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: 56



Work Step by Step

Given: $\int^{\pi/2}_{0} 2\sinh(\sin \theta) \cos \theta d\theta$ Plug in: $\sin \theta=t$ and $dt= \cos \theta d\theta$ Thus, the limits of integration will also get changed. Then, $\int^{\pi/2}_{0} 2\sinh(\sin \theta) \cos \theta d\theta=\int^{1}_{0} 2 \sinh t dt$ or, $=[2 \cosh h t]^{1}_{0}$ or, $=2 \cosh (1) -2$ Since, $\cosh \theta=\dfrac{e^{\theta} + e^{-\theta}}{2}$ Thus, we have, $= 2[\dfrac{e^{1} + e^{-1}}{2}]-2$ or, $=e+e^{-1}-2$
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