integral $ \int_0^{\frac{\pi}{3}}\mathrm{ln}\left(\frac{\mathrm{sin}(x)}{\mathrm{sin}(x+\frac{\pi}{3})}\right)\ \mathrm{d}x$ We want to evaluate
$ \displaystyle \int_0^{\frac{\pi}{3}}\mathrm{ln}\left(\frac{\mathrm{sin}(x)}{\mathrm{sin}(x+\frac{\pi}{3})}\right)\ \mathrm{d}x$.
We tried contour integration which was not helpful. Then we started trying to use symmetries, this also doesn't work.


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 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
&\bbox[10px,#ffd]{\ds{%
\int_{0}^{\pi/3}\ln\pars{\sin\pars{x} \over \sin\pars{x + \pi/3}}\,\dd x}} =
\int_{-\pi/6}^{\pi/6}\ln\pars{\sin\pars{x + \pi/6} \over \cos\pars{x}}\,\dd x
\\[5mm] = &\
\int_{-\pi/6}^{\pi/6}\ln\pars{{\root{3} \over 2}\,\tan\pars{x} + { 1 \over 2}}
\,\dd x
\,\,\,\,\,\,\stackrel{\large x\ =\ \arctan\pars{2t - 1 \over \root{3}}}{=}\,\,\,
\,\,\,
{\root{3} \over 2}\int_{0}^{1}{\ln\pars{t} \over t^{2} - t + 1}\,\dd t
\\[5mm] = &\
{\root{3} \over 2}\int_{0}^{1}{\ln\pars{t} \over \pars{t - r}\pars{t - \bar{r}}}\,\dd t\label{1}\tag{1}
\end{align}

where $\ds{r \equiv {1 \over 2} + {\root{3} \over 2}\,\ic = \exp\pars{{\pi \over 3}\,\ic}}$

\eqref{1} is reduced to
\begin{align}
&\bbox[10px,#ffd]{\ds{%
\int_{0}^{\pi/3}\ln\pars{\sin\pars{x} \over \sin\pars{x + \pi/3}}\,\dd x}} =
{\root{3} \over 2}\int_{0}^{1}\ln\pars{t}
\pars{{1 \over t - r} - {1 \over t - \bar{r}}}{1 \over r - \bar{r}}\,\dd t
\\[5mm] = &\
{\root{3} \over 2}\,{1 \over 2\ic\,\Im\pars{r}}\,2\ic\,\Im\int_{0}^{1}{\ln\pars{t} \over t - r}\,\dd t =
-\,\Im\int_{0}^{1}{\ln\pars{t} \over r - t}\,\dd t =
-\,\Im\int_{0}^{1/r}{\ln\pars{rt} \over 1 - t}\,\dd t
\\[5mm] = &\
-\,\Im\int_{0}^{\large\bar{r}}{\ln\pars{1 - t} \over t}\,\dd t =
\Im\int_{0}^{\large\bar{r}}\mrm{Li}_{2}'\pars{t}\,\dd t\qquad
\pars{~\mrm{Li}_{s}\ \mbox{is the}\ PolyLogarithm\ Function~}
\\[5mm] \implies &\
\bbx{\int_{0}^{\pi/3}\ln\pars{\sin\pars{x} \over \sin\pars{x + \pi/3}}\,\dd x = \Im\mrm{Li}_{2}\pars{\expo{-\pi\ic/3}} \approx -1.0149}
\end{align}
