Integration involving $\arcsin$ How to integrate the following function:
$$\int_0^a \arcsin\sqrt{a \over a+x}dx$$
By using the substitution $x = a\tan^2\theta$, I managed to write the integral as:
$$2a\int_0^{\pi \over 4}\theta \frac{\sin\theta}{\cos^3\theta}d\theta$$
How would I proceed? Should I use by parts method?

HINT(OTHER APPROACH)
Let  $u=a+x$, then we have
\begin{align}
\int{arcsin\sqrt{\frac{x-u}{u}}}du
\end{align}
Using Integration By Parts\, let $f=arcsin\sqrt{\frac{x-u}{u}}du$ and $dg=du$
We obtain
\begin{align}
=x arcsin(\sqrt{\frac{a}{a+x}})+\frac{{\sqrt{\frac{ax}{(a+x)^2}}(a+x)(\sqrt x-\sqrt a\arctan(\sqrt{\frac{\sqrt x}{\sqrt a}})}}{\sqrt x}+C
\end{align}
