If I₁ = ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{xsin}\left(\mathrm{x}\right)\mathrm{dx}$, I₂ = ${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\mathrm{xcos}\left(\mathrm{x}\right)\mathrm{dx}$, then whi ch one f the followin is true ?

• I₁ = I₂

• I₁ + I₂ = 0

• ${\mathrm{I}}_1 = \frac{\mathrm{\pi }}{2} . {\mathrm{I}}_2$

• ${\mathrm{I}}_1 + {\mathrm{I}}_2 = \frac{\mathrm{\pi }}{2}$

The value of ${\int }_{- 1}^2\frac{\left|\mathrm{x}\right|}{\mathrm{x}}\mathrm{dx}$, is

• 0

• 1

• 2

• 3

${\int }_0^{\mathrm{\pi }}\frac{{\cos }^4\left(\mathrm{x}\right)}{{\cos }^4\left(\mathrm{x}\right) + {\sin }^4\left(\mathrm{x}\right)}\mathrm{dx}$ is equal to

• $\frac{\mathrm{\pi }}{4}$

• $\frac{\mathrm{\pi }}{2}$

• $\frac{\mathrm{\pi }}{8}$

• $\mathrm{\pi }$

If f(x) = $\mathrm{f}\left(\mathrm{\pi } + \mathrm{e} - \mathrm{x}\right) \mathrm{and} {\int }_{\mathrm{e}}^{\mathrm{\pi }}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} = \frac{2}{\mathrm{e} +\hspace{0.17em}\mathrm{\pi }}, \mathrm{then} {\int }_{\mathrm{e}}^{\mathrm{\pi }}\mathrm{xf}\left(\mathrm{x}\right)\mathrm{dx}$ is equal to

• $\mathrm{\pi } - \mathrm{e}$

• $\frac{\mathrm{\pi } + \mathrm{e}}{2}$

• 1

• $\frac{\mathrm{\pi } - \mathrm{e}}{2}$

If linear function f(x) and g(x) satisfy $\int \left[\left(3\mathrm{x} - 1\right)\cos \left(\mathrm{x}\right) + \left(1 - 2\mathrm{x}\right)\sin \left(\mathrm{x}\right)\right]\mathrm{dx}$ = $\mathrm{f}\left(\mathrm{x}\right)\cos \left(\mathrm{x}\right) + \mathrm{g}\left(\mathrm{x}\right)\sin \left(\mathrm{x}\right) + \mathrm{C}$, then

• f(x) = 3(x - 1)

• f(x) = 3x - 5

• g(x) = 3(x - 1)

• g(x) = 3 + x

The value of the integral ${\int }_{- \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\sec \left(\mathrm{\theta }\right) - \tan \left(\mathrm{\theta }\right)\right)\mathrm{d\theta }$ is

• 0

• $\frac{\mathrm{\pi }}{4}$

• $\mathrm{\pi }$

• $\frac{\mathrm{\pi }}{2}$

$\int \frac{\sin \left(2\mathrm{x}\right)}{{\sin }^2\left(\mathrm{x}\right) + 2{\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}$ is equal to

• $- \log \left(1 + {\sin }^2\left(\mathrm{x}\right)\right) + \mathrm{C}$

• $\log \left(1 + {\cos }^2\left(\mathrm{x}\right)\right) + \mathrm{C}$

• $- \log \left(1 + {\cos }^2\left(\mathrm{x}\right)\right) + \mathrm{C}$

• $\log \left(1 + {\tan }^2\left(\mathrm{x}\right)\right) + \mathrm{C}$

$\int \frac{1}{{\mathrm{x}}^2{\left({\mathrm{x}}^4 + 1\right)}^{{\displaystyle \frac{3}{4}}}}\mathrm{dx}$ is equal to

• $\frac{- {\left(1 + {\mathrm{x}}^4\right)}^{{\displaystyle \frac{1}{4}}}}{2\mathrm{x}} + \mathrm{C}$

• $\frac{- {\left(1 + {\mathrm{x}}^4\right)}^{{\displaystyle \frac{1}{4}}}}{\mathrm{x}} + \mathrm{C}$

• $\frac{- {\left(1 + {\mathrm{x}}^4\right)}^{{\displaystyle \frac{3}{4}}}}{\mathrm{x}} + \mathrm{C}$

• $\frac{- {\left(1 + {\mathrm{x}}^4\right)}^{{\displaystyle \frac{1}{4}}}}{{\mathrm{x}}^2} + \mathrm{C}$

${\int }_0^{\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\log \left(\frac{\sin \left(\mathrm{x}\right) + \cos \left(\mathrm{x}\right)}{\cos \left(\mathrm{x}\right)}\right)\mathrm{dx}$ is equal to

• ${\int }_0^{\raisebox{1ex}{${\displaystyle \mathrm{\pi }}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle 4}$}\right.}\log \left(\frac{{\displaystyle \sin \left(\mathrm{x}\right) + \cos \left(\mathrm{x}\right)}}{{\displaystyle \cos \left(\mathrm{x}\right)}}\right)\mathrm{dx}$

• $\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 4}}\log \left(2\right)$

• $\log \left(2\right)$

• $\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 2}}\log \left(2\right)$

520.

$\int \frac{{\sin }^2\left(\mathrm{x}\right)}{1 + \cos \left(\mathrm{x}\right)}\mathrm{dx}$ is equal to

• $\sin \left(\mathrm{x}\right) + \mathrm{C}$

• $\mathrm{x} + \sin \left(\mathrm{x}\right) + \mathrm{C}$

• $\cos \left(\mathrm{x}\right) + \mathrm{C}$

• $\mathrm{x} - \sin \left(\mathrm{x}\right) + \mathrm{C}$