Show that $ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, [\arcsin{\frac{x}{a}} / \arccos{\frac{x}{a}}]= \frac{\pi^2}{8} $? I found this very interesting result:

$$ \int_0^a  \frac{dx}{\sqrt{a^2-x^2}} \, \arcsin{\frac{x}{a}} = \int_0^a  \frac{dx}{\sqrt{a^2-x^2}} \, \arccos{\frac{x}{a}} = \frac{\pi^2}{8}, $$

for $a>0.$ I wonder how one could prove this?

We have that
 $$\int_0^a  \frac{dx}{\sqrt{a^2-x^2}} \, 
\arcsin{\frac{x}{a}}=\int_0^1  \frac{dt}{\sqrt{1-t^2}} \, \arcsin{t}\\
=\int_0^1  \arcsin(t)\ d(\arcsin{t})=\frac{1}{2}\left[\arcsin^2{t}\right]_0^1=\frac{\pi^2}{8}.$$
For the other one you can integrate in a similar way or use the identity $\arcsin(t)+\arccos(t)=\frac{\pi}{2}$ for $t>0$.
