integration - Integral $int_0^frac{pi}{2} arcsin(sqrt{sin x}) dx$

I am trying to calculate $$I=\int_0^\frac{\pi}{2} \arcsin(\sqrt{\sin x}) dx$$ So far I have done the following. First I tried to let $\sin x= t^2$ then:

$$I=2\int_0^1 \frac{x\arcsin x}{\sqrt{1-x^4}}dx =\int_0^1 (\arcsin^2 x)'\frac{x}{\sqrt{1+x^2}}dx $$
$$=\frac{\pi^2}{8}-\int_0^1 \frac{\arcsin^2 x}{(1+x^2)^{3/2}}dx$$ We can expand into power series the integral, we have: $\arcsin^2z=\sum\limits_{n\geq1}\frac {2^{2n-1}z^{2n}}{n^2\binom {2n}n}$ and using the binomial series for $(1+x^2)^{-3/2}$ will result in: $$\sum_{n\geq1}\frac{2^{2n-1}x^{2n}}{n^2\binom {2n}n}\sum_{k\ge 0}\binom{-3/2}{k}x^{2k}$$ But I dont know how to simplify this. I tried one more thing, letting $\sin x= \sin^2 t$ gives:
$$I=2\int_0^\frac{\pi}{2}\frac{x\sin x}{\sqrt{1+\sin^2 x}}dx$$ Since $\int \frac{\sin x}{\sqrt{1+\sin^2x}}dx=-\arcsin\left(\frac{\cos x}{\sqrt 2} \right)+C$ we can integrate by parts to obtain: $$I=2\int_0^\frac{\pi}{2}\arcsin\left(\frac{\cos x}{\sqrt 2}\right)dx=2\int_0^\frac{\pi}{2}\arcsin\left(\frac{\sin x}{\sqrt 2}\right)dx$$ But I am stuck, so I would appreciate some help.

Edit: By letting $\frac{\sin x}{\sqrt 2} =t $ We get: $$I=2\int_0^\frac1{\sqrt{2}} \frac{\arcsin x}{\sqrt{\frac12-x^2}}dx=2\text{Li}_2\left(\frac1{\sqrt 2}\right)-\frac{\pi^2}{24}+\frac{\ln^2 2}{4}$$ Where the latter integral was evaluated with wolfram. I would love to see a proof for that.

Answer

Write
$$
I(t)=\int_0^{\frac{1}{\sqrt{2}}} \frac{2\arcsin(tx)}{\sqrt{\frac{1}{2}-x^2}} \, {\rm d}x
$$

and calculate
\begin{align}
I'(t) &= \int_0^{\frac{1}{\sqrt{2}}} \frac{2x}{\sqrt{\left(\frac{1}{2}-x^2\right)\left(1-(tx)^2\right)}} \, {\rm d}x \\
&= \frac{\log\left(\sqrt{2}+t\right)-\log\left(\sqrt{2}-t\right)}{t} \\
&= \frac{{\rm Li}_1 \left(\frac{t}{\sqrt{2}}\right) - {\rm Li}_1 \left(-\frac{t}{\sqrt{2}}\right)}{t}\, .
\end{align}

Then
\begin{align}
I(1) &= \int_0^1 I'(t) \, {\rm d}t \\

&={\rm Li}_2 \left(\frac{1}{\sqrt{2}}\right) - {\rm Li}_2 \left(-\frac{1}{\sqrt{2}}\right) \, .
\end{align}