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Beta-Gamma: ∫ cos⁴3θ sin²6θ dθ = 5π/192

Short answer

Let φ=3θ, then sin²6θ = 4 sin²φ cos²φ, giving I = (4/3) ∫₀^{π/2} sin²φ cos⁶φ dφ. Using Beta-Gamma, this integral equals 5π/256, so I = 5π/192.

Use Gamma function to prove that ∫₀^{π/6} cos⁴3θ sin²6θ dθ = 5π/192.

Let I = ∫₀^(π/6) cos⁴3θ sin²6θ dθ. Put φ = 3θ, dθ = dφ/3; limits: θ = 0 → φ = 0, θ = π/6 → φ = π/2.

sin 6θ = sin 2φ = 2 sin φ cos φ, so sin²6θ = 4 sin²φ cos²φ.

I = (1/3) ∫₀^(π/2) cos⁴φ · 4 sin²φ cos²φ dφ
  = (4/3) ∫₀^(π/2) sin²φ cos⁶φ dφ

Use ∫₀^(π/2) sin^p φ cos^q φ dφ = Γ((p+1)/2)Γ((q+1)/2) / (2Γ((p+q+2)/2)) with p = 2, q = 6:

∫₀^(π/2) sin²φ cos⁶φ dφ = Γ(3/2)Γ(7/2) / (2Γ(5)).

Values: Γ(3/2) = √π/2; Γ(7/2) = (5/2)(3/2)(1/2)√π = 15√π/8; Γ(5) = 24.

= (√π/2)(15√π/8) / 48 = (15π/16)/48 = 5π/256.

I = (4/3)(5π/256) = 20π/768

I = 5π/192. Proved.

✓ Answer checked by Claude Opus 5.5 (high effort)

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