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: 43


$$\tanh (x/2) +C$$

Work Step by Step

Given $$\int \frac{d x}{1+\cosh x}$$ Let $u=\tanh (x/2) $ $$\cosh x=\frac{1+u^{2}}{1-u^{2}}, \quad \sinh x=\frac{2 u}{1-u^{2}}, \quad d x=\frac{2 d u}{1-u^{2}} $$ Then \begin{aligned} \int \frac{d x}{1+\cosh x} &=\int \frac{1-u^{2}}{2}\left(\frac{2 d u}{1-u^{2}}\right) \\ &=\int d u \\ &=u+C\\ &=\tanh (x/2) +C \end{aligned}
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