Question

Evaluate :
${\int }_0^{\pi }\frac{xtanx}{secx+tanx}$

Answer 1

$I={\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx$
$I={\int }_0^{\pi }\frac{(\pi -x)\tan x}{\sec x+\tan x}dx$
$2I={\int }_0^{\pi }\frac{\pi \tan x}{\sec x+\tan x}dx$
$2I=\pi {\int }_0^{\pi }\frac{\sin x}{1+\sin x}dx$
$2I=\pi {\int }_0^{\pi }\frac{1+\sin x-1}{1+\sin x}dx$
$2I=\pi [\pi -{\int }_0^{\pi }\frac{1}{1+\sin x}dx]$
$2I={\pi }^2-\pi {\int }_0^{\pi }\frac{1-\sin x}{1-{\sin }^{2}x}dx$
$2I={\pi }^2-\pi {\int }_0^{\pi }({\sec }^{2}x-\sec x\tan x)dx$
$2I={\pi }^2-\pi {\left[\tan x-\sec x\right]}_0^{\pi }$
$\Rightarrow I=\pi (\frac{\pi }{2}-1)$