Answer
$$\int sech^2(1-2x) dx
=-\frac{1}{2}\tanh (1-2x)+c.$$
Work Step by Step
Let $ u=1-2x $, then $ du=-2dx $ and hence
$$\int sech^2(1-2x) dx =-\frac{1}{2}\int sech^2udu =-\frac{1}{2}\tanh u+c\\
=-\frac{1}{2}\tanh (1-2x)+c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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