Show ∫ x sin x cos x/(cos⁴x+sin⁴x) dx = π²/16

Let I = ∫₀^(π/2) x sin x cos x/(cos⁴x + sin⁴x) dx. Replace x by π/2 − x (the denominator and sin x cos x are unchanged):

I = ∫₀^(π/2) (π/2 − x) sin x cos x/(cos⁴x + sin⁴x) dx

Add: 2I = (π/2) ∫₀^(π/2) sin x cos x/(sin⁴x + cos⁴x) dx.

Evaluate K = ∫₀^(π/2) sin x cos x/(sin⁴x + cos⁴x) dx. Put t = sin²x, dt = 2 sin x cos x dx. Then cos²x = 1 − t, sin⁴x + cos⁴x = t² + (1 − t)² = 2t² − 2t + 1. Limits 0 to 1.

K = (1/2) ∫₀¹ dt/(2t² − 2t + 1)
  = (1/4) ∫₀¹ dt/((t − 1/2)² + 1/4)
  = (1/4) · 2 [tan⁻¹(2t − 1)]₀¹
  = (1/2)(π/4 + π/4)
  = π/4

So 2I = (π/2)(π/4) = π²/8.

I = π²/16. Shown.