University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 418: 42


$5 \cosh(\dfrac{ x}{5})+C$

Work Step by Step

Given: $\int \sinh (\dfrac{x}{5}) dx$ Use the identity: $\int \sinh x dx=\cosh x +C$ Thus, $\int \sinh (\dfrac{x}{5}) dx=\dfrac{ \cosh(\dfrac{ x}{5})}{1/5}+C= 5 \cosh(\dfrac{ x}{5})+C$
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