Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 42


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

Work Step by Step

As we are given that $\int \sinh (\dfrac{x}{5}) dx$ Since, $\int \sinh x dx=\cosh x +C$ Therefore, $\int \sinh (\dfrac{x}{5}) dx=\dfrac{ \cosh(\dfrac{ x}{5})}{\dfrac{1}{5}}+C$ or, $= 5 \cosh(\dfrac{ x}{5})+C$
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