why doesn't wolfram alpha evaluate this integral? i am trying to compute the integral given below using Wolfram alpha
$$\int_0^{\sqrt2}\sin^{-1}\left(\frac{\sqrt{2-x^2}}{2}\right)dx$$
However it can be solved by numerical method but i don't know why Wolfram alpha doesn't compute it. could somebody explain or help me solve this integral? thank u very much 

The most direct way of calculating this integral starts as already explained by @Claude Leibovici: We integrate by parts and observe that the boundary term vanishs, so we are down to 
$$
I=\int_0^{\sqrt{2}}dx\frac{x^2}{2\sqrt{1-(x^4/4)}}
$$
now setting $r=x^4/4$ we obtain
$$
I=\frac{1}{2\sqrt{2}}\int_0^{1}dr\frac{1}{r^{1/4}(1-r)^{1/2}}
$$
this integrals equals  an representation of the Eulerian Beta function which, may in turn be expressed in terms of Gamma functions
$$
I=\frac{1}{2\sqrt{2}}\frac{\Gamma(1/2)\Gamma(3/4)}{\Gamma(5/4)}
$$
using Gamma duplication as well as $\Gamma(1/2)=\sqrt{\pi}$ this boils down to 

$$
I=\frac{1}{2\sqrt{2}}\frac{4 \sqrt{2}\pi^{3/2}}{\Gamma(1/4)^2}=\frac{2\pi^{3/2}}{\Gamma(1/4)^2}
$$

as expected
