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: 46


$$\sqrt 3 \ln | \sinh (\dfrac{\theta}{\sqrt 3})| +C= \sqrt 3 \ln |e^{\theta/\sqrt{3}}-e^{-\theta/\sqrt{3}}| +C$$

Work Step by Step

Given: $\int \coth (\dfrac{\theta}{\sqrt 3}) d\theta$ Plug in $\dfrac{\theta}{\sqrt 3}=t$ and $d \theta =\sqrt 3 dt$ Thus, $\int \coth (\dfrac{\theta}{\sqrt 3}) d\theta=\sqrt 3 \int \coth t dt=\sqrt 3 \int \dfrac{\cosh t}{\sinh t} dt$ Now, plug in $\sinh t =u \implies \cosh t dt =du$ Thus, we have: $=\sqrt 3 \int \dfrac{du}{u}$ or, $= \sqrt 3 \ln |u| +C$ $= \sqrt 3 \ln | \sinh (\dfrac{\theta}{\sqrt 3})| +C$ $= \sqrt 3 \ln | \frac{e^{\theta/\sqrt{3}}-e^{-\theta/\sqrt{3}}}{2}| +C$ $= \sqrt 3 \ln |e^{\theta/\sqrt{3}}-e^{-\theta/\sqrt{3}}| +C$
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