By using the properties of definite integrals,evaluate the integral $\int_{0}^{\pi} \log (1+\cos x) d x$.

(D) Let $I = \int_{0}^{\pi} \log (1+\cos x) d x$ ..... $(1)$
Using the property $\int_{0}^{a} f(x) d x = \int_{0}^{a} f(a-x) d x$,we get:
$I = \int_{0}^{\pi} \log (1+\cos(\pi-x)) d x = \int_{0}^{\pi} \log (1-\cos x) d x$ ..... $(2)$
Adding $(1)$ and $(2)$:
$2I = \int_{0}^{\pi} \{\log(1+\cos x) + \log(1-\cos x)\} d x$
$2I = \int_{0}^{\pi} \log(1-\cos^2 x) d x = \int_{0}^{\pi} \log(\sin^2 x) d x$
$2I = 2 \int_{0}^{\pi} \log(\sin x) d x \Rightarrow I = \int_{0}^{\pi} \log(\sin x) d x$
Using $\int_{0}^{2a} f(x) d x = 2 \int_{0}^{a} f(x) d x$ if $f(2a-x) = f(x)$:
$I = 2 \int_{0}^{\pi/2} \log(\sin x) d x$ ..... $(3)$
Also,$I = 2 \int_{0}^{\pi/2} \log(\cos x) d x$ ..... $(4)$
Adding $(3)$ and $(4)$:
$2I = 2 \int_{0}^{\pi/2} (\log(\sin x) + \log(\cos x)) d x$
$I = \int_{0}^{\pi/2} \log(\sin x \cos x) d x = \int_{0}^{\pi/2} \log\left(\frac{\sin 2x}{2}\right) d x$
$I = \int_{0}^{\pi/2} \log(\sin 2x) d x - \int_{0}^{\pi/2} \log 2 d x$
Let $2x = t$,then $2 dx = dt$. When $x=0, t=0$; when $x=\pi/2, t=\pi$:
$I = \frac{1}{2} \int_{0}^{\pi} \log(\sin t) d t - \frac{\pi}{2} \log 2$
$I = \frac{1}{2} (2 \int_{0}^{\pi/2} \log(\sin t) d t) - \frac{\pi}{2} \log 2 = I - \frac{\pi}{2} \log 2$ (Wait,this implies $I = I - \frac{\pi}{2} \log 2$,which is incorrect. Re-evaluating: $I = \frac{1}{2} I - \frac{\pi}{2} \log 2$ is wrong. The correct step is $I = \frac{1}{2} (2I) - \frac{\pi}{2} \log 2$ is not right. Actually,$\int_{0}^{\pi/2} \log(\sin 2x) d x = \frac{1}{2} \int_{0}^{\pi} \log(\sin t) d t = \int_{0}^{\pi/2} \log(\sin t) d t = I/2$.)
Thus,$I = I/2 - \frac{\pi}{2} \log 2 \Rightarrow I/2 = -\frac{\pi}{2} \log 2 \Rightarrow I = -\pi \log 2$.