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: 57



Work Step by Step

Given: $\int^{2}_{1} \dfrac{\cosh (\ln t)}{t} dt$ Plug in: $\ln t= a$ and $da=\dfrac{dt}{t}$ Thus, the limits of integration will also get changed. Then, $\int^{2}_{1} \dfrac{\cosh (\ln t)}{t} dt=\int^{\ln 2}_{0} \cosh a da$ or, $=[ \sinh h a]^{\ln 2}_{0}$ or, $= \sinh (\ln 2) -0$ Since, $\sinh \theta=\dfrac{e^{\theta} -e^{-\theta}}{2}$ Thus, we have, $= \dfrac{e^{\ln 2} -e^{-\ln 2}}{2}$ or, $=\dfrac{3}{4}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.