Exactly...
When you do that, your integral becomes
\(\displaystyle \L \int (\sec t)^2\left[ (\sec t)^2 - 1 \right] (\sec t \tan t) dt\)
and letting \(\displaystyle u = \sec t\) gives you
\(\displaystyle \L \int u^2 \left[ u^2 - 1 \right] du\)
and I'm sure you can integrate that, right?