Evaluate ∫₀^(π/2) sin⁶ x cos⁴ x dx using reduction formula.

Evaluate ∫₀^(π/2) sin⁶ x cos⁴ x dx using reduction formula.

• A. 3π/512
• B. 3π/1024
• C. π/256
• D. π/512

Answer: B) 3π/1024

Explanation: I = ∫₀^(π/2) sin⁶ x cos⁴ x dx. Using Beta function: m−1=6 → m=7; n−1=4 → n=5. B(7,5) = Γ(7)Γ(5)/Γ(12) = 6! 4! / 11! = (720×24)/39916800 = 17280/39916800 = 1/2310? Actually ∫₀^(π/2) sin^(2p) x cos^(2q) x dx = Γ(p+1/2)Γ(q+1/2)/(2Γ(p+q+1)). Here 2p=6 → p=3; 2q=4 → q=2. I = Γ(3.5)Γ(2.5)/(2 Γ(6)). Γ(3.5) = (5/2)(3/2)(1/2)√π = 15√π/8. Γ(2.5) = (3/2)(1/2)√π = 3√π/4. Product = (45π/32). Γ(6) = 120. I = (45π/32)/(240) = 45π/(7680) = 3π/512. That gives option 0. m=6, n=4. B(7/2, 5/2) = Γ(7/2)Γ(5/2)/Γ(6). Γ(7/2) = (5/2)(3/2)(1/2)√π = 15√π/8. Γ(5/2) = (3/2)(1/2)√π = 3√π/4. Product = 45π/32. Γ(6) = 120. B = 45π/(3840) = 3π/256. Then I = (1/2) B = 3π/512. So option 0 is 3π/512. But I had earlier computed 3π/512. We check option 1: 3π/1024 is half of that. The correct is 3π/512. So index 0.