Integrals Test ...

$\displaystyle \int_{0}^{\pi/4}{\dfrac{x.\sin x}{\cos^{3}x}dx}$ equals to :

The value of $\cfrac { \int _{ 0 }^{ \pi /2 }{ { \left( \sin { x }  \right)  }^{ \sqrt { 3 } +1 } } dx }{ \int _{ 0 }^{ \pi /2 }{ { \left( \sin { x }  \right)  }^{ \sqrt { 3 } -1 } }  }$ is

The value of $\displaystyle \int _{-1}^{1} \log{\left(\dfrac{2-x}{2+x}\right)}\sin^{2}{x}dx$

$\displaystyle\int^1_0\tan^{-1}\left[\dfrac{2x-1}{1+x-x^2}\right]dx=?$

$\displaystyle\int _{ -1 }^{ 1 }{ x\ell n\left( 1+{ e }^{ x } \right) dx } =$

$\displaystyle\int^{2+\sqrt{3}}_{2-\sqrt{3}}\dfrac{xdx}{(1+x)(1+x^2)}=?$

$\displaystyle\int^{2\pi}_0[\sin x]dx$.

$\int _{ \pi /4 }^{ 3\pi /4 }{ \dfrac { dx }{ 1+\cos { x }  }  }$ is equal to 

Evaluate: $\displaystyle \int \dfrac { 1 } { x ^ { 2 } \left( x ^ { 4 } + 1 \right) ^ { \frac { 3 } { 4 } } } d x ; x = 0$

If $\displaystyle \int _{0}^{1}\cot^{-1}(1+x^{2}-x)dx=k\left(\dfrac {\pi}{4}-\log_{e}\sqrt {2}\right)$, then the value of $k$ is equal to