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: 40

Answer

$$\int \tanh 3x \, sech\, 3x dx \ =-\frac{1}{3}sech \, 3x+c.$$

Work Step by Step

Let $ u=3x $, then $ du=3dx $ and hence $$\int \tanh 3x \, sech\, 3x dx =\frac{1}{3}\int \tanh u \, sech\,u dxdu =-\frac{1}{3}sech \, u+c\\ =-\frac{1}{3}sech \, 3x+c.$$
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