Answer
$$1 - \frac{{\sqrt 2 }}{2}$$
Work Step by Step
$$\eqalign{
& \int_0^{\sqrt \pi /2} {\frac{x}{{1 + \sin {x^2}}}} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx,{\text{ }}dx = \frac{{du}}{{2x}} \cr
& {\text{The new limits of integration are}} \cr
& x = \frac{{\sqrt \pi }}{2} \to u = {\left( {\frac{{\sqrt \pi }}{2}} \right)^2} = \frac{\pi }{4} \cr
& x = 0 \to u = {\left( 0 \right)^2} = 0 \cr
& {\text{Substituting}} \cr
& \int_0^{\sqrt \pi /2} {\frac{x}{{1 + \sin {x^2}}}} dx = \int_0^{\pi /4} {\frac{x}{{1 + \sin u}}\left( {\frac{1}{{2x}}} \right)} du \cr
& = \frac{1}{2}\int_0^{\pi /4} {\frac{1}{{1 + \sin u}}} du \cr
& {\text{Integrate by tables }}\int {\frac{1}{{1 + \sin u}}du = \tan u - \sec u + C,{\text{ then}}} \cr
& \frac{1}{2}\int_0^{\pi /4} {\frac{1}{{1 + \sin u}}} du = \frac{1}{2}\left[ {\tan u - \sec u} \right]_0^{\pi /4} \cr
& = \frac{1}{2}\left[ {\tan \left( {\frac{\pi }{4}} \right) - \sec \left( {\frac{\pi }{4}} \right)} \right] - \frac{1}{2}\left[ {\tan \left( 0 \right) - \sec \left( 0 \right)} \right] \cr
& = \frac{1}{2}\left[ {1 - \sqrt 2 } \right] - \frac{1}{2}\left[ {0 - 1} \right] \cr
& = \frac{1}{2} - \frac{{\sqrt 2 }}{2} + \frac{1}{2} \cr
& = 1 - \frac{{\sqrt 2 }}{2} \cr} $$
