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: 32

Answer

$$\frac{{{{\left( {\tanh x} \right)}^2}}}{2} + C$$

Work Step by Step

$$\eqalign{ & \int {{{\operatorname{sech} }^2}x\tanh x} dx \cr & {\text{substitute }}u = \tanh x,{\text{ }}du = {\operatorname{sech} ^2}xdx \cr & \int {{{\operatorname{sech} }^2}x\tanh x} dx = \int u du \cr & {\text{find the antiderivative by the power rule}} \cr & = \frac{{{u^2}}}{2} + C \cr & {\text{replacing}}\,\,\,u = \tanh x \cr & = \frac{{{{\left( {\tanh x} \right)}^2}}}{2} + C \cr} $$
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