Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 4 - Applications of the Derivative - 4.9 Antiderivatives - 4.9 Exercises - Page 238: 40

Answer

$$\tan 2v + C$$

Work Step by Step

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