Show that

Solution:

Let $I=\int_{0}^{\pi} \log (1+\cos x) d x$            (1)

$\Rightarrow I=\int_{1}^{\pi} \log (1+\cos (\pi-x)) d x$               

$\Rightarrow I=\int_{0}^{\pi}\log (1-\cos x) d x$                      ...(2)

Adding (1) and (2), we obtain

$2 I=\int_{0}^{\pi}\{\log (1+\cos x)+\log (1-\cos x)\} d x$

$\Rightarrow 2 I=\int_{0}^{\pi} \log \left(1-\cos ^{2} x\right) d x$

$\Rightarrow 2 I=\int_{0}^{\pi} \log \sin ^{2} x d x$

$\Rightarrow 2 I=2 \int_{0}^{\pi} \log \sin x d x$

$\Rightarrow I=\int_{0}^{\pi} \log \sin x d x$                    ...(3)

$\sin (\pi-x)=\sin x$

$\therefore I=2 \int_{0}^{\frac{\pi}{2}} \log \sin x d x$      ...(4)

$\Rightarrow I=2 \int_{0}^{\frac{\pi}{2}} \log \sin \left(\frac{\pi}{2}-x\right) d x=2 \int_{1}^{\frac{\pi}{2}} \log \cos x d x$    ...(5)

Adding (4) and (5), we obtain

$2 I=2 \int_{0}^{\frac{\pi}{2}}(\log \sin x+\log \cos x) d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}}(\log \sin x+\log \cos x+\log 2-\log 2) d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}}(\log 2 \sin x \cos x-\log 2) d x$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \log \sin 2 x d x-\int_{0}^{\frac{\pi}{2}} \log 2 d x$

Let $2 x=t \Rightarrow 2 d x=d t$

When $x=0, t=0$

and when $x=\frac{\pi}{2}, t=\pi$

$\therefore I=\frac{1}{2} \int_{0}^{\pi} \log \sin t \mathrm{~d} t-\frac{\pi}{2} \log 2$

$\Rightarrow I=\frac{I}{2}-\frac{\pi}{2} \log 2 \quad[$ from 3$]$

$\Rightarrow \frac{I}{2}=-\frac{\pi}{2} \log 2$

$\Rightarrow I=-\pi \log 2$