For $K_n$ = integral $tan^n$(x) dx: Write $tan^n$(x) = tan^(n-2)(x) * $tan^2$(x) = tan^(n-2)(x)($sec^2$(x)-1). $K_n$ = integral tan^(n-2)(x)$sec^2$(x)dx - K_(n-2) = tan^(n-1)$\frac{x}{(n-1)}$ - K_(n-2)

For $L_n$ = integral $sec^n$(x) dx: Use parts: u = sec^(n-2)(x), dv = $sec^2$(x)dx. Result: $L_n$ = sec^(n-2)(x)tan$\frac{x}{(n-1)}$ + (n-2)L_$\frac{n-2}{(n-1)}$

Base cases:

• $K_0$ = x, $K_1$ = -ln|cos x| = ln|sec x|
• $K_2$ = tan(x) - x
• $L_0$ = x, $L_1$ = ln|sec x + tan x|
• $L_2$ = tan(x)
• $L_3$ = (sec(x)tan(x) + ln|sec x + tan x|)/2

Key observation: tan and sec reduction formulas reduce by 2, like sin and cos.