Evaluate $\int\limits_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} dx.$

$\frac{\pi}{2}(\pi - 2)$

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

Let $I = \int\limits_{0}^{\pi} \frac{x \tan x}{\sec x + \tan x} dx \dots(i)$

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

$\Rightarrow I = \int\limits_{0}^{\pi} \frac{(\pi-x) \tan(\pi-x)}{\sec(\pi-x) + \tan(\pi-x)} dx$

$\Rightarrow I = \int\limits_{0}^{\pi} \frac{(\pi-x)(-\tan x)}{-\sec x - \tan x} dx$

$\Rightarrow I = \int\limits_{0}^{\pi} \frac{(\pi-x) \tan x}{\sec x + \tan x} dx \dots(ii)$

(i) + (ii) $\Rightarrow 2I = \int\limits_{0}^{\pi} \frac{(x + \pi - x) \tan x}{\sec x + \tan x} dx$

$\Rightarrow I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \sin x} dx$

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

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

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

$= \frac{\pi}{2} \int\limits_{0}^{\pi} [\sec x \tan x - \sec^2 x + 1] dx$

$= \frac{\pi}{2} [\sec x - \tan x + x]_{0}^{\pi}$

$= \frac{\pi}{2} [(-1 - 0 + \pi) - (1 - 0 + 0)]$

$∴I = \frac{\pi}{2} [\pi - 2]$