<s>[INST] 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
 [/INST] We want to calculate

$$
I=\int_0^1 dx\frac{\arcsin(x)\arcsin\left(\frac{x}{\sqrt{2}}\right)}{\sqrt{2-x^2}}
$$

Let us start by a substitution $x=\sqrt{2}\sin(y)$ and integrate by parts so we can write $I$ as
$$
\int_0^{\pi/4}dy y \arcsin(\sqrt{2}\sin(y))\underbrace{=
}_{i.b.p}\frac{\pi^3}{64}-\frac{1}{2}\underbrace{\int_0^{\pi/4}dy\frac{y^2\cos(y)}{ \sqrt{1-2\sin(y)^2}}}_{J}$$
but since $1-2\sin(y)^2=\cos(2y)$ and by another substitution $2y =q$ 
$$
J=\frac{1}{8}\int_0^{\pi/2}dq\frac{ q^2 \cos(q/2) }{\sqrt{\cos(q)}}
$$
Now we perform small trick (credit to Mr. Feynman) by rewriting 
$$
J=-\frac{1}{8}\left[\frac{d^2}{da^2}\int_0^{\pi/2} dq\cos(q)^{-1/2}\cos(aq)\right]_{a=1/2}
$$
which equals according to Tunk-Fey's answer here 
$$
J=-\frac{1}{8}\frac{d^2}{da^2}\left[\frac{2\pi}{\sqrt{2}}\frac{1}{B\left(\frac{3/2+a}{2},\frac{3/2-a}{2}\right)}\right]_{a=1/2}
$$
Or 

$$
I=\frac{\pi^3}{64}+\frac{\pi}{8\sqrt{2}}\frac{d^2}{da^2}\left[\frac{1}{B\left(\frac{3/2+a}{2},\frac{3/2-a}{2}\right)}\right]_{a=1/2}
$$

Since i have no CAS at hand today and are way to lazy to do the above derivative by hand that's it for the moment. But everything from here on should be straightforward so i leave that to you... 
</s>