Finding $\int_0^{\frac{\pi}{2}}\arctan\left(\sin x\right)dx$ 
$$\int_0^{\frac{\pi}{2}}\arctan\left(\sin x\right)dx$$

I try to solve it, but failed. Who can help me to find it?
I encountered this integral when trying to solve $\displaystyle{\int_0^\pi\frac{x\cos(x)}{1+\sin^2(x)}\,dx}$.
 A: Using the integral definition of the arctangent function, we may write $$\arctan{\left(\sin{x}\right)}=\int_{0}^{1}\mathrm{d}y\,\frac{\sin{x}}{1+y^2\sin^2{x}},$$
thus, transforming the integral into a double integral. Changing the order of integration, we find:
$$\begin{align}
\mathcal{I}
&=\int_{0}^{\frac{\pi}{2}}\arctan{\left(\sin{x}\right)}\,\mathrm{d}x\\
&=\int_{0}^{\frac{\pi}{2}}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{\sin{x}}{1+y^2\sin^2{x}}\\
&=\int_{0}^{1}\mathrm{d}y\int_{0}^{\frac{\pi}{2}}\mathrm{d}x\,\frac{\sin{x}}{1+y^2\sin^2{x}}\\
&=\int_{0}^{1}\mathrm{d}y\int_{0}^{\frac{\pi}{2}}\mathrm{d}x\,\frac{\sin{x}}{1+y^2\left(1-\cos^2{x}\right)}\\
&=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\frac{\mathrm{d}t}{1+y^2-y^2t^2}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{\tanh^{-1}{\left(\frac{y}{\sqrt{1+y^2}}\right)}}{y\sqrt{1+y^2}}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{\sinh^{-1}{\left(y\right)}}{y\sqrt{1+y^2}}\\
&=\int_{0}^{\sinh^{-1}{(1)}}\mathrm{d}u\,\frac{u}{\sinh{(u)}}\\
&=-\sinh^{-1}{(1)}^2-\int_{0}^{\sinh^{-1}{(1)}}\mathrm{d}u\,\ln{\left(\tanh{\frac{u}{2}}\right)}\\
&=-\sinh^{-1}{(1)}^2-2\int_{0}^{\frac{1}{1+\sqrt{2}}}\mathrm{d}w\,\frac{\ln{\left(w\right)}}{1-w^2}.\\
\end{align}$$
At this point, the resolution of the last integral in terms of dilogarithms is a straightforward matter.
A: Hint: Letting $t=\sin x$, the integral becomes $F(1)$, where $F(a)=\displaystyle\int_0^1\frac{\arctan(at)}{\sqrt{1-t^2}}dx$. Evaluate $F'(a)$.
A: 
Actually I didn't post my answer as I was unable to continue it where I left but after Pranav provided link I'm posting it

$$\sin x=u\iff \frac{\,\mathrm du}{\sqrt{1-u^2}}=\,\mathrm dx$$
$$\int_0^{{\pi/2}}\arctan\left(\sin x\right)\,\mathrm dx=\int_0^{1}\frac{\arctan(u)}{\sqrt{1-u^2}}\,\mathrm du$$
Now consider parametric integral
$$I(\alpha)=\int_0^{1}\frac{\arctan(\alpha t)}{\sqrt{1-t^2}}\,\mathrm dt$$
Then
$$\begin{align}
I'(\alpha)&= \int_{0}^{1} \frac{t}{(1+\alpha^{2}t^{2})\sqrt{1-t^{2}}} \,\mathrm dt\\
&= \frac{1}{\alpha \sqrt{1+\alpha^{2}}} \text{artanh} \left[ \frac{\alpha}{\sqrt{1+\alpha^{2}}} \right]\\
&= \frac{1}{\alpha \sqrt{1+\alpha^{2}}} \text{arsinh}(\alpha) .\end{align}$$
Now for further process see this
A: 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
$$
\int_0^{\pi/2}\arctan(\sin(x))\,\mathrm{d}x
=\frac{\pi^2}8-\frac12\mathrm{arcsinh}^2(1)\tag{5}
$$
