Evaluate $\int\limits_{0}^{\pi} \log(1 + \cos x) dx.$

$-\pi \log 2$

The correct answer is Option (3) → $-\pi \log 2$

Let $I = \int\limits_{0}^{\pi} \log(1 + \cos x) dx \dots(i)$

$\Rightarrow I = \int\limits_{0}^{\pi} \log[1 + \cos(\pi - x)] dx$

$\Rightarrow I = \int\limits_{0}^{\pi} \log(1 - \cos x) dx \dots(ii)$

Adding eqs. (i) and (ii)

$2I = \int\limits_{0}^{\pi} \log[(1 + \cos x)(1 - \cos x)] dx$

$\Rightarrow I = \frac{1}{2} \int\limits_{0}^{\pi} \log(1 - \cos^2 x) dx$

$\Rightarrow I = \frac{1}{2} \int\limits_{0}^{\pi} \log \sin^2 x dx$

$= \int\limits_{0}^{\pi} \log \sin x dx$

Since, $\log[\sin(\pi - x)] = \log \sin x$

$∴I = 2 \int\limits_{0}^{\pi/2} \log \sin x dx \dots(iii)$

$\Rightarrow I = 2 \int\limits_{0}^{\pi/2} \log \sin \left( \frac{\pi}{2} - x \right) dx$

$\Rightarrow I = 2 \int\limits_{0}^{\pi/2} \log \cos x dx \dots.(iv)$

Adding eqs. (iii) and (iv)

$2I = 2 \int\limits_{0}^{\pi/2} \log \sin x \cos x dx$

$\Rightarrow I = \int\limits_{0}^{\pi/2} \log \left( \frac{\sin 2x}{2} \right) dx$

$= \int\limits_{0}^{\pi/2} \log \sin 2x dx - \log 2 \int_{0}^{\pi/2} 1.dx$

$\Rightarrow I = I_1 - \log 2 [x]_{0}^{\pi/2} = I_1 - \frac{\pi}{2} \log 2 \dots(v)$

Now $I_1 = \int\limits_{0}^{\pi/2} \log \sin 2x dx$

Let $2x = t \Rightarrow 2dx = dt$

$= \int\limits_{0}^{\pi} \log \sin t \frac{dt}{2}$

$= \frac{1}{2} \int\limits_{0}^{\pi} \log \sin x dx \text{ (changing } t \text{ to } x)$

$= \frac{1}{2} \times 2 \int\limits_{0}^{\pi/2} \log \sin x dx$

$\Rightarrow I_1 = \frac{1}{2} I$

From eq. (v)

$\Rightarrow I = \frac{1}{2} I - \frac{\pi}{2} \log 2$

$\Rightarrow \frac{1}{2} I = - \frac{\pi}{2} \log 2$

$∴I = -\pi \log 2.$