I want to use the Maclaurin series of $\arctan (x)$ to show that
\begin{align}\int\_{0}^{\pi/ 2} \arctan (a \sin x) \, \mathrm dx &= 2 \sum\_{k=0}^{\infty} \frac{\left(\frac{\sqrt{1 + a^{2}}- 1}{a}\right)^{2k+1}}{(2k+1)^{2}} \\ &= \operatorname{Li}\_{2} \left(\frac{\sqrt{1+a^{2}}-1}{a} \right) - \operatorname{Li}\_{2} \left(\frac{1-\sqrt{1+a^{2}}}{a} \right).
\end{align}

I guess we should first impose the restriction $ |a| \le 1$.

Then we have

$$ \begin{align} \int\_{0}^{\pi/ 2} \arctan (a \sin x) \ dx &= \int\_{0}^{\pi /2} \sum\_{k=0}^{\infty} (-1)^{k} \, \frac{(a \sin x)^{2k+1}}{2k+1} \, \mathrm dx \\ &= \sum\_{k=0}^{\infty} (-1)^{k} \, \frac{a^{2k+1}}{2k+1} \int\_{0}^{\pi /2} \sin^{2k+1} (x) \, \mathrm dx \\ &= \sum\_{k=0}^{\infty} (-1)^{k} \frac{a^{2k+1}}{(2k+1)^{2}} \frac{2^{2k}}{\binom{2k}{k}}. \end{align}$$

But how do we proceed from here?