Calculus: Early Transcendentals (2nd Edition)

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

Chapter 6 - Applications of Integration - 6.10 Hyperbolic Functions - 6.10 Exercises: 31

Answer

$$\frac{1}{2}\sinh 2x + C$$

Work Step by Step

$$\eqalign{ & \int {\cosh 2x} dx \cr & {\text{substitute }}u = 2x,{\text{ }}du = 2dx{\text{ and }}dx = \frac{{du}}{2} \cr & \int {\cosh 2x} dx = \int {\cosh u} \frac{{du}}{2} \cr & = \frac{1}{2}\int {\cosh udu} \cr & {\text{find the antiderivative}} \cr & = \frac{1}{2}\sinh u + C \cr & {\text{ with}}\,\,\,u = 2x \cr & = \frac{1}{2}\sinh 2x + C \cr} $$
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