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 

I do not know why  Wolfram alpha does not compute this integral.
Let use consider the antiderivative $$I=\int\sin^{-1}\left(\frac{\sqrt{2-x^2}}{2}\right)dx$$ Integrating by parts lead to $$I=x \sin ^{-1}\left(\frac{\sqrt{2-x^2}}{2}\right)+\int\frac{x^2}{\sqrt{4-x^4}}dx$$ The second term involves elliptic integrals $$\int\frac{x^2}{\sqrt{4-x^4}}dx=\sqrt{2} \left(E\left(\left.\sin
   ^{-1}\left(\frac{x}{\sqrt{2}}\right)\right|-1\right)-F\left(\left.\sin
   ^{-1}\left(\frac{x}{\sqrt{2}}\right)\right|-1\right)\right)$$ Concerning the definite integral, the first term cancels because of the given bounds  and $$J=\int_0^t\frac{x^2}{\sqrt{4-x^4}}dx=\sqrt{2} \left(E\left(\left.\sin
   ^{-1}\left(\frac{t}{\sqrt{2}}\right)\right|-1\right)-F\left(\left.\sin
   ^{-1}\left(\frac{t}{\sqrt{2}}\right)\right|-1\right)\right)$$ which tends to $$\sqrt{2} \left(E(-1)-K(-1)\right)$$ when $t \to \sqrt{2}$. In fact this can simplify to $$\frac{2 \pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}\approx 0.847213$$
