Calculus
posted by Bill .
find the integral of x^2  sec3xtan3x

well spit it into 2 integrals
integral x^2 dx = x^3/3
sec^3 x tan x^3 dx
= (1/cos^3)(sin^3/cos^3) dx
=sin^3/cos^6 dx
I can give you a recursion formula for this
int dx sin ^m x/cos^n = sin^(m+1) x/[(n1)cos^(n1)x]  [(mn+2)/(n1)] int dx sin^m x/ cos^(n2) x
do that until you get to sin^3 x
now int dx sin^3 x is cos x + (1/3) cos^3 x 
or if you really meant
sec(3x)tan(3x)
the derivative of secx
= (secx)tanx
so the integral of sec3xtan3x = (1/3)sec3x
so the integral of
x^2  sec3xtan3x
is
(1/3)x^3  (1/3)sec3x + C
You have to be careful how you type these, as you can see they can be interpreted in different ways unless you use brackets to clearly state what you want.