Question

How do I evaluate `intsec^2(x) tan^2(x) dx`?

Answer 1

The answer is `=(tan^3x)/3+C`

Explanation:

Perform this integral by substitution

Let `u=tanx`, `=>`, `du=sec^2xdx`

The integral is

`I=intsec^2xtan^2xdx`

`=intu^2du`

`=u^3/3`

`=(tan^3x)/3+C`