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: 39


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

Work Step by Step

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