Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.7 - Hyperbolic Functions - Exercises 7.7 - Page 430: 55

Answer

$e-e^{-1}$

Work Step by Step

As we are given that $\int^{\pi/4}_{-\pi/4} \cosh(\tan \theta)sec^2 \theta d\theta$ Now, consider $\tan \theta=t$ and $dt= \sec^2 \theta d\theta$ Thus, $\int^{\pi/4}_{-\pi/4} \cosh(\tan \theta)sec^2 \theta d\theta=\int^{1}_{-1} \cosh t dt=[\sin h t]^{1}_{-1}$ and, $=2 \sinh (1)$ As we know $\sinh \theta=\dfrac{e^{\theta} -e^{-\theta}}{2}$ Then $= 2[\dfrac{e^{1} -e^{-1}}{2}]$ Hence, $\int^{\pi/4}_{-\pi/4} \cosh(\tan \theta)sec^2 \theta d\theta=e-e^{-1}$
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