Reduction for $sin^n$ x on [0, pi/2]: $I_n$ = $\frac{(n-1}{n}$) * I_(n-2) $I_0$ = pi/2, $I_1$ = 1

Reduction for integral$\frac{0 to pi}{2}$ $sin^m$ x $cos^n$ x dx: = [(m-1)(m-3)...(n-1)(n-3)...] / [(m+n)(m+n-2)...] * K where K = pi/2 if both m,n are even, K = 1 otherwise.

integral(0 to pi) $sin^n$ x dx = 2$I_n$* (for all n, since $sin^n$ x is symmetric about pi/2 via King's Rule)

integral$\frac{0 to pi}{2}$ $tan^n$ x dx = integral$\frac{0 to pi}{2}$ $cot^n$ x dx (by King's Rule with pi/2)