Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.3 Trigonometric Integrals - 7.3 Exercises: 20

Answer

$$\cos \theta + \frac{1}{{\cos \theta }} + C$$

Work Step by Step

$$\eqalign{ & \int {{{\sin }^3}\theta {{\cos }^{ - 2}}\theta d\theta } \cr & {\text{split off }}\sin \theta \cr & = \int {{{\sin }^2}\theta {{\cos }^{ - 2}}\theta } \sin \theta d\theta \cr & {\text{pythagorean identity}} \cr & = \int {\left( {1 - {{\cos }^2}\theta } \right){{\cos }^{ - 2}}\theta } \sin \theta d\theta \cr & u = \cos \theta ,{\text{ }}du = - \sin \theta d\theta \cr & = - \int {\left( {1 - {u^2}} \right){u^{ - 2}}} d\theta \cr & = \int {\left( {{u^0} - {u^{ - 2}}} \right)} du \cr & = \int {\left( {1 - {u^{ - 2}}} \right)} du \cr & {\text{evaluate the integral}} \cr & = u - \frac{{{u^{ - 1}}}}{{ - 1}} + C \cr & = u + \frac{1}{u} + C \cr & {\text{replace }}u = \cos \theta \cr & = \cos \theta + \frac{1}{{\cos \theta }} + C \cr} $$
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