Key observation: In definite integrals from 0 to pi/2, the boundary terms in reduction formulas vanish:

• sin^(n-1)(x)cos(x) evaluated at 0 and pi/2 gives 0 - 0 = 0.
• cos^(n-1)(x)sin(x) evaluated similarly gives 0.

So for definite integrals: W_n = (n-1)/n * W_(n-2) (pure recurrence without boundary terms).

Computing W_6 = integral(0 to pi/2) sin^6(x) dx: W_6 = (5/6)W_4 = (5/6)(3/4)W_2 = (5/6)(3/4)(1/2)(pi/2) = 5pi/32.

integral(0 to pi) sin^n(x) dx = 2*W_n for all n (since sin^n is symmetric about pi/2).

integral(0 to 2pi) sin^n(x) dx:

• n even: 4*W_n (positive in all quadrants when n even)
• n odd: 0 (cancellation between [0,pi] and [pi,2pi])