Evaluate $\int\limits_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} dx$.

$\frac{\pi}{2}$

The correct answer is Option (2) → $\frac{\pi}{2}$

Let $I = \int\limits_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} dx \quad \dots(i)$

$I = \int\limits_{0}^{\pi} \frac{e^{\cos(\pi-x)}}{e^{\cos(\pi-x)} + e^{-\cos(\pi-x)}} dx$

Using $\int\limits_{0}^{a} f(x) dx = \int\limits_{0}^{a} f(a-x) dx$:

$I = \int\limits_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x} + e^{\cos x}} dx \quad \dots(ii)$

On adding eqn. $(i)$ and $(ii)$, we get

$2I = \int\limits_{0}^{\pi} \frac{e^{\cos x} + e^{-\cos x}}{e^{\cos x} + e^{-\cos x}} dx$

$= \int\limits_{0}^{\pi} 1 dx$

$ = [x]_{0}^{\pi} = \pi$

$I = \frac{\pi}{2}$