Beta-Gamma: ∫ sin⁴x cos²x dx = (3π−4)/192

The Beta-Gamma formula applies directly to limits 0 to π/2, so first change the variable so that the limits become 0 and π/2.

Rewrite the integrand. sin⁴x cos²x = sin²x (sin x cos x)² = sin²x · (sin²2x)/4, and sin²x = (1 − cos 2x)/2:

sin⁴x cos²x = (1/8)(1 − cos 2x) sin²2x.

Put u = 2x, dx = du/2. Limits: x = 0 → u = 0; x = π/4 → u = π/2.

I = ∫₀^(π/4) sin⁴x cos²x dx = (1/16) ∫₀^(π/2) (1 − cos u) sin²u du
  = (1/16) [ ∫₀^(π/2) sin²u du − ∫₀^(π/2) sin²u cos u du ]

Use ∫₀^(π/2) sin^p u cos^q u du = Γ((p+1)/2) Γ((q+1)/2) / (2 Γ((p+q+2)/2)).

• ∫ sin²u du (p = 2, q = 0): Γ(3/2)Γ(1/2)/(2Γ(2)) = (√π/2)(√π)/2 = π/4.
• ∫ sin²u cos u du (p = 2, q = 1): Γ(3/2)Γ(1)/(2Γ(5/2)) = (√π/2)/(2 · 3√π/4) = 1/3.

I = (1/16)(π/4 − 1/3) = (1/16)(3π − 4)/12

I = (3π − 4)/192. Proved.