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.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
\int_{0}^{1}{\arcsin\pars{x} \over x}\,\dd x &
\,\,\,\stackrel{\mbox{i.b.p.}}{=}\,\,\,
-\int_{0}^{1}{\ln\pars{x} \over \root{1 - x^{2}}}\,\dd x
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
-\,{1 \over 4}\int_{0}^{1}{x^{-1/2}\ln\pars{x} \over \root{1 - x}}\,\dd x
\\[5mm] & =
\left.-\,{1 \over 4}\,\partiald{}{\mu}\int_{0}^{1}x^{\mu}\pars{1 - x}^{-1/2}\,\dd x\,\right\vert_{\ \mu\ =\ -1/2}
\\[5mm] & =
\left.-\,{1 \over 4}\,\partiald{}{\mu}\bracks{\Gamma\pars{\mu + 1}\Gamma\pars{1/2} \over \Gamma\pars{\mu + 3/2}}\,\right\vert_{\ \mu\ =\ -1/2}
\\[5mm] & =
-\,{1 \over 4}\,\root{\pi}\,
{\Gamma'\pars{1/2}\Gamma\pars{1} - \Gamma'\pars{1}\Gamma\pars{1/2} \over \Gamma^{2}\pars{1}}
\\[5mm] & =
-\,{1 \over 4}\,\root{\pi}\,\bracks{%
\Gamma\pars{1 \over 2}\Psi\pars{1 \over 2} - \Gamma\pars{1}\Psi\pars{1}\Gamma\pars{1 \over 2}}
\\[5mm] & =
-\,{1 \over 4}\,\pi\bracks{\Psi\pars{1 \over 2} + \gamma} =
-\,{1 \over 4}\,\pi\bracks{-2\ln\pars{2}} = 
\bbox[#ffe,10px,border:1px dotted navy]{\ds{{1 \over 2}\,\pi\ln\pars{2}}}
\end{align}
A: 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,
$$2I=I-\frac{\pi}{2}\ln 2$$
$$I=-\frac{\pi}{2}\ln 2$$
A: A real-analytic solution. Through the substitution $x=\sin\theta$ and integration by parts, our integral becomes
$$ I = \int_{0}^{\pi/2}\theta\cot(\theta)\,d\theta = -\int_{0}^{\pi/2}\log\sin(\theta)\,d\theta \tag{1}$$
and since
$$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right) = \frac{2n}{2^n}\tag{2} $$
is a well-known identity, by Riemann sums
$$ \int_{0}^{\pi/2}\log\sin(\theta)\,d\theta = \frac{\pi}{2}\lim_{n\to +\infty}\frac{1}{n}\log\left(\frac{2n}{2^n}\right) = -\frac{\pi}{2}\log(2).\tag{3}$$
A: \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
$$\displaystyle \int_{0}^{1} \frac{\arcsin x}{x} dx = -\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^{2}}} dx =-I^{\prime}(0)= \frac{\pi}{2} \ln 2$$
A: Transform our integral by letting $\theta=\arcsin x$, then
$$
\begin{aligned}
\int_{0}^{1} \frac{\arcsin x}{x} d x &=\int_{0}^{\frac{\pi}{2}} \frac{\theta \cos \theta d \theta}{\sin \theta} \\
&=\int_{0}^{\frac{\pi}{2}} \theta  d(\ln (\sin \theta)) \\
&=-\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d \theta
\end{aligned}
$$
By my post, $\int_{0}^{\frac{\pi}{2}} \ln (\sin \theta) d \theta=-\frac{\pi}{2}\ln 2$, we can now conclude that
$$\boxed{\int_{0}^{1} \frac{\arcsin x}{x} d x= \frac{\pi}{2}\ln 2}$$
A: $$\int_{0}^{1} \frac{\sin^{-1}x}{x} dx 
= \int_{0}^{1} \int_{0}^{1}  \frac{\sqrt{1-x^2}}{1-x^2+x^2y^2} dy\ dx \overset{x=\sin t} 
=\int_0^1 \frac{\pi}{2(1+y)}dy=\frac{\pi}{2}\ln2
$$
