Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 63


$$\int \frac{dx}{\sqrt{e^{2x}-1}} =\cos^{-1}e^{-x}+c.$$

Work Step by Step

Since $ u=e^{-x}$, then $ du=- e^{-x}dx $ and hence $$\int \frac{dx}{\sqrt{e^{2x}-1}}=-\int \frac{du}{u\sqrt{(1/u^2)-1}} =-\int \frac{du}{ \sqrt{1-u^2}} \\ =\cos^{-1 }u+c=\cos^{-1}e^{-x}+c.$$ (We could also have integrated to inverse sine by excluding the negative sign.)
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