TITLE: Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$
QUESTION [10 upvotes]: 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.

REPLY [3 votes]: Integration by parts reduces the integral to,
$$\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^2}} dx$$
And the substitution $x=\sin u$ reduces the integral to,
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\sin u) du$$
And the substitution $v=\frac{\pi}{2}-x$ reduces the integral to,
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos v) dv$$
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos u) du$$
Now adding the integrals and noting properties of logarithms we have,
$$2I=\int_{0}^{\frac{\pi}{2}} \left( \ln (2 \sin x \cos x)-\ln 2\right) dx$$
Double angle,
$$2I=\int_{0}^{\frac{\pi}{2}} \ln (\sin 2x) dx -\frac{\pi}{2} \ln 2$$
The substitution $s=2x$ gives
$$2I=\frac{1}{2}\int_{0}^{\pi} \ln (\sin s) ds -\frac{\pi}{2} \ln 2$$
But 
$$\int_{0}^{\pi} \ln (\sin s) ds=2I$$
Follows from the substitution  $w=\frac{\pi}{2}-s$  and the evenness of the function $f(w)=\ln (\cos  w)$:
$$\int_{0}^{\pi} \ln (\sin s) ds$$
$$=-\int_{\frac{\pi}{2}}^{-\frac{\pi}{2}} \ln (\cos w) dw$$
$$=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln (\cos w)dw $$
$$=2 \int_{0}^{\frac{\pi}{2}} \ln (\cos w) dw=2I$$
So,
$$2I=I-\frac{\pi}{2}\ln 2$$
$$I=-\frac{\pi}{2}\ln 2$$