Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 41


$$\int \tanh x \, sech^2\, x dx =\frac{1}{2}\tanh^2 x+c.$$

Work Step by Step

Let $ u=\tanh x $, then $ du=sech^2 x dx $ and hence $$\int \tanh x \, sech^2\, x dx = \int u du =\frac{1}{2} u^2+c\\ =\frac{1}{2}\tanh^2 x+c.$$
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