Qus : 51
1
The value of $\displaystyle \int_{-\pi/2}^{\pi/2} \frac{\sin^{2}x}{1+2^{x}},dx$ is :
1
$\displaystyle \frac{\pi}{4}$
2
$\displaystyle \frac{\pi}{8}$
3
$\displaystyle \frac{\pi}{2}$
✓ Solution
Qus : 52
2
The integral
$\int_{0}^{\tfrac{\pi}{2}} \dfrac{1}{3 + 2 \sin x + \cos x} , dx$
is equal to :
2
$\ \tan^{-1}(2) - \tfrac{\pi}{4}$
3
$\ \tfrac{1}{2}\tan^{-1}(2) - \tfrac{\pi}{8}$
✓ Solution
Qus : 53
4
The function $I(x)=\int e^{\sin^{2}x},(\cos x\sin 2x-\sin x),dx$ with $I(0)=1$. Then $I!\left(\dfrac{\pi}{3}\right)$ is equal to:
✓ Solution
Qus : 54
4
If $f(\alpha)=\int\limits_{1}^{\alpha} \frac{\log _{10} \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}, \alpha>0$ then $f\left(\mathrm{e}^{3}\right)+f\left(\mathrm{e}^{-3}\right)$ is equal to :
3
$\frac{9}{\log _{e}(10)}$
4
$\frac{9}{2 \log _{e}(10)}$
✓ Solution
Qus : 55
1
Let, for some function $y=f(x)$, $\displaystyle \int_{0}^{x} t,f(t),dt = x^{2}f(x)$ for $x>0$ and $f(2)=3$. Then $f(6)$ is equal to
✓ Solution
Qus : 56
1
The value of the integral $\displaystyle \int_{0}^{1} \frac{\sqrt{x}\,dx}{(1+x)(1+3x)(3+x)}$ is:
1
$\dfrac{\pi}{8}\left(1-\dfrac{\sqrt{3}}{2}\right)$
2
$\dfrac{\pi}{4}\left(1-\dfrac{\sqrt{3}}{6}\right)$
3
$\dfrac{\pi}{8}\left(1-\dfrac{\sqrt{3}}{6}\right)$
4
$\dfrac{\pi}{4}\left(1-\dfrac{\sqrt{3}}{2}\right)$
✓ Solution
Qus : 57
4
For $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, if
$y(x)=\displaystyle\int \frac{\csc x+\sin x}{\csc x\sec x+\tan x\sin^2 x}\,dx$,
and $\displaystyle\lim_{x\to \left(\frac{\pi}{2}\right)} y(x)=0$, then $y\!\left(\dfrac{\pi}{4}\right)$ is equal to:
1
$-\dfrac{1}{\sqrt{2}}\tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right)$
2
$\tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right)$
3
$\dfrac{1}{2}\tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right)$
4
$\dfrac{1}{\sqrt{2}}\tan^{-1}\!\left(-\dfrac{1}{2}\right)$
✓ Solution
Qus : 58
2
$\displaystyle \int_{0}^{\pi/4}\frac{\cos^{2}x,\sin^{2}x}{\big(\cos^{3}x+\sin^{3}x\big)^{2}},dx$ is equal to:
✓ Solution
Qus : 59
4
If ${b_n} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx} \over {\sin x}}dx,\,n \in N} $, then
1
$b_3 - b_2,\ b_4 - b_3,\ b_5 - b_4$ are in A.P. with common difference $-2$
2
$\dfrac{1}{b_3 - b_2},\ \dfrac{1}{b_4 - b_3},\ \dfrac{1}{b_5 - b_4}$ are in A.P. with common difference $2$
3
$b_3 - b_2,\ b_4 - b_3,\ b_5 - b_4$ are in a G.P.
4
$\dfrac{1}{b_3 - b_2},\ \dfrac{1}{b_4 - b_3},\ \dfrac{1}{b_5 - b_4}$ are in A.P. with common difference $-2$
✓ Solution
Qus : 60
2
The integral $\int\limits_1^e {\left\{ {{{\left( {{x \over e}} \right)}^{2x}} - {{\left( {{e \over x}} \right)}^x}} \right\}} \,$ loge x dx is equal to :
1
$-\frac{1}{2} + \frac{1}{e} - \frac{1}{2e^{2}}$
2
$\frac{3}{2} - e - \frac{1}{2e^{2}}$
3
$\frac{1}{2} - e - \frac{1}{e}$
4
$\frac{3}{2} - \frac{1}{e} - \frac{1}{2e^{2}}$
✓ Solution