Compute $\int_0^1 \frac{\arcsin(x)}{x}dx$ $$\int_0^1 \frac{\arcsin(x)}{x}dx$$
This is a proposed for a Calculus II exam, and I have absolutely no idea how to solve it. Tried using Frullani or Lobachevsky integrals, or beta and gamma functions, but I can't even find a way to start it. Wolfram Alpha gives a kilometric solution, but I know that cannot be the only answer. Any help appreciated! 
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$\ds{\int_{0}^{1}{\arcsin\pars{x} \over x}\,\dd x:\ {\large ?}}$

\begin{align}&\color{#66f}{\large\int_{0}^{1}{\arcsin\pars{x} \over x}\,\dd x}
=-\int_{0}^{1}{\ln\pars{x} \over \root{1 - x^{2}}}\,\dd x
=-{1 \over 4}\int_{0}^{1}x^{-1/2}\ln\pars{x}\pars{1 - x}^{-1/2}\,\dd x
\\[3mm]&=-{1 \over 4}\lim_{\mu \to -1/2}\partiald{}{\mu}
\int_{0}^{1}x^{\mu}\pars{1 - x}^{-1/2}\,\dd x
=-{1 \over 4}\lim_{\mu \to -1/2}\partiald{}{\mu}
\bracks{\Gamma\pars{\mu + 1}\Gamma\pars{1/2} \over \Gamma\pars{\mu + 3/2}}
\\[3mm]&=-{1 \over 4}\,\Gamma\pars{\half}\braces{%
{\Gamma\pars{1/2} \over \Gamma\pars{1}}\,\bracks{%
\overbrace{\Psi\pars{\half}}^{\ds{-\gamma - 2\ln\pars{2}}}\ -\
\overbrace{\Psi\pars{1}}^{\ds{-\gamma}}}}\ =\
\half\,\Gamma^{2}\pars{\half}\ln\pars{2}
\\[3mm]&=\color{#66f}{\large\half\,\pi\ln\pars{2}}
\quad\mbox{with}\quad\Gamma\pars{\half} = \root{\pi}\quad\mbox{and}\quad\Gamma\pars{1} = 1.
\end{align}

A: Let $y=\arcsin x\;\Rightarrow\;\sin y =x\;\Rightarrow\;\cos y\ dy=dx$, then
$$
\int_0^1 \frac{\arcsin(x)}{x}dx=\int_0^{\Large\frac\pi2}y\cot y\ dy.
$$
Now use IBP by taking $u=y$ and $dv=\cot y\ dy\;\Rightarrow\;v=\ln(\sin x)$, then
\begin{align}
\int_0^{\Large\frac\pi2}y\ \cot y\ dy&=\left.y\ln(\sin y)\right|_0^{\Large\frac\pi2}-\int_0^{\Large\frac\pi2}\ln(\sin y)\ dy\\
&=-\int_0^{\Large\frac\pi2}\ln(\sin y)\ dy.
\end{align}
The last integral can be evaluated by using property
$$
\int_a^b f(x)\ dx=\int_a^b f(a+b-x)\ dx.
$$
We obtain
$$
\int_0^{\Large\frac\pi2}\ln(\sin y)\ dy=-\frac\pi2\ln2,
$$
where
$$
\int_0^{\Large\frac\pi2}\ln(\sin y)\ dy=\int_0^{\Large\frac\pi2}\ln(\cos y)\ dy\quad\Rightarrow\quad\text{by symmetry}.
$$
Thus
$$
\int_0^1 \frac{\arcsin(x)}{x}dx=\large\color{blue}{\frac\pi2\ln2}.
$$
