The solution for $$x$$ of the equation $$\int\limits_{\sqrt 2 }^x {{{dt} \over {t\sqrt {{t^2} - 1} }} = {\pi \over 2}} $$ is

Let $$I = \int\limits_0^1 {{{\sin x} \over {\sqrt x }}dx} $$ and $$J = \int\limits_0^1 {{{\cos x} \over {\sqrt x }}dx} .$$ Then which one of the following is true?

$$\int\limits_0^\pi {xf\left( {\sin x} \right)dx} $$ is equal to

$$\int\limits_{ - {{3\pi } \over 2}}^{ - {\pi \over 2}} {\left[ {{{\left( {x + \pi } \right)}^3} + {{\cos }^2}\left( {x + 3\pi } \right)} \right]} dx$$ is equal to