I found the following integral as a by product of another one.
It has a nice closed form.
$$ \int_{0}^{\pi} \arctan\left(\ln\left(\sin x \right) \over x\right)\,{\rm d}x $$
Mathematica and Maple fail to give the answer. Could you find it?
Hint 1:
The closed form is
$$ -\pi\arctan \left(2\ln 2 \over \pi\right) $$
Hint 2:
The following integral may help
$$ \int_{0}^{\pi}{x \over x^{2} + \ln^{2}\left(\alpha\sin x \right)} \,{\rm d}x $$
Consider the integral of more general form $$I(\alpha)=\int_0^\pi \arctan\left(\frac{\ln(\alpha \sin x)}{x}\right)\,dx,\qquad 0<\alpha\leq 1.$$ Then for $\alpha\in(0,1)$ $$I'(\alpha)=\frac{1}{\alpha}\int_0^\pi \frac{x}{x^2+\ln^2(\alpha\sin x)}\,dx,$$ as in the hint. To calculate the last integral we use the following identity, mentioned by Jack D'Aurizio in his comment to this question: $$\frac{b}{a^2+b^2}=\int_0^{+\infty} e^{-ay}\sin by \,dy,\quad a>0.\quad (*)$$ Setting $a=-\ln(\alpha\sin x)>0$ and $b=x$, we get $$I'(\alpha)=\frac{1}{\alpha}\int_0^\pi \left(\int_0^{+\infty} e^{y\ln(\alpha\sin x)}\sin xy \,dy\right)\,dx=\frac{1}{\alpha}\int_0^{+\infty}\left(\int_0^\pi (\alpha\sin x)^y\sin xy \,dx\right)\,dy.$$ (Changing order of integration is legitimate, since $|e^{y\ln(\alpha\sin x)}\sin xy|\leq e^{y\ln \alpha}$, so integral $\int_0^{+\infty} e^{y\ln(\alpha\sin x)}\sin xy \,dy$ converges uniformly by $x\in[0,\pi]$.) Therefore $$I'(\alpha)=\frac{1}{\alpha}\int_0^{+\infty}\left(\frac{\alpha}{2}\right)^y\left(\int_0^\pi (2\sin x)^y\sin xy \,dx\right)\,dy.$$ Now we need to deal with $$J=\int_0^\pi (2\sin x)^y\sin xy \,dx.$$ Changing the variable $x=t+\frac{\pi}{2}$ yields $$J=\int_{-\pi/2}^{\pi/2}(2\cos x)^y\left(\sin ty\cos\frac{\pi y}{2}+\cos ty\sin\frac{\pi y}{2}\right)\,dt=\sin\frac{\pi y}{2}\int_{-\pi/2}^{\pi/2}(2\cos t)^y\cos ty\,dt.$$ For the last integral we observe that $$\int_{-\pi/2}^{\pi/2}(2\cos t)^y\cos ty\,dt=\int_{-\pi/2}^{\pi/2}e^{y(\ln(2\cos t)-it)}\,dy=\int_{-\pi/2}^{\pi/2}(1+e^{-2it})^y\,dt=$$ $$=-\frac{1}{2i}\int_{-\pi/2}^{\pi/2}(1+e^{-2it})^y\frac{de^{-2it}}{e^{-2it}}=-\frac{1}{2i}\int_{C^-}\frac{(1+z)^y}{z}\,dz,$$ where $C^-$ is the unit circle (clockwise). Since $f(z)=\frac{(1+z)^y}{z}$ has just one simple pole $z=0$ inside $C^-$ with residue $\mathop{\mathrm{Res}}\limits_{z=0}f(z)=\lim\limits_{z\to 0}(1+z)^y=1$, we get $$-\frac{1}{2i}\int_{C^-}\frac{(1+z)^y}{z}\,dz=-\frac{1}{2i}(-2\pi i\cdot 1)=\pi$$ and $$J=\pi\sin\frac{\pi y}{2}.$$ Using $(*)$ one more time, we get $$I'(\alpha)=\frac{\pi}{\alpha}\int_0^{+\infty}\left(\frac{\alpha}{2}\right)^y\sin\frac{\pi y}{2}\,dy=\frac{\pi^2}{2\alpha(\ln^2\frac{\alpha}{2}+\frac{\pi^2}{4})}.$$ Now we can restore $I(\alpha)$ from its derivative: $$I(\alpha)=\frac{\pi^2}{2}\int\frac{d\alpha}{\alpha(\ln^2\frac{\alpha}{2}+\frac{\pi^2}{4})}=\frac{\pi^2}{2}\frac{2}{\pi}\arctan\left(\frac{2}{\pi}\ln\frac{\alpha}{2}\right)+c=\pi\arctan\left(\frac{2}{\pi}\ln\frac{\alpha}{2}\right)+c.$$
The next step is to show that $c=0$. For that purpose we observe that $$\ln(\alpha\sin x)\leq\ln\alpha\quad\Rightarrow\quad -\frac{\pi}{2}<\arctan\left(\frac{\ln(\alpha \sin x)}{x}\right)\leq \arctan\left(\frac{\ln\alpha}{x}\right)\quad\Rightarrow$$ $$-\frac{\pi^2}{2}\leq I(\alpha)\leq \int_0^\pi \arctan\left(\frac{\ln\alpha}{x}\right)\,dx\to -\frac{\pi^2}{2}$$ as $\alpha\to 0+$, so $\lim\limits_{\alpha\to 0+}I(\alpha)=-\frac{\pi^2}{2}$. Also it has to equal to $$\lim\limits_{\alpha\to 0+}\left(\pi\arctan\left(\frac{2}{\pi}\ln\frac{\alpha}{2}\right)+c\right)=-\frac{\pi^2}{2}+c,\quad\Rightarrow \quad c=0.$$ By now we have established that $$I(\alpha)=\int_0^\pi \arctan\left(\frac{\ln(\alpha \sin x)}{x}\right)\,dx=\pi\arctan\left(\frac{2}{\pi}\ln\frac{\alpha}{2}\right),\qquad 0<\alpha<1.$$ Letting $\alpha\to1-0$ we get the desired value $$I(1)=-\pi\arctan\left(\frac{2}{\pi}\ln2\right).$$ (we can change here limit and integral, since our integral is just proper).
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\pi}\arctan\pars{\ln\pars{\sin\pars{x}} \over x}\,\dd x =-\pi\,\arctan\pars{2\ln\pars{2} \over \pi}}$
With $\ds{{\large\tt 0 < \mu < 1}}$: \begin{align} \mbox{Lets define}\quad{\cal F}\pars{\mu}&\equiv \int_{0}^{\pi}\arctan\pars{\ln\pars{\mu\sin\pars{x}} \over x}\,\dd x\quad \mbox{such that} \\[3mm]{\cal F}'\pars{\mu}&=\int_{0}^{\pi} {1 \over \bracks{\ln\pars{\mu\sin\pars{x}}/x}^{2} + 1} \,{1 \over x}\,{1 \over \mu\sin\pars{x}}\,\sin\pars{x}\,\dd x \\[3mm]&={1 \over \mu}\int_{0}^{\pi}{x \over \ln^{2}\pars{\mu\sin\pars{x}} + x^{2}} \,\dd x =-\,{1 \over \mu}\,\Im\int_{0}^{\pi}{\dd x \over \ln\pars{\mu\sin\pars{x}} + x\ic} \end{align}
$$ \mbox{We are interested in}\quad{\cal F}\pars{1^{-}}:\ {\large ?}. \quad\mbox{Note that}\quad{\cal F}\pars{0^{+}} = -\,{\pi^{2} \over 2}\tag{1} $$
\begin{align} {\cal F}'\pars{\mu}&= -\,{1 \over \mu}\,\Im \int_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi}} {1 \over \ln\pars{\mu\bracks{z^{2} - 1}/\bracks{2\ic z}} + \ln\pars{z}}\,{\dd z \over \ic z} \\[3mm]&={1 \over \mu}\,\Re \int_{\verts{z}\ =\ 1\atop{\vphantom{\Huge A}0\ <\ {\rm Arg}\pars{z}\ <\ \pi}} {1 \over \ln\pars{\mu\bracks{1 - z^{2}}\ic/2}}\,{\dd z \over z} \end{align}
We 'close' the contourn with the line segment $\ds{\braces{\pars{x,0}\ \mid\ x \in \pars{-1,1}}}$. The segment is indented, with arcs of radius $\ds{\epsilon}$ such that $\ds{0 < \epsilon < 1}$, around $\ds{z = -1}$, $\ds{z = 0}$ and $\ds{z = 1}$. It turns out that the contributions from the 'indented points' at $\ds{z = \pm 1}$ vanishes out in the limit $\ds{\epsilon \to 0^{+}}$. We are left with a principal value along $\ds{\pars{-1,1}}$ and the contribution from the 'indented point' at $\ds{z = 0}$. The above mentioned principal value vanishes out $\ds{\pars{~\mbox{its integrand is odd in}\ \pars{-1,1}~}}$ such that the whole contribution to $\ds{{\cal F}'\pars{\mu}}$, in the limit $\ds{\epsilon \to 0^{+}}$, arises 'curiously and amusing' just from the 'indented point' at $\ds{z = 0}$.
It's shown as follows: \begin{align} {\cal F}'\pars{\mu}&=\left.-\,{1 \over \mu}\,\Re\int_{\pi/2}^{0} {1 \over \ln\pars{\mu\bracks{1 - z^{2}}\ic/2}}\,{\dd z \over z} \right\vert_{\,z\ =\ -1\ +\ \epsilon\expo{\ic\theta}} \\[3mm]&\phantom{=}-\,{1 \over \mu}\,\Re\int_{-1 + \epsilon}^{\epsilon} {1 \over \ln\pars{\mu\bracks{1 - x^{2}}\ic/2}}\,{\dd x \over x} \left.-\,{1 \over \mu}\,\Re\int_{\pi}^{0} {1 \over \ln\pars{\mu\bracks{1 - z^{2}}\ic/2}}\,{\dd z \over z} \right\vert_{\,z\ =\ \epsilon\expo{\ic\theta}} \\[3mm]&\phantom{=}-\,{1 \over \mu}\,\Re\int_{\epsilon}^{1 - \epsilon} {1 \over \ln\pars{\mu\bracks{1 - x^{2}}\ic/2}}\,{\dd x \over x} \\[3mm]&\phantom{=}\left.-\,{1 \over \mu}\,\Re\int_{\pi}^{\pi/2} {1 \over \ln\pars{\mu\bracks{1 - z^{2}}\ic/2}}\,{\dd z \over z} \right\vert_{\,z\ =\ 1\ +\ \epsilon\expo{\ic\theta}} \end{align}
\begin{align} {\cal F}'\pars{\mu}&= -\,{1 \over \mu}\,\Re\pp\ \overbrace{\int_{-1}^{1} {1 \over \ln\pars{\mu\bracks{1 - x^{2}}\ic/2}}\,{\dd x \over x}}^{\ds{=\ 0}}\ -\,{1 \over \mu}\,\Re\int_{\pi}^{0}{\ic\,\dd\theta \over \ln\pars{\mu\ic/2}} \\[3mm]&=-\,{\pi \over \mu}\,\Im\bracks{1 \over \ln\pars{\mu\ic/2}} \end{align}
By using the boundary condition $\pars{1}$: \begin{align} {\cal F}\pars{1^{-}}& =\int_{0}^{\pi}\arctan\pars{\ln\pars{\sin\pars{x}} \over x}\,\dd x =-\,{\pi^{2} \over 2} -\pi\,\Im\int_{0^{+}}^{1^{-}}{1 \over \mu}\,{\dd\mu \over \ln\pars{\mu\ic/2}} \\[3mm]&=-\,{\pi^{2} \over 2} -\pi\,\Im\int_{0^{+}}^{1^{-}}{\dd\mu/\mu \over \ln\pars{\mu/2} + \pi\ic/2} =-\,{\pi^{2} \over 2} -\pi\,\Im\int_{-\infty}^{-\ln\pars{2^{+}}}{\dd t \over t + \pi\ic/2} \\[3mm]&=-\,{\pi^{2} \over 2} +\pi\int_{-\infty}^{-\ln\pars{2^{+}}}{\pi\,\dd t/2 \over t^{2} + \pars{\pi/2}^{2}} =-\,{\pi^{2} \over 2} +\left.\pi\arctan\pars{2t \over \pi}\right\vert_{\,-\infty}^{\,-\ln\pars{2^{+}}} \\[3mm]&=-\,{\pi^{2} \over 2} + \pi\bracks{% \arctan\pars{-\,{2\ln\pars{2} \over \pi}} + {\pi \over 2}} \end{align}
$$\color{#66f}{\large% \int_{0}^{\pi}\arctan\pars{\ln\pars{\sin\pars{x}} \over x}\,\dd x =-\pi\,\arctan\pars{2\ln\pars{2} \over \pi}} \approx {\tt -1.3055} $$
$\def\Im{\mathrm{Im}}$ I hope it is not too late to present an alternative solution.
Let \begin{align} I(s)=\int^\pi_0\arctan\left(\frac{\ln(s\sin{x})}{x}\right)\mathrm{d}x \end{align} Differentiate under the integral sign to get \begin{align} I'(s) &=\frac{1}{s}\int^\pi_0\frac{x}{x^2+\ln^2(s\sin{x})}\mathrm{d}x\\ &=-\frac{1}{s}\Im\int^\pi_0\frac{1}{\ln\left(\frac{se^{i2x}-s}{2i}\right)}\mathrm{d}x\\ &=-\frac{1}{s}\Im\int_{|z|=1}\frac{1}{\ln\left(\frac{sz-s}{2i}\right)}\frac{\mathrm{d}z}{2iz}\\ &=-\frac{1}{s}\Im\frac{\pi}{\ln\left(-\frac{s}{2i}\right)}\\ &=-\frac{1}{s}\Im\frac{\pi}{\ln\left(\frac{s}{2}\right)+\frac{\pi i}{2}}\frac{\ln\left(\frac{s}{2}\right)-\frac{\pi i}{2}}{\ln\left(\frac{s}{2}\right)-\frac{\pi i}{2}}\\ &=\frac{1}{2s}\frac{\pi^2}{\ln^2\left(\frac{s}{2}\right)+\frac{\pi^2}{4}} \end{align} where the fourth equality follows from the residue theorem and the fact that the indent around the branch point $z=1$ produces no contribution as $\epsilon\to0$.
Integrating back, \begin{align} I(1) &=I(\infty)+\frac{\pi^2}{2}\int^{s=1}_{s=\infty}\frac{1}{\ln^2\left(\frac{s}{2}\right)+\frac{\pi^2}{4}}\mathrm{d}\ln\left(\frac{s}{2}\right)\\ &=\frac{\pi^2}{2}+\left.\frac{\pi^2}{2}\frac{2}{\pi}\arctan\left(\frac{2\ln\left(\frac{s}{2}\right)}{\pi}\right)\right|^1_\infty\\ &=-\pi\arctan\left(\frac{2\ln{2}}{\pi}\right) \end{align} as desired.
