#### Answer

$$\tan 2v + C$$

#### Work Step by Step

$$\eqalign{
& \int {2{{\sec }^2}2v} dv \cr
& = 2\int {{{\sec }^2}2v} dv \cr
& {\text{use the formula for indefinite integrals of trigonometric functions}} \cr
& \int {{{\sec }^2}ax} dx = \frac{1}{a}\tan ax + C \cr
& = 2\left( {\frac{1}{2}\tan 2v} \right) + C \cr
& = \tan 2v + C \cr
& {\text{check by differentiation}} \cr
& {\text{ = }}\frac{d}{{dv}}\left( {\tan 2v + C} \right) \cr
& = {\sec ^2}2v\left( 2 \right) \cr
& = 2{\sec ^2}2v \cr} $$