Answer
$$\ \int \frac{\cosh x}{\sinh^2 x} dx
=- \frac{1}{\sinh u}+c $$
Work Step by Step
Let $ u=\sinh x $, then $ du=\cosh x dx $ and hence
$$\ \int \frac{\cosh x}{\sinh^2 x} dx =\int \frac{du}{u^2} = - \frac{1}{u}+c
=- \frac{1}{\sinh u}+c $$
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: 46
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