Answer
$$\frac{{{{\tan }^2}\theta }}{2} + \ln \left| {\cos \theta } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^3}\theta } d\theta \cr
& {\text{split integrand}} \cr
& = \int {\tan \theta {{\tan }^2}\theta } d\theta \cr
& {\text{use }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\tan \theta \left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& = \int {\left( {\tan \theta {{\sec }^2}\theta - \tan \theta } \right)} d\theta \cr
& = \int {\tan \theta {{\sec }^2}\theta } d\theta - \int {\tan \theta } d\theta \cr
& {\text{find antiderivatives}} \cr
& = \frac{{{{\tan }^2}\theta }}{2} - \left( { - \ln \left| {\cos \theta } \right|} \right) + C \cr
& = \frac{{{{\tan }^2}\theta }}{2} + \ln \left| {\cos \theta } \right| + C \cr} $$
