Question

$I={\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx=\frac{\pi }{C}(\pi -2).$
What is C?

Answer 1

$I={\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx$ ...(1)
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$\therefore I={\int }_0^{\pi }\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x)+\tan (\pi -x)}dx={\int }_0^{\pi }\frac{(\pi -x)\tan x}{\sec x+\tan x}$ ...(2)
Adding (1) and (2)
$2I=\pi {\int }_0^{\pi }\frac{\tan xdx}{\sec x+\tan x}=\pi {\int }_0^{\pi }\frac{\tan x(\sec x-\tan x)}{1}dx$
$\therefore I=\frac{\pi }{2}{\int }_0^{\pi }[\sec x\tan x-({\sec }^{2}x-1)]dx=\frac{\pi }{2}{[x-\tan x+\sec x]}_0^{\pi }$
$=\frac{\pi }{2}[(\pi -0-1)-(0-0+1)]=\frac{\pi }{2}(\pi -2)$