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: 42

Answer

$$\sec \theta - \tan \theta + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{\sin \theta - 1}}{{{{\cos }^2}\theta }}} d\theta \cr & {\text{split the numerator}} \cr & = \int {\left( {\frac{{\sin \theta }}{{{{\cos }^2}\theta }} - \frac{1}{{{{\cos }^2}\theta }}} \right)} d\theta \cr & = \int {\left( {\frac{1}{{\cos \theta }}\frac{{\sin \theta }}{{\cos \theta }} - \frac{1}{{{{\cos }^2}\theta }}} \right)} d\theta \cr & {\text{use trigonometric identities}} \cr & = \int {\left( {\sec \theta \tan \theta - {{\sec }^2}\theta } \right)} d\theta \cr & {\text{sum rule}} \cr & = \int {\sec \theta \tan \theta } d\theta - \int {{{\sec }^2}\theta } d\theta \cr & {\text{use the formula for indefinite integrals of trigonometric functions}} \cr & = \sec \theta - \tan \theta + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{d\theta }}\left( {\sec \theta - \tan \theta } \right) \cr & {\text{ = }}\frac{d}{{d\theta }}\left( {\sec \theta } \right) - \frac{d}{{d\theta }}\left( {\tan \theta } \right) + \frac{d}{{d\theta }}\left( C \right) \cr & = \sec \theta \tan \theta - {\sec ^2}\theta \cr} $$
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