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

Let

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

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

$={\int }_0^{\pi }\left\{\frac{\left(\mathrm{\pi }-x\right)\left(-\tan x\right)}{-\sec x-\tan x}\right\}dx$

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

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

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

$\Rightarrow 2I=\mathrm{\pi }{\int }_0^{\pi }\left\{\frac{\tan x}{\sec x+\tan x}\right\}dx$

$\Rightarrow 2I=\mathrm{\pi }{\int }_0^{\pi }\left\{\frac{\left({\displaystyle \frac{\sin x}{\cos x}}\right)}{\left({\displaystyle \frac{1}{\cos x}}\right)+\left({\displaystyle \frac{\sin x}{\cos x}}\right)}\right\}dx$

$\Rightarrow 2I=\mathrm{\pi }{\int }_0^{\pi }\left\{\frac{\sin x}{1+\sin x}\right\}dx$

 $\Rightarrow 2I=\mathrm{\pi }{\int }_0^{\pi }\left\{\frac{\sin x\left(1-\sin x\right)}{\left(1-{\sin }^{2}x\right)}\right\}dx$

 $\Rightarrow 2I=\mathrm{\pi }{\int }_0^{\pi }\left\{\frac{\sin x}{{\cos }^{2}x}-\frac{{\sin }^{2}x}{{\cos }^{2}x}\right\}dx$

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

 $\Rightarrow 2I=\mathrm{\pi }{\left[\sec x-\tan x+x\right]}_0^{\mathrm{\pi }}$

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

$\Rightarrow I=\frac{\mathrm{\pi }\left(\mathrm{\pi }-2\right)}{2}$.