Let I = $\int\limits_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}}$ dx Using $\int\limits_{0}^{a}$ f (x) = $\int\limits_{0}^{a}$ f (a - x) dx I = $\int\limits_{0}^{\pi} \frac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)}+e^{-\cos(\pi-x)}}$ dx 2I = $\int\limits_{0}^{\pi}$ $\frac{e^{-\cos x}+e^{\cos x}}{e^{\cos x}+e^{-\cos x}}$ dx I = $\frac{1}{2}\int\limits_{0}^{\pi}$ dx = $\frac{1}{2}$ [π - 0] = $\frac{\pi}{2}$ OR I = $\int\limits_{0}^{\pi/2}$ (2 log sin x - log sin 2x) dx I = $\int\limits_{0}^{\pi/2}$ $\left( \log \frac{\sin^2 x}{2\sin x \cos x} . dx \right)$ I = $\int\limits_{0}^{\pi/2}$ log $\left( \frac{\tan x}{2} \right)$ . dx ... (i) Using property $\int\limits_{0}^{a}$ f (x) dx = $\int\limits_{0}^{a}$ f (a - x) dx We get, I = $\int\limits_{0}^{\pi/2}$ log $\left( \frac{\tan(\pi/2 - x)}{2} \right)$ dx ⇒ I = $\int\limits_{0}^{\pi/2}$ log $\left( \frac{\cot x}{2} \right)$ dx ... (ii) Additing (i)&(ii) 2I = $\int\limits_{0}^{\pi/2}$ $\left[ \log\left(\frac{\tan x}{2}\right) + \log\left(\frac{\cot x}{2}\right) \right]$ dx ⇒ 2I = $\int\limits_{0}^{\pi/2}$ log $\left[ \left(\frac{\tan x}{2}\right) \left(\frac{\cot x}{2}\right) \right]$ dx ⇒ I = $\frac{1}{2}\int\limits_{0}^{\pi/2}$ log $\left( \frac{1}{4} \right)$ dx ⇒ I = $\frac{1}{2}$ log $\left( \frac{1}{4} \right) \times \left( \frac{\pi}{2} \right)$ ⇒ I = $\frac{1}{2}$ log $\left( \frac{1}{4} \right)^{1/2}$ × $\left( \frac{\pi}{2} \right)$ ⇒ I = log $\left( \frac{1}{2} \right) \times \left( \frac{\pi}{2} \right)$ ⇒ I = $\frac{\pi}{2} \log \frac{1}{2}$