Let $\begin{aligned} I &=\int_{0}^{\pi} \log (1+\cos x) d x...(i) \\ I &=\int_{0}^{\pi} \log \{1+\cos (\pi-x)\} d x \\ &=\int_{0}^{\pi} \log (1-\cos x) d x...(ii) \end{aligned}$
On adding Eqs. (i) and (ii), we get $\begin{aligned} 2 I &=\int_{0}^{\pi}\{\log (1+\cos x)+\log (1-\cos x)\} d x \\ I &=\frac{1}{2} \int_{0}^{\pi} \log \left(1-\cos ^{2} x\right) d x \\ &=\frac{1}{2} \int_{0}^{\pi} \log \sin ^{2} x d x \\ &=\int_{0}^{\pi} \log \sin x d x \\ &=2 \int_{0}^{\pi / 2} \log \sin x d x \end{aligned}$
$\begin{aligned}\left\{\because \int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x,\right.\\ &\text { if } f(2 a-x)=f(x)\} \\=2\left\{-\frac{\pi}{2} \log 2\right\} \\\left(\because \int_{0}^{\pi / 2} \log \sin x d x=-\frac{\pi}{2} \log 2\right) \\=\pi \log \frac{1}{2} \end{aligned}$