Evaluate: `int_0^pi(xtanx)/(secx+tanx)\ dx`

Explanation

Given:`int_0^pi(xtanx)/(secx+tanx)\ dx`

`I= int_0^pi(xtanx)/(secx+tanx)\ dx.....(i)`

Now replace `x` by `(pi-x)` we get

`I= int_0^pi((pi-x)tan(pi-x))/(sec(pi-x)+tan(pi-x))dx`

`I= int_0^pi(-(pi-x)tanx)/(-(secx+tanx)) dx`

`I= int_0^pi((pi-x)tanx)/(secx+tanx)dx....(ii)`

Now add (i) and (ii) we get

`2I= int_0^pi(pi tanx)/(secx+tanx)\ dx`

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