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: 53


$\dfrac{3}{32}+\ln 2$

Work Step by Step

Given: $\int^{-\ln 2}_{-\ln 4} 2e^{\theta} \cosh \theta d\theta$ Since, $\cosh \theta=\dfrac{e^{\theta}+e^{-\theta}}{2}$ The given integral can be re-written as: $\int^{-\ln 2}_{-\ln 4} 2e^{\theta} [\dfrac{e^{\theta}+e^{-\theta}}{2}] d\theta$ or, $=\int^{-\ln 2}_{-\ln 4}(e^{2\theta} +1) d\theta$ or, $=[\dfrac{1}{2}e^{2\theta}+\theta]^{-\ln 2}_{-\ln 4}$ or, $=[\dfrac{1}{8}-\ln 2]-[\dfrac{1}{32}- 2 \ln 2]$ or, $=\dfrac{3}{32}+\ln 2$
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