Answer
$$\frac{{{\pi ^2}}}{{32}}$$
Work Step by Step
$$\eqalign{
& \int_0^{1/\sqrt 2 } {\frac{{\arcsin x}}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{Let }}u = \arcsin x,{\text{ }}du = \frac{1}{{\sqrt {1 - {x^2}} }}dx,{\text{ }}dx = \sqrt {1 - {x^2}} du \cr
& {\text{The new limits of integration are}} \cr
& x = \frac{1}{{\sqrt 2 }},{\text{ }}u = \arcsin \left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{1}{4}\pi \cr
& x = 0,{\text{ }}u = \arcsin \left( 0 \right) = 0 \cr
& {\text{Substituting}} \cr
& \int_0^{1/\sqrt 2 } {\frac{{\arcsin x}}{{\sqrt {1 - {x^2}} }}} dx = \int_0^{\frac{\pi }{4}} {\frac{u}{{\sqrt {1 - {x^2}} }}} \left( {\sqrt {1 - {x^2}} } \right)du \cr
& = \int_0^{\frac{\pi }{4}} u du \cr
& {\text{Integrating}} \cr
& {\text{ = }}\left[ {\frac{{{u^2}}}{2}} \right]_0^{\pi /4} \cr
& {\text{ = }}\frac{1}{2}\left[ {{{\left( {\frac{\pi }{4}} \right)}^2} - {{\left( 0 \right)}^2}} \right] \cr
& = \frac{1}{2}\left( {\frac{{{\pi ^2}}}{{16}}} \right) \cr
& = \frac{{{\pi ^2}}}{{32}} \cr} $$
