Integrate $I=\int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx$ How to prove 
\begin{align}
 I &= \int_0^1\frac{\arcsin{(x)}\arcsin{(x\sqrt\frac{1}{2})}}{\sqrt{2-x^2}}dx \\
   &= \frac{\pi}{256}\left[ \frac{11\pi^4}{120}+2{\pi^2}\ln^2{2}-2\ln^4{2}-12\zeta{(3)}\ln{2} \right]
\end{align}
 By asking  $$x=\sqrt{2}y$$ then using integration by parts, we have
 $$I=\frac{\pi^5}{2048}-\frac{1}{4}\int_0^1{\arcsin^4\left( \frac{z}{\sqrt{2}}\right) }\frac{dz}{\sqrt{1-x^2}}$$
But how to calculate this integral? I would appreciate your help

A slight detour from Ron Gordon's answer, using the square of the $\arcsin$ series:
$$\begin{align*}
I &= \int_0^1 \frac{\arcsin(x) \arcsin\left(\frac x{\sqrt2}\right)}{\sqrt{2-x^2}} \, dx \\[1ex]
&= \frac{\pi^3}{64} - \frac12 \int_0^1 \arcsin^2\left(\frac x{\sqrt2}\right) \, \frac{dx}{\sqrt{1-x^2}} \tag{1} \\[1ex]
&= \frac{\pi^3}{64} - \frac14 \sum_{n=1}^\infty \frac{2^n}{n^{2} \binom{2n}n} \int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}} \, dx \tag{2} \\[1ex]
&= \frac{\pi^3}{64} - \frac18 \sum_{n=1}^\infty \frac{2^n}{n^2 \binom{2n}n} \int_0^1 x^{n-\frac12} (1-x)^{-\frac12} \, dx \tag{3} \\[1ex]
&= \frac{\pi^3}{64} - \frac18 \sum_{n=1}^\infty \frac{2^n}{n^2 \binom{2n}n} \frac{\Gamma\left(n+\frac12\right)\Gamma\left(\frac12\right)}{\Gamma(n+1)} \tag{4} \\[1ex]
&= \frac{\pi^3}{64} - \frac\pi8 \sum_{n=1}^\infty \frac1{2^nn^2} \tag{5} \\[1ex]
&= \frac{\pi^3}{64} - \frac\pi8 \operatorname{Li}_2\left(\frac12\right) \tag{6} \\[1ex]
&= \frac{\pi^3}{64} - \frac\pi8 \left(\frac{\pi^2}{12} - \frac{\log^2(2)}2\right) \\[1ex]
&= \boxed{\frac{\pi^3}{192} + \frac{\pi\log^2(2)}{16}}
\end{align*}$$


*

*$(1)$ : integrate by parts


*$(2)$ : power series of $\arcsin^2(x)$


*$(3)$ : substitute $x\mapsto\sqrt x$


*$(4)$ : beta function


*$(5)$ : simplify gammas and central binomial coefficient; in particular, we have the identity $\dfrac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)\Gamma(n+1)} = \dfrac1{4^n} \dbinom{2n}n$


*$(6)$ : dilogarithm

I suspect one could adapt Jack D'Aurizio's general solution for a related integral... an exercise for the reader.
