Evaluating Integral $\int_{0}^{2\pi} \ln \left(|e^{ix}-1| \right)\,dx$

Question

Consider the integral $$\int_{0}^{2\pi} \ln \left(|e^{ix}-1| \right)\,dx$$how do I solve this integral? I have a feeling that the answer is a multiple of $2\pi$. I have tried setting $e^{ix}=\cos(x)+i \sin(x)$, which turns the integral into $$\int_{0}^{2\pi}\ln \left(|\cos(x)+i \sin(x)-1| \right)\,dx$$ but this did not make the evaluation any easier because I am not sure what to do with the logarithm function.

Answer 1

$$\begin{align}I&=\int_{0}^{2\pi}\ln(|e^{ix}-1|)dx\\ \\ &=\int_{0}^{2\pi}\ln(|\cos(x)+i\sin(x)-1|)dx\\ \\ &=\int_{0}^{2\pi}\ln(\sqrt{2-2\cos x})dx\\ \\ &=\frac{1}2\int_{0}^{2\pi}\ln\left(4\sin^2 \frac{x}2\right)dx\\ \\ &=2\pi\ln2+\frac{1}2\int_{0}^{2\pi}\ln\left(\sin^2 \frac{x}2\right)dx,~~~~~~~~~x=2t\\ \\ &=2\pi\ln2+\int_{0}^{\pi}\ln\left(\sin^2 t\right)dt\\ \\ &=2\pi\ln2+2\int_{0}^{\pi}\ln\left(\sin t\right)dt\\ \\ &=2\pi\ln2-2\pi\ln2\\ \\ &=0 \end{align}$$

The last integral can be found here