Show ∫ x sin x/(1+cos²x) dx = π²/4

Let I = ∫₀^π x sin x/(1 + cos²x) dx. Using ∫₀^a f(x)dx = ∫₀^a f(a − x)dx:

I = ∫₀^π (π − x) sin x/(1 + cos²x) dx

Add the two forms:

2I = π ∫₀^π sin x/(1 + cos²x) dx

Put t = cos x, dt = −sin x dx. Limits: x = 0 → t = 1; x = π → t = −1.

∫₀^π sin x dx/(1 + cos²x) = ∫_(−1)^1 dt/(1 + t²)
                          = [tan⁻¹ t]_(−1)^1
                          = π/4 + π/4 = π/2

So 2I = π · π/2, giving I = π²/4. Shown.