Derivation for $I_n$ = integral $sin^n$(x) dx: Write $sin^n$(x) = sin^(n-1)(x) * sin(x). Use parts: u = sin^(n-1)(x), dv = sin(x)dx. Result: $nI_n$ = -sin^(n-1)(x)cos(x) + (n-1)I_(n-2)

So: $I_n$ = -sin^(n-1)(x)cos$\frac{x}{n}$ + (n-1)I_$\frac{n-2}{n}$

Similarly for $cos^n$(x): $J_n$ = cos^(n-1)(x)sin$\frac{x}{n}$ + (n-1)J_$\frac{n-2}{n}$

Base cases:

• $I_0$ = $J_0$ = x + C
• $I_1$ = -cos(x) + C
• $J_1$ = sin(x) + C
• $I_2$ = x/2 - sin$\frac{2x}{4}$ + C
• $J_2$ = x/2 + sin$\frac{2x}{4}$ + C

For definite integrals (Wallis): The boundary terms vanish at 0 and pi/2: integral$\frac{0 to pi}{2}$ $sin^n$(x) dx = $\frac{n-1}{n}$ * $\frac{n-3}{(n-2)}$ * ... * {pi/2 if n even, 1 if n odd}