Calculate integral using the method of parameter derivation or integration $$F( \alpha ) = \int^{ \pi }_{ 0} \ln
 ( 1 - 2 \alpha \cos x + \alpha^{2} ) dx$$
Should I derive the inner function? But I can't process the derived outcome.
 A: Let's generalize the problem. We want to evaluate
$$
\int_0^{\large\pi}\ln\left(a^2-2ab\cos x+b^2\right)\ dx\tag1
$$
instead of
$$
\int_0^{\large\pi}\ln\left(1-2ab\cos x+a^2\right)\ dx.\tag2
$$
Rewrite $(1)$ as
\begin{align}
\int_0^{\large\pi}\ln\left(a^2-2ab\cos x+b^2\right)\ dx&=\int_0^{\large\pi}\ln \left[a^2\left(1-\frac {2b}a\cos x+\frac {b^2}{a^2}\right)\right]\ dx\\
&=\int_0^{\large\pi}\ln a^2\ dx+\int_0^{\large\pi}\ln \left(1-\frac {2b}a\cos x+\frac {b^2}{a^2}\right)\ dx\\
&=2\pi\ln a+\int_0^{\large\pi}\ln \left(1-\frac {2b}a\cos x+\frac {b^2}{a^2}\right)\ dx.\tag3\\
\end{align}
Now let $c=\dfrac ba$, then the integral in the RHS $(3)$ turns out to be
$$
\int_0^{\large\pi}\ln \left(1-\frac {2b}a\cos x+\frac {b^2}{a^2}\right)\ dx=\int_0^{\large\pi}\ln \left(1-2c\cos x+c^2\right)\ dx.\tag4
$$
Since this is a duplicate question, I wouldn't continue my work and the rest part, $(4)$, can be seen here. I will only give you the final result and leave it to you as exercise.
$$
\int_0^{\large\pi}\ln\left(a^2-2ab\cos x+b^2\right)\ dx= \left\{ 
\begin{array}{l l}
2\pi\ln a &, \quad \text{if $0<b\le a$}\\
\\
2\pi\ln b &, \quad \text{if $0<a\le b$.}
  \end{array} \right.
$$
