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This is a Wallis' integral. For an integral of the form ∫₀^(π/2) sinⁿ x dx (or cosⁿ x dx), when n is an even integer, the formula is: I_n = [(n-1)!! / n!!] * (π/2). In this case, n = 6 (even). So, I₆ = [ (6-1)!! / 6!! ] * (π/2) = [ 5!! / 6!! ] * (π/2). Calculate the double factorials: 5!! = 5 * 3 * 1 = 15. 6!! = 6 * 4 * 2 = 48. Substitute these values: I₆ = [15 / 48] * (π/2) = (5/16) * (π/2) = 5π/32.