$\displaystyle \int \frac {1-tanx^2}{secx^2} ~dx$

I figure you have to break it up first:

$\displaystyle \int cosx^2 ~dx - \int \frac {tanx^2}{secx^2} ~dx$

But from here, I am somewhat stuck. Would $\displaystyle \int \frac {tanx^2}{secx^2} ~dx$ turn into $\displaystyle \int sinx^2 ~dx$?

If so, then I guess the next part of the question is, how do you take the anti derivative of $\displaystyle cosx^2$?