Answer
$$ - \ln \left| {\cos \theta } \right| - \frac{{{{\tan }^2}\theta }}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^3}\theta d\theta } \cr
& {\text{split off }}\tan \theta \cr
& = \int {{{\tan }^2}\theta \tan \theta d\theta } \cr
& {\text{pythagorean identity}} \cr
& = \int {\left( {1 - {{\sec }^2}\theta } \right)} \tan \theta d\theta \cr
& {\text{Split }} \cr
& = \int {\tan \theta } d\theta - \int {{{\sec }^2}\theta } \tan \theta d\theta \cr
& {\text{evaluate the integral}} \cr
& = - \ln \left| {\cos \theta } \right| - \frac{{{{\tan }^2}\theta }}{2} + C \cr} $$
