- 2068 Chaitra
Gamma function: ∫ x⁶√(1−x²) dx
Short answer
Substitute x² = t to obtain a Beta integral (1/2)B(7/2, 3/2) = (1/2)Γ(7/2)Γ(3/2)/Γ(5), which evaluates to 5π/256.
Use Gamma function to evaluate ∫₀¹ x⁶ √(1−x²) dx.
Let I = ∫₀¹ x⁶ √(1 − x²) dx. Put x² = t, so x = t^(1/2), dx = (1/2) t^(−1/2) dt. Limits stay 0 to 1.
I = ∫₀¹ t³ (1 − t)^(1/2) · (1/2) t^(−1/2) dt
= (1/2) ∫₀¹ t^(5/2) (1 − t)^(1/2) dt
= (1/2) B(7/2, 3/2) [m − 1 = 5/2, n − 1 = 1/2]
= (1/2) Γ(7/2) Γ(3/2) / Γ(5)
Values: Γ(7/2) = 15√π/8, Γ(3/2) = √π/2, Γ(5) = 24.
I = (1/2) · (15√π/8)(√π/2) / 24 = (1/2)(15π/16)/24 = 15π/768
I = 5π/256.
✓ Answer checked by Claude Opus 5.5 (high effort)
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