'Ln'tegral

$\begin{aligned} I & = \int_0^{2\pi} \ln (1-\cos x)\ dx & \small \color{#3D99F6} \text{Since }\cos x \text{ is symmetrical about }\pi. \\ & = 2 \int_0^\pi \ln (1-\cos x)\ dx & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \int_0^\pi \left(\ln (1-\cos x)+ \ln (1+\cos x)\right) dx \\ & = \int_0^\pi \ln (1-\cos^2 x)\ dx & \small \color{#3D99F6} \text{Note that }\cos^2 \theta = \frac {1+\cos (2\theta)}2 \\ & = \int_0^\pi \ln \left(\frac {1-\cos 2x}2 \right) dx \\ & = \int_0^\pi \left(\ln(1-\cos 2x) - \ln 2 \right) dx & \small \color{#3D99F6} \text{Let }\theta = 2x \implies d\theta = 2\ dx \\ & = \frac 12 \int_0^{2\pi} \ln(1-\cos \theta) \ d\theta - \int_0^\pi \ln 2 \ dx \\ & = \frac 12 I - \pi \ln 2 \\ & = - 2 \pi \ln 2 \approx \boxed{-4.355} \end{aligned}$