Answer
$$\int \cosh 3x dx=\frac{1}{3}\sinh 3x+c $$
Work Step by Step
Let $ u=3x $, then $ du=3dx $ and hence
$$\int \cosh 3x dx=\frac{1}{3}\int \cosh udu =\frac{1}{3}\sinh u+c
\\=\frac{1}{3}\sinh 3x+c.$$
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: 35
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