<s>[INST] Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$ I stumbled upon the interesting definite integral
\begin{equation}
\int\limits_0^1 \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2
\end{equation}
Here is my proof of this result. 
Let $u=\sin^{-1}(x)$ then integrate by parts,
\begin{align}
\int \frac{\sin^{-1}(x)}{x} dx &= \int u \cot(u) du \\
&= u \ln\sin(u) - \int \ln\sin(u) du
\tag{1}
\label{eq:20161030-1}
\end{align}
\begin{align}
\int \ln\sin(u) du &= \int \ln\left(\frac{\mathrm{e}^{iu} - \mathrm{e}^{-iu}}{i2} \right) du \\
&= \int \ln\left(\mathrm{e}^{iu} - \mathrm{e}^{-iu} \right) du \,- \int \ln(i2) du \\
&= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \int \ln\mathrm{e}^{iu} du \,-\, u\ln(i2) \\
&= \int \ln\left(1 - \mathrm{e}^{-i2u} \right) du + \frac{i}{2}u^{2} -u\ln2 \,-\, ui\frac{\pi}{2}
\tag{2}
\label{eq:20161030-2}
\end{align}
To evaluate the integral above, let $y=\mathrm{e}^{-i2u}$
\begin{equation}
\int \ln\left(1 - \mathrm{e}^{-i2u} \right) du = \frac{i}{2} \int \frac{\ln(1-y)}{y} dy
= -\frac{i}{2} \operatorname{Li}_{2}(y) = -\frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2u}
\tag{3}
\label{eq:20161030-3}
\end{equation}
Now we substitute equation \eqref{eq:20161030-3} into equation \eqref{eq:20161030-2}, then substitute that result into equation \eqref{eq:20161030-1}, switch variables back to (x), and apply limits,
\begin{align}
\int\limits_{0}^{1} \frac{\sin^{-1}(x)}{x} dx
&= \sin^{-1}(x)\ln(x) + \sin^{-1}(x)\left(\ln2 + i\frac{\pi}{2}\right) \\
&- \frac{i}{2}[\sin^{-1}(x)]^{2} + \frac{i}{2} \operatorname{Li}_{2}\mathrm{e}^{-i2\sin^{-1}(x)} \Big|_0^1 \\
&= \frac{\pi}{2}\ln2
\end{align}
I would be interested in seeing other solutions.
 [/INST] $$\int_{0}^{1} \frac{\sin^{-1}x}{x} dx 
= \int_{0}^{1} \int_{0}^{1}  \frac{\sqrt{1-x^2}}{1-x^2+x^2y^2} dy\ dx \overset{x=\sin t} 
=\int_0^1 \frac{\pi}{2(1+y)}dy=\frac{\pi}{2}\ln2
$$
</s>