Evaluation of $\int_{0}^{\pi}\log(2+\cos x)dx$ How can I evaluate the definite integral $$\int_{0}^{\pi}\log(2+\cos x)dx$$
without using integral with parameters? This question is not difficult , if we use the acknowledge of integral with parameters.Actually, I want find a series of skillful substitutions to deal with it concisely.
Applying the integral with parameters,my answer is $
\pi\log(1+\frac{\sqrt{3}}{2})$.
 A: Here is to integrate with just substitutions. Let $a=2-\sqrt3$ and rewrite the integral as
$$\int_{0}^{\pi}\ln(2+\cos x)dx=2\pi \ln\frac{1+\sqrt3}2 + \int_{0}^{\pi}\ln(1+a^2-2a\cos \gamma)d\gamma\tag 1
$$
Substitute $c^2= 1+a^2-2a\cos \gamma$ while recognizing that $c$, $a$ and $1$ represent the sides of a triangle with the angles $\gamma = \pi-\beta(\gamma) - \alpha (\gamma) $. Then
\begin{align}
& \frac12\int_{0}^{\pi}\ln(1+a^2-2a\cos \gamma)d\gamma 
= -\int_{0}^{\pi}\ln c(\gamma) \>( d\beta(\gamma)+ d\alpha(\gamma) )\\ 
= &-\int_{0}^{\pi}\ln \frac{1\cdot \sin\gamma}{\sin\beta(\gamma) }\>d\beta  -\int_{0}^{\pi}\ln \frac{a\sin\gamma}{\sin\alpha(\gamma) }\>d\alpha\\
= &
\int_{0}^{\pi}\ln \sin\gamma\>d\gamma +\int_{0}^{\pi}\ln \sin\beta(\gamma) \>\overset{ t= \beta(\gamma) }{d\beta(\gamma)}+ \int_{0}^{\pi}\ln \frac{\sin\alpha(\gamma) }a \>\overset{ s= \alpha(\gamma)} {d\alpha(\gamma)}\\
= &\int_{0}^{\pi}\ln \sin\gamma\>d\gamma +\int_{\pi}^{0}\ln \sin tdt+ \int_{0}^{0}\ln \frac{\sin s}ads =0 \\
\end{align}
Thus, the integral (1) is
$$\int_{0}^{\pi}\ln(2+\cos x)dx=2\pi \ln\frac{1+\sqrt3}2 $$
A: In General
For $|\alpha|\lt1$,
$$
\begin{align}
\log\left(1+\alpha^2+2\alpha\cos(\theta)\right)
&=\log\left(1+\alpha e^{i\theta}\right)+\log\left(1+\alpha e^{-i\theta}\right)\tag1\\[9pt]
&=\sum_{k=1}^\infty(-1)^{k-1}\frac{\alpha^k}ke^{ik\theta}+\sum_{k=1}^\infty(-1)^{k-1}\frac{\alpha^k}ke^{-ik\theta}\tag2\\
&=\sum_{k=1}^\infty(-1)^{k-1}2\frac{\alpha^k}k\cos(k\theta)\tag3
\end{align}
$$
Explanation:
$(1)$: $1+\alpha^2+2\alpha\cos(\theta)=\left(1+\alpha e^{i\theta}\right)\left(1+\alpha e^{-i\theta}\right)$
$(2)$: power series for $\log(1+x)$
$(3)$: $e^{ik\theta}+e^{-ik\theta}=2\cos(k\theta)$
Thus, integrating $(3)$ over $[0,\pi]$ gives
$$
\begin{align}
0
&=\int_0^\pi\log\left(1+\alpha^2+2\alpha\cos(\theta)\right)\,\mathrm{d}\theta\tag4\\
&=\pi\log(2\alpha)+\int_0^\pi\log\left(\frac{1+\alpha^2}{2\alpha}+\cos(\theta)\right)\,\mathrm{d}\theta\tag5\\
&=\pi\log\left(\frac2{\beta+\sqrt{\beta^2-1}}\right)+\int_0^\pi\log\left(\beta+\cos(\theta)\right)\,\mathrm{d}\theta\tag6
\end{align}
$$
Explanation:
$(4)$: $(3)$ and $\int_0^\pi\cos(k\theta)\,\mathrm{d}\theta=0$
$(5)$: pull $\pi\log(2\alpha)$ out of the integral
$(6)$: $\alpha=\frac1{\beta+\sqrt{\beta^2-1}}$ is the root of $\frac{1+\alpha^2}{2\alpha}=\beta$ with $|\alpha|\lt1$
Therefore,
$$
\int_0^\pi\log\left(\beta+\cos(\theta)\right)\,\mathrm{d}\theta
=\pi\log\left(\frac{\beta+\sqrt{\beta^2-1}}2\right)\tag7
$$
Substitute $\beta\mapsto\frac\beta\gamma$ and add $\pi\log(\gamma)$:
$$
\int_0^\pi\log\left(\beta+\gamma\cos(\theta)\right)\,\mathrm{d}\theta
=\pi\log\left(\frac{\beta+\sqrt{\beta^2-\gamma^2}}2\right)\tag8
$$

Applied to this problem
$$
\int_0^\pi\log(2+\cos(x))\,\mathrm{d}x=\pi\log\left(\frac{2+\sqrt3}2\right)\tag9
$$
