Find the value of the integral $\int_0^{2\pi}\ln|a+b\sin x|dx$ where $0\lt a\lt b$ 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. 
 A: This is just for $0<b < a $
\begin{align*}
I &= \int_0^{2\pi}\log(a+\sqrt{a^2 - b^2}+be^{i\theta})d\theta +  \int_0^{2\pi}\log(a+\sqrt{a^2 - b^2}+be^{-i\theta})d\theta\\ 
 &= \int_0^{2\pi} \log \left(2a^2 +2a\sqrt{a^2 - b^2} + (a + \sqrt {a^2 - b^2}) b (e^{i\theta} + e^{-i\theta}) \right )d\theta\\ 
 &= \int_0^{2\pi} \log (2 (a + \sqrt{a^2 - b^2}))d\theta + \int_0^{2\pi }\log(a + b \cos\theta)d\theta\\
\end{align*}
By Gauss MVT, the value of top two integral is 
$$4\pi\log(a + \sqrt{a^2 -b^2})$$
Hence the for $0<b<a$ is 
$$\int_0^{2\pi} \log(a+ b\cos\theta)d\theta = 2 \pi \log \left(a + \sqrt{a^2 - b^2}\over 2\right)$$
EDIT:: The integral for $\log (a + b \sin \theta )$ is equal to that of $\log (a + b \cos \theta )$ as $\theta$ goes around $2 \pi$. It can also be done by changing $be^{-i\theta}$ to $-b e^{-i\theta}$ is above derivation.
For $0<a < b$, the integral is equal to real part of the above solution as mentioned by @Lucian in comments.
A: $\newcommand{\+}{^{\dagger}}
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$\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 ?.
