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/[(n-1)cos^(n-1)x] - [(m-n+2)/(n-1)] int dx sin^m x/ cos^(n-2) 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.