Question

Solve : ${\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx$

Answer 1

$I$=${\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}={\int }_0^{\pi }\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x)+\tan (\pi -x)}dx$
${\int }_0^{\pi }\frac{\pi \tan (\pi -x)}{-(\sec x+\tan x)}-\frac{x\tan (\pi -x)}{-(\sec x+\tan x)}$
$I={\int }_0^{\pi }\frac{\pi \tan x}{\sec x+\tan x}dx-\frac{x\tan x}{\sec x\tan x}dx$
$2I={\int }_0^{\pi }\frac{\pi \tan x}{\sec x+\tan x}dx={\int }_0^{\pi }\pi \times \left(\frac{\sin x}{1+\sin x}\right)dx$
$={\int }_0^{\pi }\pi \left(\frac{\sin x(1-\sin x)}{1-{\sin }^{2}x}\right)={\int }_0^{\pi }\pi \frac{\sin x(1-\sin x)}{{\cos }^{2}x}$
$=\pi {\int }_0^{\pi }(\sec x\tan x-{\tan }^{2}x)dx={\int }_0^{\pi }\pi (\sec x\tan x-{\sec }^{2}x+1)dx$
$=({\pi \sec x]}_0^{\pi }-\pi \tan x{]}_0^{\pi }+x{]}_0^{\pi })=(-1-1)-(0-0)+\pi )\pi$
$=(\pi -2)\pi$
$I=\frac{\pi }{2}(\pi -2)$