calculus - Evaluating $int_{0}^{pi}ln (1+cos x), dx$

The problem is

$$\int_{0}^{\pi}\ln (1+\cos x)\ dx$$

What I tried was using standard limit formulas like changing $x$ to $\pi - x$ and I also tried integration by parts on it to no avail. Please help. Also this is my first question so please tell if I am wrong somewhere.

Answer

$$\begin{eqnarray*}\int_{0}^{\pi}\log(1+\cos x)\,dx &=& \int_{0}^{\pi/2}\log(1+\cos x)\,dx+\int_{0}^{\pi/2}\log(1+\cos(\pi-x))\,dx\\ &=& \int_{0}^{\pi/2}\log(\sin^2 x)\,dx=\int_{0}^{\pi}\log(\sin x)\,dx \tag{1}\end{eqnarray*}$$
And by a notorious identity:
$$\prod_{k=1}^{n-1}\sin\frac{k\pi}{n} = \frac{2n}{2^n},\tag{2}$$
hence the RHS of $(1)$ can be computed as a Riemann sum:

$$\int_{0}^{\pi}\log(\sin x)\,dx = \lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n}=\color{red}{-\pi \log 2}.\tag{3}$$
There is also a well-known proof through symmetry:
$$\begin{eqnarray*}I=\int_{0}^{\pi}\log(\sin x)&=&2\int_{0}^{\pi/2}\log(\sin(2t))\,dt=2\int_{0}^{\pi/2}\log(2\sin t\cos t)\,dt\\&=&\pi\log 2+2\int_{0}^{\pi/2}\log(\sin t)\,dt+2\int_{0}^{\pi/2}\log(\cos t)\,dt\\&=&\pi \log 2 + 2I\tag{4}\end{eqnarray*}$$
from which $I=-\pi\log 2$ immediately follows.