Evaluate $\int\limits_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} dx$.
Answer & explanation
Correct answer: option 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$
$I = \frac{\pi}{2}$