Question

The value of the definite integral $\underset{0}{\overset{\pi }{\int }}\frac{\pi \tan x}{\sec x+\tan x}dx$ is equal to

• A. $\pi (1-\pi )$
• B. $\pi (\pi -2)$
• C. $\pi (2-\pi )$
• D. $\pi (\pi -1)$

Answer 1

The correct option is B $\pi (\pi -2)$

$\frac{\tan x}{\sec x+\tan x}=\tan x(\sec x-\tan x)$
Let $I=\underset{0}{\overset{\pi }{\int }}\frac{\pi \tan x}{\sec x+\tan x}dx$
$\Rightarrow I=\underset{0}{\overset{\pi }{\int }}\pi (\sec x\tan x-{\tan }^{2}x)dx$
$=\underset{0}{\overset{\pi }{\int }}\pi (\sec x\tan x+1-{\sec }^{2}x)dx$
$=\pi [\sec x+x-\tan x{]}_0^{\pi }$
$=\pi (\pi -2)$