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: 52


$\dfrac{1}{2}\ln (\dfrac{17}{8})$

Work Step by Step

Given: $\int^{\ln 2}_{0} \tanh 2x dx$ This can be re-written as:$\int^{\ln 2}_{0} \dfrac{\sinh 2x}{\cosh 2x} dx$ Plug in: $\cosh 2x=t$ and $dt=2 \sinh 2x dx$ Thus, the limits of integration will also get changed. Then, $\int^{\ln 2}_{0} \dfrac{\sinh 2x}{\cosh 2x} dx=\dfrac{1}{2}\int^{17/8}_{1} \dfrac{dt}{t} =\dfrac{1}{2}[\ln |t|]^{17/8}_{1}=\dfrac{1}{2}[\ln \dfrac{17}{8}-\ln(1)] $ or, $=\dfrac{1}{2}\ln (\dfrac{17}{8})$
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