TITLE: Finding $\int_0^{\frac{\pi}{2}}\arctan\left(\sin x\right)dx$
QUESTION [18 upvotes]: $$\int_0^{\frac{\pi}{2}}\arctan\left(\sin x\right)dx$$

I try to solve it, but failed. Who can help me to find it?
I encountered this integral when trying to solve $\displaystyle{\int_0^\pi\frac{x\cos(x)}{1+\sin^2(x)}\,dx}$.

REPLY [10 votes]: Integrating by parts, we get
$$
\int_0^{\pi/2}\sin^{2k+1}(x)\,\mathrm{d}x
=\frac{2k}{2k+1}\int_0^{\pi/2}\sin^{2k-1}(x)\,\mathrm{d}x\tag{1}
$$
Therefore, by induction, we have
$$
\int_0^{\pi/2}\sin^{2k+1}(x)\,\mathrm{d}x
=\frac{2^k\,k!}{(2k+1)!!}\tag{2}
$$
Thus,
$$
\begin{align}
\int_0^{\pi/2}\arctan(\sin(x))\,\mathrm{d}x
&=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\int_0^{\pi/2}\sin^{2k+1}(x)\,\mathrm{d}x\tag{3a}\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\frac{2^k\,k!}{(2k+1)!!}\tag{3b}\\
&=\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}\frac{4^k}{\binom{2k}{k}}\tag{3c}
\end{align}
$$
Explanation:
$\text{(3a)}$: use the series for $\arctan(x)$
$\text{(3b)}$: use $(2)$
$\text{(3c)}$: rewrite $\text{(3b)}$ using central binomial coefficients
In this answer, it is shown that
$$
\sum_{n=0}^\infty\frac{(-4)^n}{(2n+1)^2\binom{2n}{n}}
=\frac{\pi^2}8-\frac12\mathrm{arcsinh}^2(1)\tag{4}
$$
Therefore, combining $(3)$ and $(4)$, we have
$$
\int_0^{\pi/2}\arctan(\sin(x))\,\mathrm{d}x
=\frac{\pi^2}8-\frac12\mathrm{arcsinh}^2(1)\tag{5}
$$