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: 44


$\dfrac{4}{3} \sinh (3x -\ln 2) +C$

Work Step by Step

Given: $\int 4 \cosh ( 3x -\ln 2) dx$ Use the identity: $\int \cosh x dx=\sinh x +C$ Plug in $3x -\ln 2 = a \implies 3 dx=da$ or, $dx=\dfrac{1}{3} da$ Then: $\int 4 \cosh ( 3x -\ln 2) dx=\dfrac{4}{3} \int \cosh a da= \dfrac{4}{3} \sinh a +C$ or, $=\dfrac{4}{3} \sinh (3x -\ln 2) +C$
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