Let $\mathrm{I}=\int_0^\pi \log (1+\cos x) d x \quad \ldots(i)$
Then, $\mathrm{I}=\int_0^\pi \log [1+\cos (\pi-x)] d x$
$\mathrm{I}=\int_0^\pi \log (1-\cos x) d x \quad \ldots(ii)$
Adding (i) and (ii), we get
$2 \mathrm{I}=\int_0^\pi \log \left(1-\cos ^2 x\right) d x=2 \int_0^\pi \log \sin x d x$
$\Rightarrow \mathrm{I}=\int_0^\pi \log \sin x d x=2 \int_0^{\pi / 2} \log \sin x d x=2 \mathrm{I}_1$
Now, $\mathrm{I}_1=\int_0^{\pi / 2} \log \sin x d x \quad \ldots(iii)$
$\mathrm{I}_1=\int_0^{\pi / 2} \log \sin \left(\frac{\pi}{2}-x\right) d x=\int_0^{\pi / 2} \log \cos x d x \text {...(iv) }$
Adding (iii) and (iv), we get
$\begin{aligned}
&2 \mathrm{I}_1=\int_0^{\pi / 2} \log \left(\frac{\sin 2 x}{2}\right) d x \\
&=\int_0^{\pi / 2} \log \sin 2 x d x-\int_0^{\pi / 2} \log 2 d x \\
&=\int_0^{\pi / 2} \log \sin 2 x d x-\frac{\pi}{2} \log 2=\mathrm{I}_2-\frac{\pi}{2} \log _2 \ldots \text { (v) }
\end{aligned}$
Put $2 x=t$, so that $2 d x=d t$
When $x=0, t=0$; when $x=\frac{\pi}{2}, t=\pi$
$\therefore \mathrm{I}_2=\frac{1}{2} \int_0^\pi \log \sin t d t=\int_0^{\pi / 2} \log \sin x d x=\mathrm{I}_1$
$\therefore \quad$ From $(v)$, we get; $\mathrm{I}_1=-\frac{\pi}{2} \log 2$
$\therefore \quad \mathrm{I}=2 \times\left(-\frac{\pi}{2} \log 2\right)=-\pi \log 2$