Let $I=\int\limits_{0}^{\pi}\log(1+\cos x)dx$ ...(i) $I=\int\limits_{0}^{\pi}\log[1+\cos (\pi-x)]dx$ $\left[\because \int\limits_{0}^{a}f(x)dx=\int\limits_{0}^{a}f(a-x)dx\right]$ $=\int\limits_{0}^{\pi}\log(1-\cos x)dx$ ...(ii) Adding (i) and (ii), we get $2I= \int\limits_{0}^{\pi}[\log(1+\cos x) +\log(1-\cos x)]dx$ $= \int\limits_{0}^{\pi}\log(1-\cos^2 x)dx$ $= \int\limits_{0}^{\pi}\log\sin^2 x dx$ $=2 \int\limits_{0}^{\pi}\log\sin x dx$ $\Rightarrow I= \int\limits_{0}^{\pi}\log\sin x dx$ $= 2\int\limits_{0}^{\frac{\pi}{2}}\log\sin x dx =2I_1$where $I_1= \int\limits_{0}^{\frac{\pi}{2}}\log\sin x dx$...(iii)Then, $I_1= \int\limits_{0}^{\frac{\pi}{2}}\log\sin \left(\frac{\pi}{2}-x\right)dx$$I_1= \int\limits_{0}^{\frac{\pi}{2}}\log\cos x dx$...(iv)Adding (iii) and (iv), we get$2I_1= \int\limits_{0}^{\frac{\pi}{2}}\log\sin x dx+ \int\limits_{0}^{\frac{\pi}{2}}\log\cos x dx$$= \int\limits_{0}^{\frac{\pi}{2}}\log(\sin x \cos x)dx$$= \int\limits_{0}^{\frac{\pi}{2}}\log\left(\frac{2\sin x \cos x}{2}\right)dx$$= \int\limits_{0}^{\frac{\pi}{2}}\log\left(\frac{\sin 2x}{2}\right)dx$$= \int\limits_{0}^{\frac{\pi}{2}}\log(\sin 2x)dx- \int\limits_{0}^{\frac{\pi}{2}}\log 2 dx$ $= \int\limits_{0}^{\frac{\pi}{2}}\log(\sin 2x)dx- (\log 2)[x]^{\frac{\pi}{2}}_{0}$ $= \int\limits_{0}^{\frac{\pi}{2}}\log(\sin 2x)dx- \frac{\pi}{2}\log 2$ $= I_2- \frac{\pi}{2}\log 2$...(v) Where $I_2=\int\limits_{0}^{\frac{\pi}{2}}\log(\sin 2x)dx$Put $2x = t$ $\Rightarrow 2dx = dt$; When $x = 0, t = 0$; when $x =\frac{\pi}{2}, t=\pi$$I_2=\frac{1}{2}\int\limits_{0}^{\pi}\log\sin t dt$ $=\frac{1}{2} \cdot 2 = \frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\log\sin t dt$$= \int\limits_{0}^{\frac{\pi}{2}}\log\sin x dx=I_1$∴ From (v), we get$2I_1=I_2-\frac{\pi}{2}\log 2$$\Rightarrow 2I_1=I_1-\frac{\pi}{2}\log 2$$\Rightarrow I_1=-\frac{\pi}{2}\log 2$$\therefore I=2\times \left(-\frac{\pi}{2}\log 2\right)$ $=-\pi\log 2$