Question

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

Answer 1

$I={\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx$

$={\int }_0^{\pi }\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x)+\tan (\pi -x)}.dx$

$={\int }_0^{\pi }\frac{(\pi -x)(-\tan x)}{-(\sec +\tan x)}$

$I={\int }_0^{\pi }\frac{\pi .\tan x}{\sec x+\tan x}dx-{\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx$

$I={\int }_0^{\pi }\frac{\pi \tan x\times \sec x-\tan x}{(\sec x+\tan x)(\sec x-\tan x)}-I$

$I+I={\int }_0^{\pi }\pi \tan x\left(\sec x\tan x\right)dx$ $(\because {\sec }^{2}x-{\tan }^{2}x=I)$

$2I=\pi {\int }_0^{\pi }(\tan x\sec x-{\tan }^{2}x)dx$

$=\pi {\int }_0^{\pi }\tan x.\sec x-({\sec }^{2}x-2)dx$

$2I=\pi {\int }_0^{\pi }(\tan x.\sec x-{\sec }^{2}x+1)dx$

$I=\frac{\pi }{2}[\sec x-\tan x+x{]}_0^{\pi }$

$=\frac{\pi }{2}\left[\right(\sec \pi -\tan \pi +\pi )-(\sec 0-\tan 0+0\left)\right]$

$=\frac{\pi }{2}[-1+\pi -1]$

$=\frac{\pi }{2}[\pi -2]$ Ans.