Answer
$$\frac{1}{2}sec2x + C$$
Work Step by Step
$$\eqalign{
& \int {\sec 2x\tan 2x} dx \cr
& {\text{use the formula for indefinite integrals of trigonometric functions}} \cr
& \int {\sec ax} \tan axdx = \frac{1}{a}secax + C \cr
& {\text{letting }}a = 2,{\text{ we have}} \cr
& = \frac{1}{2}sec2x + C \cr} $$
