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(x)/n + (n-1)I_(n-2)/n

Similarly for cos^n(x): J_n = cos^(n-1)(x)sin(x)/n + (n-1)J_(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(2x)/4 + C
• J_2 = x/2 + sin(2x)/4 + C

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