Evaluate ∫π0xsinx/(1+cos2x)dx

I=∫π0xsinxdx/(1+cos2x) --(1)
We know that ∫a0f(x)dx=∫a0f(a−x)dx
I=∫π0(π−x)sin(π−x)dx/(1+cos2(π−x))=∫π0(π−x)sinxdx/(1+cos2x)
⇒∫π0πsinxdx/(1+cos2x)−∫π0xsinxdx/(1+cos2x) -- (2)
Adding (1) and (2)
2I=π∫π0sinxdx/(1+cos2x)
Put cosx=t ⇒ sinxdx=−dt. Also when x=0, t=cos0=1 and x=π ⇒ t=cosπ=−1
Therefore, I=∫−11−dt/(1+t2) ⇒ π/2∫1−1dt/(1+t2) = π/2[tan−1t]1−1
I=π/2[tan−1(1)−tan−1(−1)]=π/2[π/4−(−π/4)]
I=(π2)2