Beta-Gamma: ∫ x⁵√(2ax−x²) dx

Formulas

Beta function B(m, n) = 2∫₀^(π/2) sin^(2m−1)θ cos^(2n−1)θ dθ = Γ(m)Γ(n)/Γ(m+n). Γ(n+1) = nΓ(n), Γ(1/2) = √π.

Substitution

I = ∫₀^(2a) x⁵ √(2ax − x²) dx. Put x = 2a sin²θ, so dx = 4a sin θ cos θ dθ. Limits: x = 0 → θ = 0; x = 2a → θ = π/2.

2ax − x² = x(2a − x) = 2a sin²θ · 2a cos²θ = 4a² sin²θ cos²θ, so the square root is 2a sin θ cos θ.

I = ∫ (2a)⁵ sin¹⁰θ · 2a sinθ cosθ · 4a sinθ cosθ dθ
  = 256 a⁷ ∫₀^(π/2) sin¹²θ cos²θ dθ

Using Beta function

2m − 1 = 12 gives m = 13/2; 2n − 1 = 2 gives n = 3/2.

∫₀^(π/2) sin¹²θ cos²θ dθ = (1/2) B(13/2, 3/2) = Γ(13/2)Γ(3/2) / (2Γ(8))

Values:

• Γ(13/2) = (11/2)(9/2)(7/2)(5/2)(3/2)(1/2)√π = (10395/64)√π
• Γ(3/2) = √π/2
• Γ(8) = 7! = 5040

I = 256 a⁷ · Γ(13/2)Γ(3/2) / (2Γ(8))
  = 128 a⁷ · (10395/64)√π · (√π/2) / 5040
  = 128 a⁷ · 10395 π / (128 · 5040)
  = (10395/5040) π a⁷
  = 33π a⁷/16

Answer: ∫₀^(2a) x⁵√(2ax − x²) dx = 33πa⁷/16.