Answer
$\ln \left| {\tan \theta } \right| + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\sec }^2}\theta }}{{\tan \theta }}} d\theta \cr
& {\text{Let }}u = \tan \theta ,{\text{ then }}du = {\sec ^2}\theta d\theta ,{\text{ then substituting}} \cr
& \int {\frac{{{{\sec }^2}\theta }}{{\tan \theta }}} d\theta = \int {\frac{{du}}{u}} \cr
& {\text{Integrate }} \cr
& = \ln \left| u \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}u = \tan \theta \cr
& = \ln \left| {\tan \theta } \right| + C \cr} $$