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.

\begin{array}{r}
\displaystyle \int_{0}^{1} \frac{\arcsin x}{x} d x=& \int_{0}^{1} \arcsin x d(\ln x) 
\stackrel{IBP}{=} \displaystyle  -\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} d x
\end{array}
Let $\displaystyle I(a)=\int_{0}^{1} \frac{x^{a}}{\sqrt{1-x^{2}}} d x$ and $x=\sin \theta$, then our integral becomes $$
\begin{aligned}
I(a) &=\int_{0}^{\frac{\pi}{2}} \sin ^{2\left(\frac{a+1}{2}\right)-1} \theta \cos ^{2\left(\frac{1}{2}\right)-1} \theta d \theta \\
&=\frac{1}{2} B\left(\frac{a+1}{2}, \frac{1}{2}\right) \\
&=\frac{\sqrt{\pi}}{2} \frac{\Gamma\left(\frac{a+1}{2}\right)}{\Gamma\left(\frac{a}{2}+1\right)}
\end{aligned}
$$
Using logarithmic differentiation yields
$$
\frac{I^{\prime}(a)}{I(a)}=\frac{1}{2}\left[\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a}{2}+1\right)\right]
$$
Putting $a=0$ gives
$$
\begin{aligned}
I^{\prime}(0) &=\frac{I(0)}{2} \cdot\left(\psi\left(\frac{1}{2}\right)-\psi(1)\right)=\frac{\pi}{4}(-\ln 4)=-\frac{\pi}{2} \ln 2
\end{aligned}
$$
Now we can conclude that
$$\displaystyle \int_{0}^{1} \frac{\arcsin x}{x} dx = -\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} dx =-I^{\prime}(0)= \frac{\pi}{2} \ln 2$$
