Integrals Test ...

$$\displaystyle \int_{\pi/4}^{\pi/2}{\sqrt{2+\sqrt{2+2\cos 4x}}dx}$$ is equal to 

The value $$\int _{ 3 }^{ 6 }{ \left( \sqrt { x+\sqrt { 12x-36 }  } +\sqrt { x-\sqrt { 12x-36 }  }  \right)  } dx$$ is equal to 

If $${ I }_{ m }=\overset { e }{ \underset { 1 }{ \int   }  } (lnx)^{ m }dx,$$ where $$m\epsilon N,$$then $${ I }_{ 10 }+10{ I }_{ 9 }$$ is equal to-

If $$f\left( x \right) =\int _{ 0 }^{ 1 }{ \left( xf\left( t \right) +1 \right) dt,then\int _{ 0 }^{ 3 }{ f\left( x \right) dx=12 }  } $$ 
because 
Statement-2: f(x) = 3x + 1

If for every integer n, $$\int _{ n }^{ n+1 }{ f(x)dx={ n }^{ 2 } } $$, then the value of $$\int _{ -2 }^{ 4 }{ f(x)dx } $$ is -

If I = $$\overset { 2 }{ \underset { -3 }{ \int }  }$$ (|x + 1 | + |x + 2| + |x - 1|)dx, then I equal

$$7 \left( \int_{\pi}^{0}\dfrac{x^{4}(1-x)^{4} dx}{1+x^{2}}+\pi \right)$$ is equal to 

If $$\overset { 1 }{ \underset { 0 }{\int  }  } 2^{x^2}dx,I_2=\overset { 1 }{ \underset { 0 }{\int  }  }2^{x^3}dx,I_3 =\overset { 2 }{ \underset { 1 }{\int  }  }2^{x^2}dx, $$ and $$I_4=\overset { 1 }{ \underset { 0 }{\int  }  }2^{x^3}dx $$ then-

$$\int _{ 0 }^{ 4036 }{ \dfrac { { 2 }^{ x } }{ { 2 }^{ x }+{ 1 }^{ 4036-x } }  } dx=............$$

If $${ I }_{ 1 }=\int _{ x }^{ 1 }{ \cfrac { 1 }{ 1+{ t }^{ 2 } }  } dt$$ and $${ I }_{ 2 }=\int _{ 1 }^{ 1/x }{ \cfrac { 1 }{ 1+{ t }^{ 2 } }  } dt$$ for x > 0, then