Question

How do you integrate `secx`?

Answer 1

The standard trick is to multiply by `(secx+tanx)/(secx+tanx)`, then use substitution.

Explanation:

`int sec x dx = int secx (secx+tanx)/(secx+tanx) dx`

`= int (sec^2x+secxtanx)/(secx+tanx) dx`

With `u = secx+tanx`, we have `int 1/u du`, so

`int sec x dx = ln abs(sec x + tan x) +C`