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: 43

Answer

$12 \sinh (\dfrac{x}{2} -\ln 3 )+ C$

Work Step by Step

Given: $\int 6 \cosh (\dfrac{x}{2} -\ln 3) dx$ Use the identity: $\int \cosh x dx=\sinh x +C$ Plug in $\dfrac{x}{2} -\ln 3 = a \implies \dfrac{dx}{2}=da$ or, $dx=2 da$ Then $\int 6 \cosh (\dfrac{x}{2} -\ln 3) dx=\int 6 \cosh a (2 da)= 12 \sinh a +C $ Thus, $\int 6 \cosh (\dfrac{x}{2} -\ln 3) dx=12 \sinh a +C =12 \sinh (\dfrac{x}{2} -\ln 3 )+ C$
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