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: 46

Answer

$\sqrt 3 \ln | \sinh (\dfrac{\theta}{\sqrt 3})| +C$

Work Step by Step

As we are given that $\int \coth (\dfrac{\theta}{\sqrt 3}) d\theta$ Now, plug $\dfrac{\theta}{\sqrt 3}=t$ and $d \theta =\sqrt 3 dt$ This implies: $\int \coth (\dfrac{\theta}{\sqrt 3}) d\theta=\sqrt 3 \int \coth t dt=\sqrt 3 \int \dfrac{\cosh t}{\sinh t} dt$ Now, consider $\sinh t =u$ and $ \cosh t dt =du$ Thus, $\sqrt 3 \int \dfrac{\cosh t}{\sinh t} dt=\sqrt 3 \int (\dfrac{du}{u})= \sqrt 3 [\ln |u|] +C= \sqrt 3 \ln | \sinh (\dfrac{\theta}{\sqrt 3})| +C$
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