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! 

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
$$
\int_0^1 \frac{\arcsin(x)}{x}dx=\large\color{blue}{\frac\pi2\ln2}.
$$
