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: 33

Answer

$$\ln \left( {1 + \cosh x} \right) + C$$

Work Step by Step

$$\eqalign{ & \int {\frac{{\sinh x}}{{1 + \cosh x}}} dx \cr & {\text{substitute }}u = 1 + \cosh x,{\text{ }}du = \sinh xdx{\text{ }} \cr & \int {\frac{{\sinh x}}{{1 + \cosh x}}} dx = \int {\frac{{du}}{u}} \cr & {\text{find the antiderivative}} \cr & \int {\frac{{du}}{u}} = \ln \left| u \right| + C \cr & {\text{ with}}\,\,\,u = 1 + \cosh x \cr & = \ln \left| {1 + \cosh x} \right| + C \cr & or \cr & = \ln \left( {1 + \cosh x} \right) + C \cr} $$
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