Question Detail
Question
By Wallis' formula, evaluate ∫₀^(π/2) cos⁵ x sin² x dx.
Correct Answer
Option A is the correct answer.
Detailed Explanation
Wallis' formula for ∫₀^(π/2) sinᵐ x cosⁿ x dx is:[(m-1)!! (n-1)!!] / [(m+n)!!] * Kwhere K = π/2 if both m and n are even, and K = 1 otherwise.Here, m=2 (for sin x) and n=5 (for cos x). Both are not even, so K=1.Numerator: (2-1)!! (5-1)!! = 1!! * 4!! = 1 * (4 * 2) = 8.Denominator: (2+5)!! = 7!! = 7 * 5 * 3 * 1 = 105.So, the integral evaluates to 8/105.
Hint
Apply Wallis' formula: [(m-1)!! (n-1)!!] / [(m+n)!!]. Remember to check if m and n are both even for the π/2 factor.