Answer
$$ = \frac{{{{\sec }^3}t}}{3} + C$$
Work Step by Step
$$\eqalign{
& \int {\tan t{{\sec }^3}t} dt \cr
& {\text{split exponent se}}{{\text{c}}^3}t \cr
& = \int {\tan t{{\sec }^2}t} \sec tdt \cr
& = \int {{{\sec }^2}t} \sec t\tan tdt \cr
& {\text{substitute }}u = \sec t,{\text{ }}du = \sec t\tan tdt \cr
& = \int {{u^2}} du \cr
& {\text{find the antiderivative by the power rule}} \cr
& = \frac{{{u^3}}}{3} + C \cr
& {\text{write in terms of }}t,{\text{ replace }}u = \sec t \cr
& = \frac{{{{\sec }^3}t}}{3} + C \cr} $$
