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: $W_n$ = $\frac{n-1}{n}$ * W_(n-2) (pure recurrence without boundary terms).

Computing $W_6$ = integral$\frac{0 to pi}{2}$ $sin^6$(x) dx: $W_6$ = \frac{5}{6}$$W_4 = \frac{5}{6}$$\frac{3}{4}$$W_2 = \frac{5}{6}$$\frac{3}{4}$$\frac{1}{2}$$\frac{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])