Question

The value of the integral $\int^{\pi}_0 \frac{x \tan \, x}{\sec \, x + \tan \, x} dx$ is equal to

• A. $\pi \left( \frac{\pi}{2} - 1 \right)$
• B. $\frac{\pi}{2} \left( \pi - 1 \right)$
• C. $\pi ( \pi - 1)$
• D. $\frac{\pi}{2} \left( \pi + 1 \right)$

Answer 1

Answer: (A)

$\pi \left( \frac{\pi}{2} - 1 \right)$

Answer 2

Let $I=\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x$
$I=\int_{0}^{\pi} \frac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)} d x$
$\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]$
$I=\int_{0}^{\pi} \frac{(\pi-x) \tan x}{\sec x+\tan x} d x$
$\Rightarrow 2 I=\pi \int_{0}^{\pi} \frac{\tan x}{\sec x+\tan x} d x$
$\Rightarrow 2I =\pi \int_{0}^{\pi} \tan x(\sec x-\tan x) d x$
$\Rightarrow 2 I =\pi \int_{0}^{\pi}\left(\sec x \tan x-\tan ^{2} x\right) d x$
$\Rightarrow 2 I=\pi \int_{0}^{\pi}\left(\sec x \tan x-\sec ^{2} x+1\right) d x$
$\Rightarrow 2I =\pi[\sec x-\tan x+x]_{0}^{\pi}$
$\Rightarrow I =\pi[(\sec \pi-\tan \pi+\pi)-(\sec 0-\tan 0+0)]$
$\Rightarrow I=\pi\left[\frac{\pi}{2}-1\right]$