why doesn't wolfram alpha evaluate this integral?

Question

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

Answer 1

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

Answer 2

WolframAlpha isn't really that good at reading $\TeX$; you need to simplify the input a bit.

Answer 3

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 $$\frac{2 \pi ^{3/2}}{\Gamma \left(\frac{1}{4}\right)^2}\approx 0.847213$$