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Beta-Gamma: ∫ cos²6θ sin⁴3θ dθ = 7π/192

Short answer

Substitute 3θ = φ to express the integral in terms of sin powers. Apply Beta-Gamma formula to evaluate ∫₀^{π/2} sinⁿφ dφ, yielding 7π/192.

Apply Beta and Gamma function to evaluate ∫₀^{π/6} cos²6θ sin⁴3θ dθ = 7π/192.

Let I = ∫₀^(π/6) cos²6θ sin⁴3θ dθ. Put φ = 3θ, dθ = dφ/3; limits 0 to π/2. Then cos 6θ = cos 2φ = 1 − 2 sin²φ.

I = (1/3) ∫₀^(π/2) (1 − 2 sin²φ)² sin⁴φ dφ
  = (1/3) ∫₀^(π/2) (sin⁴φ − 4 sin⁶φ + 4 sin⁸φ) dφ

Beta-Gamma formula: ∫₀^(π/2) sinⁿφ dφ = Γ((n+1)/2) Γ(1/2) / (2Γ((n+2)/2)).

  • n = 4: Γ(5/2)Γ(1/2)/(2Γ(3)) = (3√π/4)(√π)/(4) = 3π/16
  • n = 6: Γ(7/2)Γ(1/2)/(2Γ(4)) = (15√π/8)(√π)/12 = 5π/32
  • n = 8: Γ(9/2)Γ(1/2)/(2Γ(5)) = (105√π/16)(√π)/48 = 35π/256
I = (1/3) [ 3π/16 − 4(5π/32) + 4(35π/256) ]
  = (1/3) [ 12π/64 − 40π/64 + 35π/64 ]
  = (1/3)(7π/64)

I = 7π/192. Shown.

✓ Answer checked by Claude Opus 5.5 (high effort)

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