TITLE: Evaluating a trigonometric integral.
QUESTION [14 upvotes]: Show That
$$\int_{0}^{\pi}\ln (\sin x) \cos(2nx) \, dx = -\dfrac{\pi}{2n}$$

I tried to use substitutions and taylor series of $\ln (\sin x)$, but to no avail.
Any help will be appreciated.
Thanks.

REPLY [3 votes]: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\cos\pars{2nx}\,\dd x =
-\,{\pi \over2n}:\ {\large ?}\,,\qquad n \in \mathbb{Z}\setminus\braces{0}}$.

\begin{align}
&\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\cos\pars{2nx}\,\dd x =
2\,\Re\int_{0}^{\pi/2}\ln\pars{\sin\pars{x}}\expo{2\verts{n}x\ic}\,\dd x
\\[5mm] = &\
\left.2\,\Re\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2}
\ln\pars{z - 1/z \over 2\ic}z^{2\verts{n}}
\,{\dd z \over \ic z}\,\right\vert_{\ z\ =\ \exp\pars{\ic\theta}}
\\[5mm] = &\
\left.2\,\Im\int_{\theta\ =\ 0}^{\theta\ =\ \pi/2}
\ln\pars{{1 - z^{2} \over 2z}\,\ic}z^{2\verts{n} - 1}\,\,\dd z\,
\right\vert_{\ z\ =\ \exp\pars{\ic\theta}}
\\[1cm] = &\
-2\,\ \overbrace{\Im\int_{1}^{\epsilon}\ln\pars{1 + y^{2} \over 2y}y^{2\verts{n} - 1}\exp\pars{\ic\,{\pi \over 2}\bracks{2\verts{n} - 1}}\ic\,\dd y}
^{\ds{=\ 0}}
\\[5mm] - &\
2\,\Im\int_{\pi}^{-\pi}\ln\pars{\ic\expo{-\ic\theta} \over 2\epsilon}
\epsilon^{2\verts{n} - 1}\exp\pars{\bracks{2\verts{n} - 1}\theta\,\ic}\epsilon\expo{\ic\theta}\ic\,\dd\theta
\\[5mm] - &\
2\,\Im\int_{0}^{1}\bracks{\ln\pars{1 - x^{2} \over 2x} + {\pi \over 2}\,\ic}x^{2\verts{n} - 1}\,\dd x
\end{align}

In the limit $\ds{\epsilon \to 0^{+}}$:
\begin{align}
&\int_{0}^{\pi}\ln\pars{\sin\pars{x}}\cos\pars{2nx}\,\dd x =
-\,\pi\int_{0}^{1}x^{2\verts{n} - 1}\,\dd x =
\bbx{\ds{-\,{\pi \over 2\verts{n}}\,,\quad n \in \mathbb{Z}\setminus\braces{0}}}
\end{align}