrm{x}}}-\dfrac{1}{4}. \displaystyle \dfrac{1}{3^{\mathrm{x}}}) dx$

$=\left \{ 3\dfrac{\dfrac{1}{4^{x}}}{(log\dfrac{1}{4})}-\dfrac{1}{4} \dfrac{\dfrac{1}{3^{x}}}{log\left ( \dfrac{1}{3} \right )}\right \}_{0}^{1}$

$=\displaystyle \dfrac{3(\dfrac{1}{4}-1)}{-\log_{\mathrm{e}}4}-\dfrac{1}{4}\dfrac{(\dfrac{1}{3}-1)}{-\log_{\mathrm{e}}3}$

$=\displaystyle \dfrac{9}{4}\log_{4}\mathrm{e}-\dfrac{1}{6}\log_{3}\mathrm{e}$

Ans: B

Let

$\displaystyle \frac{d}{dx}F(x)=\frac{e^{{s}{m}{x}}}{x}$ ,

$x>0$ . lf

$\displaystyle \int_{1}^{4}\frac{3}{x}e^{{s}{m}{x}^{3}}dx=F(k)-F(1)$ , then one of the possible values of

${k}$ is

0%
$16$
0%
$62$
0%
$64$
0%
$15$

Explanation

Given:

$\displaystyle\int_{1}^{4} \dfrac{3}{x} e^{\sin{x^3}}dx$

Let

$z=x^3\implies dz=3x^2dx=\dfrac{3}{x}x^3dx$

$\implies dz=\dfrac{3}{x}zdx$

$\implies \dfrac{3}{x}dx=\dfrac{dz}{z}$

And when

$x=1,z=1$ and when

$x=4,z=64$

Hence, integration becomes:-

$\displaystyle\int_{1}^{64}e^{\sin{z}}\dfrac{dz}{z}$

Now it is given that:-

$\dfrac{d}{dx}F(x)=\dfrac{e^{\sin{x}}}{x}, x>0$

$\implies \displaystyle\int_{1}^{64} \dfrac{e^{\sin{z}}}{z}dz$

$=\displaystyle\int_{1}^{64} dF(z)$

$=F(64)-F(1)$

Hence,

$k=64$

$\displaystyle\int_0^\infty{f\left(x+\frac{1}{x}\right).\frac{\ln{x}}{x}dx}$ is equal to

0%
0
0%
1
0%
$\displaystyle\frac{1}{2}$
0%
cannot be evaluated

Explanation

$I=\int _{ 0 }^{ \infty }{ f\left( x+\dfrac { 1 }{ x } \right) .\dfrac { \ln { x } }{ x } dx }$
Let

$x={ e }^{ t }$

$\Rightarrow dx={ e }^{ t }dt$

$I=\int _{ -\infty }^{ \infty }{ f\left( { e }^{ t }+{ e }^{ -t } \right) tdt } =0$ since,

$f\left( { e }^{ t }+{ e }^{ -t } \right) t$ is an odd function.

Ans: 0

$\displaystyle \int_{0}^{1}\displaystyle \frac{x(2x^{2}+1)}{x^{8}+2x^{6}-x^{2}+1}dx=$

0%
$\displaystyle \frac{\pi }{3}$
0%
$\displaystyle \frac{\pi }{2\sqrt{3}}$
0%
$\displaystyle \frac{2\pi }{3}$
0%
$\displaystyle \frac{\pi }{6}$

Explanation

Let

$\displaystyle I=\int_{0}^{1}\displaystyle \frac{x(2x^{2}+1)}{x^{8}+2x^{6}-x^{2}+1}dx$

$=\int_{0}^{1}\displaystyle \frac{(2x^{3}+x)}{(x^{4}+x^{2})^{2}-(x^{4}+x^{2})+1}dx$

Put

$t=x^{4}+x^{2}$

$\Rightarrow dt =4x^{3}+2x$
Also

$t=0$ for

$x=0$ and

$t=2$ for

$x=1$

So,

$I=\displaystyle \frac{1}{2} \int_{0}^{2} \frac{dt}{t^{2}-t+1}$

$=\displaystyle \frac{1}{2}\int_{0}^{2}\displaystyle \frac{dt}{\left ( t-\displaystyle \frac{1}{2} \right )^{2}+\left ( \displaystyle \frac{\sqrt{3}}{2} \right )^{2}}$

$=\displaystyle \frac{1}{2}.\displaystyle \frac{2}{\sqrt{3}}\left \{ tan^{-1}\displaystyle \frac{2\left ( t-\displaystyle \frac{1}{2} \right )}{\sqrt{3}} \right \}_{0}^{2}$

$=\displaystyle \frac{1}{\sqrt{3}}\left ( tan^{-1} (\sqrt{3})-tan^{-1}\left ( -\displaystyle \frac{1}{\sqrt{3}} \right )\right )$

$=\displaystyle \frac{1}{\sqrt{3}}\left ( \displaystyle \frac{\pi }{3} +\displaystyle \frac{\pi }{6}\right )$

$=\displaystyle \frac{\pi }{2\sqrt{3}}$

$I_{n}=\displaystyle \int_{0}^{\infty}e^{-x}x^{n-1} dx$ , then

$\displaystyle \int_{0}^{\infty}e^{-\lambda x}x^{n-1} dx$ is equal to?

0%
$\lambda I_{\mathrm{n}}$
0%
$\displaystyle \frac{1}{\lambda}I_{\mathrm{n}}$
0%
$\displaystyle \frac{I_{\mathrm{n}}}{\lambda^{\mathrm{n}}}$
0%
$\lambda^{n}I_{n}$

Explanation

$I_n=\displaystyle \int_{0}^{\infty }e^{-x}x^{n-1}dx$

then

$I_{1}=\displaystyle \int_{0}^{\infty }e^{-x}x^{n-1}dx$

$I_{1}=\displaystyle \int_{0}^{\infty }e^{-x}\dfrac{((\lambda x)^{n-1})}{\lambda }dx$

$I_{1}=\dfrac{1}{\lambda ^{n}}\displaystyle \int_{0}^{\infty }e^{-t}t^{n-1}dt$

$I_{1}=\dfrac{I_n}{\lambda ^{n}}.$

$So, \displaystyle \int_{0}^{\infty }e^{-t}x^{n-1}dt=\dfrac{I_n}{\lambda^n }.$

The value of

$\displaystyle \int_{0}^{\pi}\dfrac {dx}{1+2sin^2x}$ is

0%
$0$
0%
$\dfrac {\pi}{2\sqrt 3}$
0%
$\dfrac {\pi}{\sqrt 3}$
0%
None of these

Explanation

$\displaystyle I=\displaystyle \int_{0}^{\pi}\frac {dx}{1+2sin^2x}$

$\displaystyle =2\displaystyle \int_{0}^{\pi /2}\frac {dx}{1+2sin^2x}$

$\because \displaystyle \int_{0}^{2a}f(x)dx=2\displaystyle \int_{0}^{a} f(x)dx$ if

$f(2a-x)=f(x)$

$\displaystyle =2\displaystyle \int_{0}^{\pi /2}\frac {\sec^2xdx}{3 \tan^2x+1}$

Let

$t=\tan x$

$\Rightarrow \sec^2 xdx=dt$

$\displaystyle I=2\displaystyle \int_{0}^{\infty}\frac {dt}{3t^2+1}$

$\displaystyle =\frac {2}{\sqrt 3}tan^{-1}\sqrt {3t}\mid_{0}^{\infty}$

$\displaystyle =\frac {\pi}{\sqrt 3}$

The value of the integral

$\displaystyle \int_{0 }^{\log5}\frac{e^x\sqrt{e^x-1}}{e^x+3}\:dx$ is

0%
$3+2\pi$
0%
$4-\pi$
0%
$2+\pi$
0%
none of these

Explanation

Putting

$e^x-1=t^2$ in the given integral, we have

$\displaystyle \int_{0 }^{\log 5}\frac{e^x\sqrt{e^x-1}}{e^x+3}dx=2\int_{0}^{2}\frac{t^2}{t^2+4}dt=2\left ( \int_{0}^{2} 1 dt -4\int_{0}^{2}\frac{dt}{t^2+4}\right )$

$\displaystyle =2\left [ \left ( t-2\tan^{-1}\left ( \frac{t}{2} \right ) \right )_0^2 \right ]$

$\displaystyle =2[(2-2\times \frac{\pi}{4})]=4-\pi$

The value of the integral

$\displaystyle \int_{0}^{\frac{1}{\sqrt{3}}}\frac{dx}{(1+x^2)\sqrt{1-x^2}}$

0%
$\displaystyle \frac{\pi }{2\sqrt{2}}$
0%
$\displaystyle \frac{\pi }{4\sqrt{2}}$
0%
$\displaystyle \frac{\pi }{8\sqrt{2}}$
0%
none of these

Explanation

Orl putting

$x= \sin \theta$ ,we get

$dx =\cos \theta d \theta$
Integral (without limits)

$\displaystyle =\int \dfrac{\cos\theta d\theta}{(1+ \sin^2\theta )(\cos \theta)}$

$\displaystyle =\int \dfrac{d\theta}{1+ \sin^2\theta}=\int\dfrac{\ cosec{^2}\theta d \theta }{2+\cot^2}$

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

$t=\cot \theta$

$=-\dfrac{1}{\sqrt{2}}\tan^{-1}\dfrac{t}{\sqrt{2}}$

$=-\dfrac{1}{\sqrt{2}}\tan^{-1}\dfrac{\cot\theta}{\sqrt{2}}$

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

$\therefore$ Definite integral

$= -\dfrac{1}{\sqrt{2}}\tan^{-1}1+\dfrac{1}{\sqrt{2}}\tan^{-1}{\infty }$

$=-\dfrac{\pi}{4\sqrt{2}}+\dfrac{\pi}{2\sqrt{2}}=\dfrac{\pi}{4\sqrt{2}}$

The value of

$\displaystyle {\int}_0^{\pi}\frac{dx}{1+2 \sin^2 x}$ is

0%
$0$
0%
$\displaystyle \frac{\pi}{2\sqrt{3}}$
0%
$\displaystyle \frac{\pi}{\sqrt{3}}$
0%
none of these

Explanation

Let

$\displaystyle I={\int}_0^{\pi} \frac{dx}{1+2 \sin^2 x}$

Since,

$\sin (\pi-x)=\sin x$

$\displaystyle =2\int _{0}^{\frac{\pi}{2}} \frac{dx}{1+2\sin^{2}x}$

$\displaystyle=2\int _{0}^{\frac{\pi}2}\frac{\sec^{2}x dx}{\sec^{2}x+2\tan^{2} x}=2\int _{0}^{\frac{\pi}2}\frac{\sec^{2}x dx}{1+3\tan^{2} x}$

Put

$\tan x=t\Rightarrow \sec^{2}x dx=dt$

So,

$\displaystyle I=2\int _{0}^{\infty} \frac{dt}{1+3t^{2}} = \frac{2}{\sqrt{3}} [\tan^{-1}{(\sqrt{3}t)}]_{0}^{\infty}$

$\displaystyle I=\frac{\pi}{\sqrt{3}}$

$\displaystyle \int_{- \dfrac{\pi}{2}}^{\dfrac{\pi}{2}}\sin^{2}x. \cos^{3} x dx$ is equal to

0%
$\displaystyle \dfrac{4\pi}{30}$
0%
$\displaystyle \dfrac{4}{15}$
0%
$\displaystyle \dfrac{\pi}{30}$
0%
$\displaystyle \dfrac{1}{15}$

Explanation

Let

$\displaystyle I=\int _{ \dfrac { -\pi }{ 2 } }^{ \dfrac { \pi }{ 2 } }{ \sin ^{ 2 }{ x\left( 1-\sin ^{ 2 }{ x } \right) \cos { x } dx } }$

Substitute

$\sin { x } =t\quad \Rightarrow \cos { xdx=dt }$

$\displaystyle I=\int _{ -1 }^{ 1 }{ \left( { t }^{ 2 }\left( 1-{ t }^{ 2 } \right) \right) } dt=\int _{ -1 }^{ 1 }{ \left( { t }^{ 2 }-{ t }^{ 4 } \right) } dt$

$\displaystyle ={ \left( \frac { { t }^{ 3 } }{ 3 } -\frac { { t }^{ 5 } }{ 5 } \right) }_{ -1 }^{ 1 }=\frac { 4 }{ 15 }$

$\displaystyle \int_{\frac{5}{2}}^{5}\frac{\sqrt{(25-x^2)^3}}{x^4}\:dx$ is equal to

0%
$\displaystyle \frac{\pi }{6}$
0%
$\displaystyle \frac{2\pi }{3}$
0%
$\displaystyle \frac{5\pi }{6}$
0%
$\displaystyle \frac{\pi }{3}$

Explanation

$\displaystyle I=\int_{\frac{5}{2}}^{5}\frac{\sqrt{(25-x^2)^3}}{x^4}dx$
Let

$x= 5\sin\theta$

$\displaystyle \therefore dx= 5 \cos\theta d\theta$

$\displaystyle \therefore I=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\frac{5^3\cos ^3\theta.5\cos\theta }{5^4 \sin^4\theta}d\theta$

$\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2 \theta(\ cosec^2\theta -1)d\theta$

$\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2 \theta\ cosec^2\theta d\theta-\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2 \theta d \theta$

$\displaystyle =\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\cot^2 \theta\ cosec^2\theta d\theta- \int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(\ cosec^2 \theta-1)$

$\displaystyle =\left [ -\frac{\cot^3\theta}{3} +\cot\theta+\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$

$\displaystyle =-0+0+\frac{\pi}{2}-\left ( -\frac{3\sqrt{3}}{3}+\sqrt{3}+\frac{\pi}{6} \right )$

$\displaystyle =\frac{\pi}{3}$

$\displaystyle \int_{\log 2}^{x}\displaystyle \frac{dx}{\sqrt{e^{x}-1}}= \displaystyle \frac{\pi }{6}$ , the value of

$x$ is

0%
$4$
0%
$\log 8$
0%
$\log 4$
0%
none of these

Explanation

Let

$\displaystyle I=\int _{ \log 2 }^{ x } \frac { dx }{ \sqrt { e^{ x }-1 } } =\frac { \pi }{ 6 }$

Put

${ e }^{ x }-1={ t }^{ 2 }$

$\displaystyle \Rightarrow 2\tan ^{ -1 }{ \sqrt { { e }^{ 2 }-1 } } -\frac { \pi }{ 2 } =\frac { \pi }{ 6 } \Rightarrow \sqrt { { e }^{ 2 }-1 } =\tan { \frac { \pi }{ 3 } } =\sqrt { 3 }$

$\Rightarrow { e }^{ x }=4\Rightarrow x=\log { 4 }$

The minimum value of the function f(x) =

$\int^x_0 \frac{d \theta}{cos \theta} + \int^{\pi/2}_x \frac{d \theta}{sin \theta}$ where

$x \in [0, \frac{\pi}{2}],$ is

0%
$2ln(\sqrt{2} + 1)$
0%
$ln(2\sqrt{2} + 2)$
0%
$ln(\sqrt{3} + 2)$
0%
$ln(\sqrt{2} + 3)$

Explanation

f(x) =

$\int^x_0 \frac{d \theta}{cos \theta} + \int^{\pi/2}_x \frac{d \theta}{sin \theta}$

$f^{'}(x) = \frac{1}{cos x} - \frac{1}{sin x}$
=

$\frac{sin x - cos x}{sin xcos x}$

$f^{'}(x)$ change sign

$-t_0 + at x = \frac{\pi}{4}$
so f(x) takes minimum value at x =

$\frac{\pi}{4}$

$f(x) = \int^{\pi /4}_0 sec \theta d\theta+ \int^{\pi /2}_{\pi /4} cosec\theta d\theta$

$= (ln(sec \theta + tan \theta))^{\pi/4}_0 + (ln(cosec \theta - cot \theta))^{\pi/2}_{\pi/4}$

$= ln(\sqrt{2} + 1) - ln(1 + 0) + ln 1 - ln(\sqrt{2} - 1)$

$= ln[\frac{\sqrt{2} + 1}{\sqrt{2} - 1}] = ln(\sqrt{2} + 1)^2 = 2 ln(\sqrt{2} + 1)$

The value of

$\int _{-\pi /2}^{\pi /2}\left (psin^3 x+qsin^4 x +r sin^5 x \right )$ does not depend on

0%
p, q, r
0%
p, r only
0%
p only
0%
q, r only

Explanation

$\overset {b}{\underset {a}{\int}}f(x) dx=\overset {b}{\underset {a}{\int}}f(a+b-x)$

$\displaystyle \int_{0}^{1} \frac{\tan^{-1} x}{x}dx$ is equal to

0%
$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{\sin x}{x}dx$
0%
$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x}dx$
0%
$\displaystyle \frac{1}{2} \int_{0}^{\frac{\pi}{2}}\frac{\sin x}{x}dx$
0%
$\displaystyle \frac{1}{2} \int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x}dx$

Explanation

$\displaystyle I= \int_{0}^{1} \frac{\tan^{-1} x}{x}dx$
Putting

$x=\tan \theta$ or

$dx= \sec^2 \theta d\theta$ , we get

$I\displaystyle =\int_{0}^{\frac{\pi}{4}}\frac{\theta}{\tan \theta}\sec^2 \theta d \theta$

$\displaystyle =\int_{0}^{\frac{\pi}{4}}\frac{2\theta}{\sin 2\theta}d\theta$
Putting

$2\theta=t$ ,i.e.,

$2d\theta=dt$ , we get

$I\displaystyle =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{t}{\sin t}dt$

$\displaystyle =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\frac{x}{\sin x}dx$

Let

$\displaystyle I_1=\int_{0}^{1}\frac{e^x dx}{1+x}$ and

$\displaystyle I_2=\int_{0}^{1}\dfrac{x^2 dx}{e^{x^3}(2-x^3)}$ .

Then

$\dfrac{I_1}{I_2}$ is equal to?

0%
$\displaystyle \dfrac{3}{e}$
0%
$\displaystyle \frac{e}{3}$
0%
$\displaystyle {3}{e}$
0%
$\displaystyle \dfrac{1}{3e}$

Explanation

$\displaystyle{ I }_{ 2 }=\int _{ 0 }^{ 1 }{ \cfrac { { x }^{ 2 }dx }{ { e }^{ { x }^{ 3 } }\left( 2-{ x }^{ 3 } \right) } }$
Substitute

$1-{ x }^{ 3 }=t\Rightarrow -3{ x }^{ 2 }dx=dt$
We get

$\displaystyle{ I }_{ 2 }=-\cfrac { 1 }{ 3 } \int _{ 1 }^{ 0 }{ \cfrac { dt }{ { e }^{ 1-t }\left( 1+t \right) } } =\cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ t }dt }{ 1+t } } =\cfrac { 1 }{ 3e } \int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ 1+x } }$
As

$\displaystyle{ I }_{ 1 }=\int _{ 0 }^{ 1 }{ \cfrac { { e }^{ x }dx }{ 1+x } }$
Therefore

$\cfrac { { I }_{ 1 } }{ { I }_{ 2 } } =3e$

The value of

$\displaystyle \int_{0}^{\infty} \frac {dx}{1 + x^4}$ is

0%
same as that of $\displaystyle \int_{0}^{\infty} \frac {x^2 + 1dx}{1 + x^4}$
0%
$\displaystyle \frac {\pi}{2\sqrt{2}}$
0%
same as that of $\displaystyle \int_{0}^{\infty} \frac {x^2 \: dx}{1 + x^4}$
0%
$\displaystyle \frac {\pi}{\sqrt{2}}$

Explanation

$\displaystyle I = \int_{0}^{\infty} \frac {dx}{1 + x^4}$ ....(1)

$\displaystyle \int_{0}^{\infty} \frac {x^2 + 1 - x^2}{1 + x^4} dx$

$\displaystyle = \int_{0}^{\infty} \frac {x^2}{1 + x^4} dx + \int_{0}^{\infty} \frac {1 - x^2}{1 + x^4} dx$ ....(2)

Let

$I_1 =\int_{0}^{\infty} \frac {x^2}{1 + x^4} dx$

and

$I_2= \int_{0}^{\infty} \frac {1 - x^2}{1 + x^4} dx$

$\displaystyle I_2 = \int_{0}^{\infty} \frac {\frac {1}{x^2}-1}{\frac {1}{x^2}+x^2} dx$

$I_{ 2 }=\int _{ 0 }^{ \infty } \displaystyle \frac { \frac { 1 }{ x^{ 2 } } -1 }{ \frac { 1 }{ x^{ 2 } } +x^{ 2 }+2-2 } dx$

$I_{ 2 }=\int _{ 0 }^{ \infty } \displaystyle \frac { \frac { 1 }{ x^{ 2 } } -1 }{ (x+\frac { 1 }{ x } )^{ 2 }-2 } dx$

Put

$\displaystyle x + \frac {1}{x} = y$

$\Rightarrow dy=(1-\frac { 1 }{ x^{ 2 } } )dx$
Also, when

$x=0\Rightarrow y=\infty$
And when

$x=\infty \Rightarrow y=\infty$

$\displaystyle \therefore I_2 = \int_{\infty}^{\infty} \displaystyle \frac {-1}{y^2 - 2} dy = 0$

$\therefore I=\int _{ 0 }^{ \infty } \displaystyle \frac { x^{ 2 }dx }{ 1+x^{ 4 } }$ .....(3)

Adding equations (1) and (3), we get

$2I=\int _{ 0 }^{ \infty } \displaystyle \frac { 1+x^{ 2 } }{ 1+x^{ 4 } } dx$

$= \int_{0}^{\infty} \displaystyle \frac {\frac {1}{x^2}+1}{\frac {1}{x^2}+x^2}dx$

$\displaystyle \:$

$=\int _{ 0 }^{ \infty } \displaystyle \frac { \frac { 1 }{ x^{ 2 } } +1 }{ \frac { 1 }{ x^{ 2 } } +x^{ 2 }+2-2 } dx$

$=\int _{ 0 }^{ \infty } \displaystyle \frac { \frac { 1 }{ x^{ 2 } } +1 }{ (x-\frac { 1 }{ x } )^{ 2 }+2 } dx$

Put

$\displaystyle x - \frac {1}{x} = t$

$\Rightarrow dt=(1+\frac { 1 }{ x^{ 2 } } )dx$

$\displaystyle 2I = \int_{-\infty}^{\infty} \frac {dt}{t^2 + 2}$

$\displaystyle = [\displaystyle\frac {1}{\sqrt {2}} tan^{-1} \frac {t}{\sqrt{2}}]_{- \infty}^{\infty} = \frac {\pi}{\sqrt {2}}$

$\therefore \displaystyle I = \frac {\pi}{2 \sqrt {2}}$

The value of the integral

$\displaystyle \int_{-\pi}^{\pi} (\cos ax - \sin b x)^2 dx$ , where

$a$ and

$b$ are integers, is

0%
$2\pi (1+a+b)$
0%
$0$
0%
$\pi$
0%
$2 \pi$

Explanation

$\displaystyle I=\int _{ -\pi }^{ \pi }{ { \left( \cos ax-\sin bx \right) }^{ 2 }dx } \ =\int _{ -\pi }^{ \pi }{ { \sin }^{ 2 }bx\>dx } -2\int _{ -\pi }^{ \pi }{ \left( \left( \cos ax \right) \left( \sin bx \right) \right) dx } +\int _{ -\pi }^{ \pi }{ { \cos }^{ 2 }ax\>dx }$
Let

$\displaystyle I={ I }_{ 1 }+{ I }_{ 2 }+{ I }_{ 3 }$ ...(1)
Such that,

$\displaystyle{ I }_{ 1 }=\int _{ -\pi }^{ \pi }{ { \sin }^{ 2 }bxdx }$
Substituting

$bx=t\Rightarrow b\>dx=dt$
We get,

$\displaystyle{ I }_{ 1 }=\cfrac { 1 }{ b } \int _{ -\pi b }^{ \pi b }{ { \sin }^{ 2 }tdt } =\cfrac { 1 }{ b } \int _{ -\pi b }^{ \pi b }{ \left( \cfrac { 1 }{ 2 } -\cfrac { 1 }{ 2 } \cos2t \right) dt }$

$\displaystyle=\cfrac { 1 }{ b } { \left[ \cfrac { t }{ 2 } -\cfrac { \sin2t }{ 4 } \right] }_{ -\pi b }^{ \pi b }=\pi -\cfrac { \sin\left( 2\pi b \right) }{ 2b }$ ...(2)

$\displaystyle{ I }_{ 2 }=-2\int _{ -\pi }^{ \pi }{ \left( \left( \cos ax \right) \left( \sin bx \right) \right) dx }$
As

$\left( \cos ax \right) \left( \sin bx \right)$ is an odd function,

${ I }_{ 2 }=0$ ...(3)

$\displaystyle{ I }_{ 3 }=\int _{ -\pi }^{ \pi }{ { \cos }^{ 2 }axdx }$
Substituting,

$ax=t\Rightarrow adx=dt$ , we get

$\displaystyle{ I }_{ 3 }=\cfrac { 1 }{ a } \int _{ -\pi a }^{ \pi a }{ { \cos }^{ 2 }tdt } =\cfrac { 1 }{ a } \int _{ -\pi a }^{ \pi a }{ \left( \cfrac { 1 }{ 2 } \cos2t+\cfrac { 1 }{ 2 } \right) dt }$

$\displaystyle=\cfrac { 1 }{ a } { \left[ \cfrac { t }{ 2 } +\cfrac { \sin2t }{ 4 } \right] }_{ -\pi a }^{ \pi a }=\cfrac { \sin\left( 2\pi a \right) }{ 2a } +\pi$ ...(4)
Substituting (2), (3) and (4) in (1), we get

$\displaystyle I=\cfrac { \sin\left( 2\pi a \right) }{ 2a } -\cfrac { \sin\left( 2\pi b \right) }{ 2b } +2\pi =0+0+2\pi =2\pi$
Hence, option 'D' is the correct answer.

Evaluate

$\displaystyle\int_0^{\displaystyle\frac{\pi}{4}}{\frac{\sin{x}+\cos{x}}{9+16\{1-{(\sin{x}-\cos{x})}^2\}}dx}$

0%
$\displaystyle\frac{1}{20}(\log{\sqrt{3}})$
0%
$\displaystyle\frac{1}{10}(\log{3})$
0%
$\displaystyle\frac{1}{20}(\log{3})$
0%
$\displaystyle\frac{1}{20}(\log{3}-\log{2})$

Explanation

Let

$\displaystyle I=\int_0^{\displaystyle\frac{\pi}{4}}{\frac{\sin{x}+\cos{x}}{9+16\{1-{(\sin{x}-\cos{x})}^2\}}dx}$

$\displaystyle\therefore I=\int_0^{\displaystyle\frac{\pi}{4}}{\frac{\sin{x}+\cos{x}}{25-16{(\sin{x}-\cos{x})}^2}dx}$ ,
Substitute

$4(\sin{x}-\cos{x})=t\implies4(\cos{x}+\sin{x})dx=dt$

$\displaystyle\frac{1}{4}\int_{\displaystyle-4}^0{\frac{dt}{25-t^2}}$

$\displaystyle I=\frac{1}{4}.\frac{1}{2(5)}{\left(\log{\left|\frac{5-t}{5+t}\right|}\right)}_{\displaystyle-4}^0$

$\displaystyle=\frac{1}{40}\left[\log{\left|\frac{5-0}{5+0}\right|}-\log{\left|\frac{5+4}{5-4}\right|}\right]$

$\displaystyle=\frac{1}{40}\left[\log{1}-\log{\left(\frac{1}{9}\right)}\right]$ , (where

$\log{1}=0$ )

$\displaystyle=\frac{1}{40}[\log{9}]=\frac{2}{40}\log{3}=\frac{1}{20}(\log{3})$

$\displaystyle \int_{1}^{2} e^{x^2} dx= a$ , then

$\displaystyle \int_{e}^{e^4}\sqrt{\ln x} \:dx$ is equal to

0%
$2e^4-2e-a$
0%
$2e^4-e-a$
0%
$2e^4-e-2a$
0%
$e^4-e-a$

Explanation

Let

$I=\int _{ e }^{ { e }^{ 4 } }{ \sqrt { \ln { x } } dx }$

Substitute

$\sqrt { \ln { x } } =t\Rightarrow \cfrac { dx }{ 2x\sqrt { \ln { x } } } =dt\Rightarrow dx=2t{ e }^{ { t }^{ 2 } }dt$

We get

$I=\int _{ 1 }^{ 2 }{ 2{ t }^{ 2 }{ e }^{ t^{ 2 } }dt }$
Now integrating by parts, we get

$I={ \left[ t{ e }^{ t^{ 2 } } \right] }_{ 1 }^{ 2 }-\int _{ 1 }^{ 2 }{ { e }^{ t^{ 2 } }dt }$

As

$\int _{ 1 }^{ 2 }{ { e }^{ { x }^{ 2 } }dx } =a$
Therefore

$I=2{ e }^{ 4 }-e-a$
Hence, option 'B' is correct.

$\displaystyle \int_{ 0 }^{\infty}\frac{ \:dx}{\left [ x+\sqrt{x^2+1} \right ]^3}$ is equal to

0%
$\displaystyle \frac{3}{8}$
0%
$\displaystyle \frac{1}{8}$
0%
$\displaystyle -\frac{3}{8}$
0%
none of these

Explanation

Putting

$x =\tan \theta$ ,we get

$\displaystyle \int_{ 0 }^{\infty}\frac{ \:dx}{\left [ x+\sqrt{x^2+1} \right ]^3}=\int_{0}^{\frac{\pi}{2} }\frac{\sec^2\theta d \theta}{(\tan\theta+\sec\theta)^3}$

$\displaystyle =\int_{0}^{\frac{\pi}{2}}\frac{\cos\theta}{(1+\sin\theta)^3}d\theta$

$\displaystyle =\left [ \frac{1}{2(1+\sin\theta)^2} \right ]_0^{\frac{\pi}{2}}=-\frac{1}{8}+\frac{1}{2}=\frac{3}{8}$

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

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Practice Class 12 Commerce Applied Mathematics Quiz Questions and Answers

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