Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions - Exercises - Page 415: 7

Answer

$$\frac{1}{2} \tanh(x)+C$$

Work Step by Step

Given $$\int \tanh (x) \operatorname{sech}^2(x) d x$$ Let $$ u=\tanh x \ \ \ \to du = \operatorname{sech}^2(x) dx$$ Then \begin{aligned} \int \tanh (x) \operatorname{sech}^2(x) d x &=\int ud u \\ &=\frac{1}{2} u^2+C\\ &=\frac{1}{2} \tanh(x)+C \end{aligned}
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