Question Detail
Question
Using the Wallis formula, the integral of ∫₀^(π/2) sin⁶x dx is...
Correct Answer
Option A is the correct answer.
Detailed Explanation
The Wallis formula for ∫₀^(π/2) sinⁿx dx (or cosⁿx dx) when n is an even integer is:∫₀^(π/2) sinⁿx dx = [(n-1)/n] * [(n-3)/(n-2)] * ... * [1/2] * (π/2)For n=6:∫₀^(π/2) sin⁶x dx = (6-1)/6 * (6-3)/(6-2) * (6-5)/(6-4) * (π/2)= (5/6) * (3/4) * (1/2) * (π/2)= (15/48) * (π/2)= (5/16) * (π/2)= 5π/32
Hint
For even powers of n, the Wallis integral for sinⁿx or cosⁿx from 0 to π/2 is (n-1)/n * (n-3)/(n-2) * ... * 1/2 * (π/2).