I’m trying to find a closed form for this integral.Any help is appreciated.Thanks $$I=\int_0^{1}\frac{{ \arcsin}({x^2})}{\sqrt{1-x^2}}dx$$
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http://www.wolframalpha.com/input/?i=integral+(arcsin(x%5E2)%2Fsqrt(1-x%5E2)) – MichaelChirico Feb 04 '16 at 21:09
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When formatted correctly, WA finds no antiderivative, and gives no simple expression for the definite integral either. I'd suspect there is none, then, though I've been proven wrong once or twice on MSE. – Ian Feb 04 '16 at 21:22
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2Related: Improper Integral $\int_0^1\frac{\arcsin^2(x^2)}{\sqrt{1-x^2}}dx$: with the same subs we could write it as $\displaystyle \int_0^{\pi/2} \left(\frac{\pi}{2} - 2\arcsin \left(\frac{\sin \alpha}{\sqrt{2}}\right)\right),d\alpha = \frac{\pi^2}{4} - 2\chi_2\left(\frac{1}{\sqrt{2}}\right)$. – r9m Feb 08 '16 at 04:51
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@r9m, Well, tanks(+1) – user178256 Feb 08 '16 at 09:25
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As already said in comments, it does not seem that the antiderivative of the integrand exists.
Concerning the integral, a CAS found something you will not like very much $$I=\frac{\pi }{4} \, _4F_3\left(\frac{1}{2},\frac{1}{2},\frac{3}{4},\frac{5}{4};1,\frac{3}{2},\frac{3}{ 2};1\right)$$ where appears the generalized hypergeometric function.
Numerically, $$I\approx 0.9552018064811796875605004$$
Claude Leibovici
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To make it more clear why this integral is so problematic, we can represent it as a double integral, using the definition of $\arcsin$:
$$\arcsin z=z \int_0^1 \frac{dy}{\sqrt{1-z^2y^2}}$$
Then the integral in the OP will have the form:
$$I=\int_0^{1}\frac{{ \arcsin}({x^2})}{\sqrt{1-x^2}}dx=\int_0^1 \int_0^1 \frac{x^2 dx dy}{\sqrt{1-x^2}\sqrt{1-y^2x^4}}=$$
$$=\int_0^1 \int_0^1 \frac{x^2 dx dy}{\sqrt{1-x^2}\sqrt{1-yx^2}\sqrt{1+yx^2}}=$$
$$=\int_0^1 \int_0^1 \frac{dx dy}{\sqrt{1-x^2}\sqrt{1-y^2x^4}}-\int_0^1 \int_0^1 \frac{\sqrt{1-x^2}dx dy}{\sqrt{1-y^2x^4}}$$
If we had $\arcsin x$ instead of $\arcsin x^2$, then we would at least be able to use complete elliptic integrals of the first and second kind.
But with $\arcsin x^2$ we don't get any nice special functions, thus, apparently, the generalized Hypergeometric function from Claude Leibovici's answer is the only way to go.
Yuriy S
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