Calculate integral using the method of parameter derivation or integration

Question

$$F( \alpha ) = \int^{ \pi }_{ 0} \ln ( 1 - 2 \alpha \cos x + \alpha^{2} ) dx$$

Should I derive the inner function? But I can't process the derived outcome.

Answer 1

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. $$ \int_0^{\large\pi}\ln\left(a^2-2ab\cos x+b^2\right)\ dx= \left\{ \begin{array}{l l} 2\pi\ln a &, \quad \text{if $0<b\le a$}\\ \\ 2\pi\ln b &, \quad \text{if $0<a\le b$.} \end{array} \right. $$