TITLE: Find the value of the integral $\int_0^{2\pi}\ln|a+b\sin x|dx$ where $0\lt a\lt b$
QUESTION [5 upvotes]: Find the value of the integral $$\int_0^{2\pi}\ln|a+b\sin x|dx$$ where $0\lt a\lt b$. What is the use of this inequality. I tried to integrate the integral by parts, but the integral of the 2nd term was quite messy.Please help.

REPLY [4 votes]: $\newcommand{\+}{^{\dagger}}
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$\ds{0 < a < b}$.

\begin{align}
&\int_{0}^{2\pi}\ln\pars{\verts{a + b\sin\pars{x}}}\,\dd x=
2\pi\ln\pars{\verts{b}}
+\color{#c00000}{\int_{0}^{2\pi}\ln\pars{\verts{\mu + \sin\pars{x}}}\,\dd x}
\\[3mm]&\mbox{where}\quad
\mu \equiv {a \over b}\,,\quad\mbox{Note than}\quad 0 < \mu < 1
\end{align}

\begin{align}
&\color{#c00000}{\int_{0}^{2\pi}\ln\pars{\verts{\mu + \sin\pars{x}}}\,\dd x}
=\int_{-\pi}^{\pi}\ln\pars{\verts{\mu - \sin\pars{x}}}\,\dd x
=\sum_{\sigma = \pm}\int_{0}^{\pi}\ln\pars{\verts{\mu + \sigma\sin\pars{x}}}\,\dd x
\\[3mm]&=\sum_{\sigma = \pm}\int_{-\pi/2}^{\pi/2}
\ln\pars{\verts{\mu + \sigma\cos\pars{x}}}\,\dd x
=2\sum_{\sigma = \pm}\int_{0}^{\pi/2}
\ln\pars{\verts{\mu + \sigma\cos\pars{x}}}\,\dd x
\\[3mm]&=2\int_{0}^{\pi/2}\ln\pars{\verts{\mu - \cos\pars{x}}}\,\dd x
+2\int_{0}^{\pi/2}\ln\pars{\mu + \cos\pars{x}}\,\dd x
\\[3mm]&=2\int_{0}^{\tilde{\mu}}\ln\pars{\cos\pars{x} - \mu}\,\dd x
+2\int^{\pi/2}_{\tilde{\mu}}\ln\pars{\mu - \cos\pars{x}}\,\dd x
+2\int_{0}^{\pi/2}\ln\pars{\mu + \cos\pars{x}}\,\dd x
\end{align}
where $\ds{\tilde{\mu} = \arccos\pars{\mu}}$.

$$
\color{#c00000}{\!\!\!\!\!\int_{0}^{2\pi}\ln\pars{\verts{\mu + \sin\pars{x}}}\,\dd x}
=
2\int_{0}^{\tilde{\mu}}\ln\pars{\cos^{2}\pars{x} - \mu^{2}}\,\dd x
+2\int^{\pi/2}_{\tilde{\mu}}\ln\pars{\mu^{2} - \cos^{2}\pars{x}}\,\dd x
$$

Can you take it from here ?.