The Student Room Group
Reply 1
First let's lose the 2x. Let u=2x, then du=2dx.
So:
∫ sec2x dx becomes:
(1/2) ∫ secu du

Now there are two ways you can proceed:
1. Manipulating the integral:
(1/2) ∫ secu * (secu+tanu/secu+tanu) du
(1/2) ∫ (sec²u + sectanu)/(tanu + secu) du
Note that the nominator is the derivative of the denominator. So the integral becomes:
(1/2) ln|tanu+secu|
= (1/2) ln|tan(2x)+sec(2x)| + C
If you don't know how the integral became like this, use the substitution v=(tanu+secu), with dv=(sec²u+sectanu)du.
2. Using the substitution t=tan(u/2). There's an example somewhere in the book that shows you how to properly use this substitution. Make sure you learn it -- it's an important substitution.