The integral $$\int\limits_0^\pi {\sqrt {1 + 4{{\sin }^2}{x \over 2} - 4\sin {x \over 2}{\mkern 1mu} } } dx$$ equals:

Statement-1 : The value of the integral
$$\int\limits_{\pi /6}^{\pi /3} {{{dx} \over {1 + \sqrt {\tan \,x} }}} $$ is equal to $$\pi /6$$

Statement-2 : $$\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^b {f\left( {a + b - x} \right)} dx.$$

The value of $$\int\limits_0^1 {{{8\log \left( {1 + x} \right)} \over {1 + {x^2}}}} dx$$ is

Let $$p(x)$$ be a function defined on $$R$$ such that $$p'(x)=p'(1-x),$$ for all $$x \in \left[ {0,1} \right],p\left( 0 \right) = 1$$ and $$p(1)=41.$$ Then $$\int\limits_0^1 {p\left( x \right)dx} $$ equals :