Evaluate $\int\limits_{0}^{\pi} \frac{x}{1 + \sin x} \, dx$

$\pi$

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

Let $I = \int\limits_{0}^{\pi} \frac{x}{1 + \sin x} \, dx \quad \dots(i)$

and $I = \int\limits_{0}^{\pi} \frac{\pi - x}{1 + \sin(\pi - x)} \, dx = \int\limits_{0}^{\pi} \frac{\pi - x}{1 + \sin x} \, dx \quad \dots(ii)$

[Applying property $\int\limits_{0}^{a} f(x) \, dx = \int\limits_{0}^{a} f(a - x) \, dx$]

On adding Eqs. (i) and (ii), we get

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

$= \pi \int\limits_{0}^{\pi} \frac{(1 - \sin x) \, dx}{(1 + \sin x)(1 - \sin x)}$

$= \pi \int\limits_{0}^{\pi} \frac{(1 - \sin x) \, dx}{\cos^2 x}$

$= \pi \int\limits_{0}^{\pi} (\sec^2 x - \tan x \cdot \sec x) \, dx$

$= \pi \int\limits_{0}^{\pi} \sec^2 x \, dx - \pi \int\limits_{0}^{\pi} \sec x \cdot \tan x \, dx$

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

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

$= \pi [\tan \pi - \sec \pi - \tan 0 + \sec 0]$

$\Rightarrow 2I = \pi [0 - (-1) - 0 + 1]$

$2I = 2\pi$

$∴I = \pi$