<s>[INST] Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$ 
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.

My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(\displaystyle\frac{1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}.
\end{align*}$$
Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But 
$$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider
$$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$
So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$
It looks too easy, is there any crucial lack?
$(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof.
 [/INST] For some reason, this problem appeared on the main page for me even though the problem was posted quite a while ago. Let me present an alternative approach which does not really require any integrating.
Let us consider your integral:
$$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos(x)+a^2\right)\ \text{d}x,\;\ a>1$$
Consider a triangle $ABC$ with angles $\alpha,\beta,\gamma$ opposite to the sides $a,b,c$, respectively:

The law of cosines states
$$c^2=a^2+b^2-2ab\cos(\gamma)$$
To keep with this notation, let $b=1$ and $\gamma=x$.
Our integral becomes
$$\begin{align}
I(a) & = \int_{0}^{\pi}\ln\left(c^2\right)\ \text{d}\gamma \\
 & = 2\int_{0}^{\pi}\ln\left(c\right)\ \text{d}\gamma \\ 
\end{align}$$
The law of sines states
$$\frac{\sin(\alpha)}{a}=\frac{\sin(\beta)}{b}=\frac{\sin(\gamma)}{c}$$
Our integral becomes
$$\begin{align}
I(a) & = 2\int_{0}^{\pi}\ln\left(a\ \frac{\sin(\gamma)}{\sin(\alpha)}\right)\ \text{d}\gamma \\
 & = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma-2\int_{0}^{\pi}\ln\left(\sin(\alpha)\right)\ \text{d}\gamma \\ 
\end{align}$$
Consider the right-hand integral. In a triangle, the three angles $\alpha,\beta,\gamma$ add up to $\pi$ radians. Let $\gamma=\pi-\alpha-\beta$ such that $\text{d}\gamma=-\text{d}\alpha-\text{d}\beta$. 
When $\gamma\rightarrow 0:\alpha\rightarrow \pi,\beta\rightarrow 0$. When $\gamma\rightarrow \pi:\alpha\rightarrow 0,\ \beta\rightarrow 0$.
Our integral becomes
$$\begin{align}
I(a) & = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma\\
 & -2\left[\int_{\pi}^{0}\ln\left(\sin(\alpha)\right)(-\text{d}\alpha)+\int_{0}^{0}\ln\left(\sin(\alpha)\right)(-\text{d}\beta)\right] \\
 & = 2\pi \ln(a)+2\int_{0}^{\pi}\ln\left(\sin(\gamma)\right)\ \text{d}\gamma-2\int_{0}^{\pi}\ln\left(\sin(\alpha)\right)\ \text{d}\alpha \\ 
\end{align}$$
Recognize that the integrals evaluate to the same value and therefore will cancel out each other.
$$I(a)=2\pi\ln(a),\;\ a>1$$
</s>