${\int }_{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\frac{\mathrm{dx}}{1 + \sqrt{\tan \left(\mathrm{x}\right)}}$ is equal to :

• $\frac{\mathrm{\pi }}{12}$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{3\mathrm{\pi }}{2}$

• $2\mathrm{\pi }$

${\int }_{- \mathrm{\pi }}^{\mathrm{\pi }}\frac{{\sin }^4\left(\mathrm{x}\right)}{{\sin }^4\left(\mathrm{x}\right) + {\cos }^4\left(\mathrm{x}\right)}$dx is equal to :

• $\mathrm{\pi }$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{3\mathrm{\pi }}{2}$

• $2\mathrm{\pi }$

583.

The value of $\frac{2^{\sin \left(\mathrm{x}\right)}}{2^{\sin \left(\mathrm{x}\right)} + 2^{\cos \left(\mathrm{x}\right)}}\mathrm{dx}$ is :

• 2

• $\mathrm{\pi }$

• $\frac{\mathrm{\pi }}{4}$

• $2\mathrm{\pi }$

If f is continuous function, then :

• ${\int }_{- 2}^2\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} = {\int }_0^2\left[\mathrm{f}\left(\mathrm{x}\right) - \mathrm{f}\left(- \mathrm{x}\right)\right]\mathrm{dx}$

• ${\int }_{- 3}^{5}2\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x} = {\int }_{- 6}^{10}\mathrm{f}\left(\mathrm{x} - 1\right)\mathrm{dx}$

• ${\int }_{- 3}^5\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} = {\int }_{- 4}^4\mathrm{f}\left(\mathrm{x} - 1\right)\mathrm{dx}$

• ${\int }_{- 3}^5\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} = {\int }_{- 2}^6\mathrm{f}\left(\mathrm{x} - 1\right)\mathrm{dx}$

If $\int \frac{\sqrt{\mathrm{x}}}{\mathrm{x} + 1}\mathrm{dx} = \mathrm{A}\sqrt{\mathrm{x}} + {\mathrm{Btan}}^{-1}\left(\sqrt{\mathrm{x}}\right) + \mathrm{c}$, then :

• A = 1, B = 1

• A = 1, B = 2

• A = 2, B = 2

• A = 2, B = - 2

$\int \frac{{\mathrm{x}}^3\sin \left[{\tan }^{-1}\left({\mathrm{x}}^4\right)\right]}{1 + {\mathrm{x}}^8}\mathrm{dx}$ is equal to :

• $\frac{1}{4}\cos \left[{\tan }^{-1}\left({\mathrm{x}}^4\right)\right] + \mathrm{c}$

• $\frac{1}{4}\sin \left[{\tan }^{-1}\left({\mathrm{x}}^4\right)\right] + \mathrm{c}$

• $- \frac{1}{4}\cos \left[{\tan }^{-1}\left({\mathrm{x}}^4\right)\right] + \mathrm{c}$

• $\frac{1}{4}{\sec }^{-1}\left[{\tan }^{-1}\left({\mathrm{x}}^4\right)\right] + \mathrm{c}$

I_n = ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\tan }^{\mathrm{n}}\left(\mathrm{x}\right)\mathrm{dx}$, then $\underset{\mathrm{n}\to \infty }{\lim }\mathrm{n}\left[{\mathrm{I}}_{\mathrm{n}} + {\mathrm{I}}_{\mathrm{n} +\hspace{0.17em}2}\right]$ equals :

• $\frac{1}{2}$

• 1

• $\infty$

• zero

If $\int \mathrm{xf}\left(\mathrm{x}\right)\mathrm{dx} = \frac{\mathrm{f}\left(\mathrm{x}\right)}{2}$, then f(x) is equal to :

• e^x

• e^{- x}

• log(x)

• $\frac{{\mathrm{e}}^{{\mathrm{x}}^2}}{2}$

${\int }_0^2\left[{\mathrm{x}}^2\right]\mathrm{dx}$ is :

• $2 - \sqrt{2}$

• $2 + \sqrt{2}$

• $\sqrt{2} - 1$

• $- \sqrt{2} - \sqrt{3} + 5$

${\int }_0^{\mathrm{\pi }}\left|\cos \left(\mathrm{x}\right)\right|\mathrm{dx}$ is equal to :

• $\frac{1}{2}$

• - 2

• 1

• - 1