Question

What is `intsec(2x)tan(2x) dx`?

Answer 1

`1/2 sec(2x) + C`

Explanation:

`int sec(2x)tan(2x) dx`
Let `u = 2x`
`du = 2dx`
`(du/2)=dx`

The integral is rewritten as

`int sec(u)tan(u) (du)/2`
`=1/2 int sec(u)tan(u) du`

`=1/2 sec(u) + C`

Substituting back for `u` we get `1/2 sec(2x) + C`