Integrals Test ...

Value of the definite integral $$\displaystyle \int _{ 0 }^{ 1 }{ \cot ^{ -1 }{ \left( 1-x+{ x }^{ 2 } \right)  } dx }$$ is:

If $$\displaystyle I=\int _{ { -\pi  }/{ 6 } }^{ { \pi  }/{ 6 } }{ \dfrac { \pi +4{ x }^{ 5 } }{ 1-\sin { \left( \left| x \right| +\dfrac { \pi  }{ 6 }  \right)  }  }  } dx$$, then I equals to

$$\displaystyle \int^{\pi/2}_{-\pi/2}\sqrt {\cos x-\cos^{3}x}dx=$$

$$For x>0$$,let  $$f\left( x \right) =\int _{ 1 }^{ 2 }{ \dfrac { \log { t }  }{ 1+t }  } dt$$,then$$f\left( x \right) +f\left( \dfrac { 1 }{ x }  \right)$$is equal to:

If, $$\int^{\frac{\pi}{2}} _0 \frac{sin^2x}{(1 + cosx^2)} dx = 0$$

$${I}_{n}=\int_{1}^{e}{\left(logx\right)^{n}dx}$$ and $${I}_{n}=A+{BI}_{n-1}$$ then
A=........., B=............

If, $$\displaystyle\int^{\displaystyle\frac{\pi}{2}}_0\frac{\sin^2x}{(1+\cos x)^2}dx=$$.

What is $$\int _{ 0 }^{ 1 }{ x{ \left( 1-x \right)  }^{ 9 }dx } $$ equal to?

If $$(-1, 2)$$ and $$(2, 4)$$ are two points on the curve $$y=f(x)$$ and if $$g(x)$$ is the gradient of the curve at point (x, y), then the value of the integral $$\displaystyle\int^{2}_{-1}g(x)dx$$, is?

The integral $$\int _{ 0 }^{ \pi  }{ \sqrt { 1+4\sin { ^{ 2 }\frac { x }{ 2 } -4 } \sin { \frac { x }{ 2 }  }  } dx }$$ is equal to