<s>[INST] 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?
 [/INST] $$ \arcsin{\frac{x}{a}}=t $$
$$ \frac{dx}{\sqrt{a^2 - x^2}}=dt$$
$$ \int_0^a  \frac{dx}{\sqrt{a^2-x^2}} \, \arcsin{\frac{x}{a}}=\frac{(\arcsin{\frac{x}{a}})^2}{2}$$
</s>