Answer
$$\frac{{{{\sec }^3}2x}}{6} + C$$
Work Step by Step
$$\eqalign{
& \int {{{\sec }^3}2x} \tan 2xdx \cr
& {\text{split se}}{{\text{c}}^3}2x \cr
& \int {{{\sec }^2}2x\sec 2x} \tan 2xdx \cr
& {\text{substitute }}u = \sec 2x,{\text{ }}du = 2\sec 2x\tan 2xdx \cr
& \int {{{\sec }^2}2x\sec 2x} \tan 2xdx = \int {{u^2}\left( {\frac{1}{2}du} \right)} \cr
& = \frac{1}{2}\int {{u^2}du} \cr
& {\text{find antiderivative }} \cr
& = \frac{{{u^3}}}{6} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{{{{\sec }^3}2x}}{6} + C \cr} $$
