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_$\frac{n-2}{n}$

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

$tan^n$(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)$\frac{x}{n-1}$ - 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*$\frac{tanx}{n-1}$ + (n-2)L_$\frac{n-2}{(n-1)}$

$x^n$*$e^x$: Parts with u = $x^n$, dv = $e^x$ dx. Reduces power by 1 each time. Result: $M_n$ = $x^n$$e^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 $W_n$ = $\frac{n-1}{n}$ * W_(n-2).