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: 55



Work Step by Step

Given: $\int^{\pi/4}_{-\pi/4} \cosh(\tan \theta)sec^2 \theta d\theta$ Plug in: $\tan \theta=t$ and $dt= \sec^2 \theta d\theta$ Thus, the limits of integration will also get changed. Then, $\int^{\pi/4}_{-\pi/4} \cosh(\tan \theta)sec^2 \theta d\theta=\int^{1}_{-1} \cosh t dt$ or, $=[\sin h t]^{1}_{-1}$ or, $=2 \sinh (1)$ Since, $\sinh \theta=\dfrac{e^{\theta} -e^{-\theta}}{2}$ Thus, we have, $= 2[\dfrac{e^{1} -e^{-1}}{2}]$ or, $=e-e^{-1}$
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