Evaluate: ∫π04xsinx1+cos2xdx

Let I = ∫π04(π−x)sin(π−x)1+cos2(π−x)dx
We know that ∫f(a−x) = ∫f(a)−∫f(x)
I = ∫π04πsinx1+cos2x−∫π04xsinx1+cos2xdx
or, I = ∫π04πsinx1+cos2x−I
2I = 4π∫π0sinx1+cos2xdx
2I = 4π×2∫π20sinx1+cos2xdx
Applying ∫2a0f(x)dx = 2∫a0f(x)dx if f(2a−x) = f(x)
I = 4π∫π20sinx1+cos2xdx
Put cosx = t, ⇒ −sinxdx = dt
As well for x=0, x=π2 t=1, t=0
Therefore, I = 4π∫01−dt1+t2
I = 4π∫10dt1+t2
∫baf(x)dx = −∫abf(x)dx
I = 4π[tan−1;t]10 = 4π[tan−1;1−tan−1;0] = 4π×π4 = π2