General strategy: Write I_n as a product suitable for integration by parts, then solve for I_n from the resulting equation.

sin^n(x): Write as sin^(n-1)sin. Parts with u = sin^(n-1), dv = sin dx. The cos^2 in the resulting integral converts to 1-sin^2, giving I_n in terms of I_(n-2). Result: I_n = -sin^(n-1)xcosx/n + (n-1)I_(n-2)/n

cos^n(x): Analogous derivation. Result: J_n = cos^(n-1)x*sinx/n + (n-1)J_(n-2)/n

tan^n(x): Write tan^n = tan^(n-2)tan^2 = tan^(n-2)(sec^2-1). The sec^2 part integrates directly. Result: K_n = tan^(n-1)x/(n-1) - K_(n-2)

sec^n(x): Parts with u = sec^(n-2), dv = sec^2 dx. Produces tan^2 = sec^2-1 leading to I_n on both sides. Result: L_n = sec^(n-2)x*tanx/(n-1) + (n-2)L_(n-2)/(n-1)

x^n*e^x: Parts with u = x^n, dv = e^x dx. Reduces power by 1 each time. Result: M_n = x^ne^x - nM_(n-1)

(ln x)^n: Parts with u = (ln x)^n, dv = dx. Result: N_n = x(ln x)^n - n*N_(n-1)

For definite integrals [0, pi/2]: Boundary terms vanish for sin^n and cos^n, giving the pure Wallis recurrence W_n = (n-1)/n * W_(n-2).