<s>[INST] 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.
 [/INST] Mathematica gives:
$$\frac{1}{24} \left(-6 \text{Li}_2\left(\frac{1}{2}-\frac{1}{\sqrt{2}}\right)+6
   \text{Li}_2\left(3-2 \sqrt{2}\right)+4 \pi ^2-3 \log ^2(2)+3 \log
   ^2\left(\sqrt{2}-1\right)+3 \log ^2\left(3+2 \sqrt{2}\right)+ \log (64) \log
   \left(\sqrt{2}-1\right)+6 \sinh ^{-1}(1)^2-12 \log \left(2
   \left(1+\sqrt{2}\right)\right) \sinh ^{-1}(1)-6 i \pi  \left(2 \sinh ^{-1}(1)-\log
   \left(3+2 \sqrt{2}\right)\right)\right)$$
which strongly suggests that hand calculation will be extremely difficult and error prone.  Incidentally, the numerical value of that expression is  $1.5122$, and the value of the numerical evaluation of the original is also $1.5122$, which strongly suggests the answer is correct.
Here is a graph of the function, and shaded value of the integral, which suggests there isn't a major error:

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