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: 39

Answer

$$\tan x - x + C$$

Work Step by Step

$$\eqalign{ & \int {\left( {{{\sec }^2}x - 1} \right)} dx \cr & {\text{sum Rule}} \cr & = \int {{{\sec }^2}x} dx - \int {dx} \cr & {\text{integrate}} \cr & = \tan x - x + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dx}}\left( {\tan x - x + C} \right) \cr & = \frac{d}{{dx}}\left( {\tan x} \right) + \frac{d}{{dx}}\left( { - x} \right) + \frac{d}{{dx}}\left( C \right) \cr & = {\sec ^2}x - 1 + 0 \cr & = {\sec ^2}x - 1 \cr} $$
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