You might be making this integration harder than it is.

INT (2 - 3sin(2x))/cos(2x) dx

2*INT sec(2x) dx - 3*INT tan(2x) dx

We are now integrating sec(2x) and tan(2x) (I broke up the fraction and simplified the terms). For each, we will need a u substitution:

Let u = 2x <--> du = 2 dx --> dx = 1/2 du

2(1/2)*INT secu du - 3(1/2)*INT tanu du

There are two basic formulas you can plug these integrations into. If you need them further explained, I can.

INT secu du = ln|secu + tanu|

INT tanu du = -ln|cosu|

Therefore, we get:

ln|sec(2x) + tan(2x)| + 3/2*ln|cos(2x)| + C

As I'm sure you noticed, this does not match the answer in the back of the book, but with a little trig and log manipulation, it will.

ln|sec(2x) + tan(2x)| = ln|1/cos(2x) + sin(2x)/cos(2x)| = ln|(1 + sin(2x))/cos(2x)| = ln|1 + sin(2x)| - ln|cos(2x)|

Therefore, we get:

ln|sec(2x) + tan(2x)| + 3/2*ln|cos(2x)| + C = ln|1 + sin(2x)| - ln|cos(2x)| + 3/2*ln|cos(2x)| + C =ln|1 + sin(2x)| + 1/2*ln|cos(2x)| + C