The value of \(\int_{0}^{\pi} \log (1+\cos x) d x\) is

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