How to evaluate $\int_{0}^{\pi} \log(2+\cos x)dx$? I tried integrating it by part but it didn't work. I konw it can be solved by the Gauss Mean Value theorem .Is there some elementary method for evaluate it or just some others? It seems there are at least four differential ways(page 4 (17)) but I cannot think out any(the link only says there are four but not give any way).
 A: Another way: differentiate under integral sign!
Let $\displaystyle\quad J(a) = \int_0^\pi \log(a + \cos x) dx,\quad$ and $t = \tan\frac{x}{2}$, we have:
$$\begin{align}
J'(a) 
= & \int_0^\pi \frac{dx}{a + \cos x} 
= \int_0^{\infty} \frac{\frac{2 dt}{1+t^2}}{a + \frac{1-t^2}{1+t^2}}
= \frac{2}{a+1}\int_0^{\infty}\frac{dt}{1 + \frac{a-1}{a+1}t^2}\\
= & \frac{2}{\sqrt{a^2-1}}\int_0^\infty \frac{ds}{1+s^2} = \frac{\pi}{\sqrt{a^2-1}}
\end{align}$$
Let $a = \cosh\theta$, and integrate over $a$, we get
$$\begin{align}
F(a) 
= & \int F'(a) da = \pi \int \frac{d\cosh\theta}{\sqrt{\cosh^2\theta-1}}\\
= & \pi \theta + C  = \pi\cosh^{-1}(a) + C = \pi \log( a + \sqrt{a^2 - 1} ) + C\tag{*1}
\end{align}$$
for some integration constant $C$. For large $a$, we have 
$$\log(a + \cos x) = \log a + \frac{\cos x}{a} + O(\frac{1}{a^2})
\quad\implies\quad 
J(a) = \pi \log a + O(\frac{1}{a})\tag{*2}
$$
Compare $(*1)$ with $(*2)$, we find $C = -\pi \log 2$ and hence:
$$J(a) = \pi\log\left(\frac{a + \sqrt{a^2-1}}{2}\right)$$
In particular, the integral we want to calculate is $\displaystyle\quad J(2) = \pi\log\left(\frac{2+\sqrt{3}}{2}\right)$.
A: It depends on what counts as elementary. One way is to consider 
\begin{align}I(a)&=\int_0^{\pi}\ln\left(\frac{a+a^{-1}}{2}+\cos x\right)dx=\\
&=
\frac12\int_0^{2\pi}\ln\left(\frac{a+a^{-1}}{2}+\cos x\right)dx=\\
&=\frac12\int_0^{2\pi}\left[-\ln 2a+\ln\left(1+ae^{ix}\right)+\ln\left(1+ae^{-ix}\right)\right]dx.
\end{align}
with $0<a<1$. The integrals of $\ln\left(1+ae^{\pm ix}\right)$ are equal to zero. To show this, just expand the logarithms in Taylor series in $a$ and integrate termwise. Therefore,
$$I(a)=-\pi\ln 2 a.$$
In your case, $a+a^{-1}=4$ so that $a=2-\sqrt{3}$.
