Question

How do you integrate `y= sin^2x/cosx`?

Answer 1

`ln (sec x + tan x) - sin x + C`

Explanation:

`y=(1-cos^2x)/cos x= secx - cos x`

`= (ln( sec x + tan x ))'-(sin x)'`.

So, `int y dx=ln( sec x + tan x) -sin x+C`.