Question

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

Answer 1

$LetI={\int }_0^{\pi }\frac{x\tan x}{\sec x+\tan x}dx...\left(i\right)I={\int }_0^{\pi }\frac{(\pi -x)\tan (\pi -x)}{\sec (\pi -x)+\tan (\pi -x)}[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx]I={\int }_0^{\pi }\frac{(\pi -x)\tan x}{\sec x+\tan x}dx$
By adding equations $\left(i\right)$ and $\left(ii\right)$, we obtain
$2I=\pi {\int }_0^{\pi }\frac{\tan x}{\sec x+\tan x}dx$
Multiplying and dividing by $(\sec x-\tan x),$ we obtain
$2I=\pi {\int }_0^{\pi }\frac{\tan x(\sec x-\tan x)}{{\sec }^{2}x-{\tan }^{2}x}=\pi {\int }_0^{\pi }(\sec x\tan x-{\tan }^{2}x)dx=\pi {\int }_0^{\pi }\sec x\tan xdx-\pi {\int }_0^{\pi }{\sec }^{2}xdx+{\int }_0^{\pi }dx=\pi {\left[\sec x\right]}_0^{\pi }-\pi {\left[\tan x\right]}_0^{\pi }+\pi {\left[x\right]}_0^{\pi }=\pi (-1-1)-0+\pi (\pi -0)=\pi (\pi -2)\Rightarrow 2I=\pi (\pi -2)\Rightarrow I=\frac{\pi }{2}(\pi -2)$