Evaluate $\int_0^{2\pi}\ln(1+a^2-2a\cos(x))dx\text{ ,with }a>0$

Question

$$\int_0^{2\pi}\ln(1+a^2-2a\cos(x))dx,\;\;\;\;\text{with }a>0$$

How to evaluate Integral of $\ln(1+a^2-2a\cos x) dx$? where $x$ from $0$ to $2\pi$ and $a>0$, $\ln$ is the natural logarithm.

Answer 1

Idea: let $$f(a)=\int_0^{2\pi}\ln(1+a^2-2a\cos(x))dx$$

Now differentiate $f$ on $a$ and then $f'(a) $ integrate on $x$ and then integrate on $a$. This works many times. Perhaps now too.

Answer 2

$$I(a)=\int_0^{2\pi}\ln\left(1+a^2-2a\cos x\right)dx$$ $$I'(a)=\int_{-\pi}^{\pi}\frac{2a-2\cos x}{1+a^2-2a\cos x}dx$$ now the most obvious choice would be to use $t=\tan\frac{x}{2}$ this transforms our integral into: $$I'(a)=2\int_{-\infty}^\infty\frac{2a-2\frac{1-t^2}{1+t^2}}{1+a^2-2a\frac{1-t^2}{1+t^2}}\frac{dt}{1+t^2}$$$$=4\int_{-\infty}^\infty\frac{a(1+t^2)-(1-t^2)}{(1+t^2)\left[(1+t^2)(1+a^2)-2a(1-t^2)\right]}\frac{dt}{1+t^2}$$$$=4\int_{-\infty}^\infty\frac{(a-1)+(a+1)t^2}{(1+t^2)^2\left[(1+t^2)(1+a^2)-2a(1-t^2)\right]}dt$$