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Evaluate int0^pi (xsinx)/(1+cos^2x)""...

Evaluate `int_0^pi (xsinx)/(1+cos^2x)""dx`

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Given: `int_0^pi (xsinx)/(1+cos^2x)dx`
`I=int_0^pi (xsinx)/(1+cos^2x)dx....(i)`
As we know `int_a^b f(x)dx=int_a^b f(a+b-x)dx`
`I=int_0^pi ((pi-x)sin(pi-x))/(1+cos^2(pi-x))dx`
`I=int_0^pi ((pi-x)sinx)/(1+cos^2x)dx.....(ii)`
Now add (i) and (ii) we get
`2I=pi int_0^pi (sinx)/(1+cos^2x)dx`
Let `cosx=t implies dt=−sinxdx`
...
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