Answer
$\frac{1}{4}{\tan ^4}\theta + C$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^2}\theta } {\tan ^3}\theta d\theta \cr
& = \int {{{\tan }^3}\theta {{\sec }^2}\theta } d\theta \cr
& {\text{Let }}u = \tan \theta ,{\text{ then }}du = {\sec ^2}\theta d\theta \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& \int {{{\tan }^3}\theta {{\sec }^2}\theta } d\theta = \int {{u^3}} du \cr
& {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr
& \int {{u^3}} du = \frac{1}{4}{u^4} + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}u = \tan \theta \cr
& = \frac{1}{4}{\left( {\tan \theta } \right)^4} + C \cr
& = \frac{1}{4}{\tan ^4}\theta + C \cr} $$
