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: 51

Answer

$\ln (\dfrac{5}{2})$

Work Step by Step

As we are given that $\int^{\ln 4}_{\ln 2} coth x dx$ Re-write:$\int^{\ln 4}_{\ln 2} \dfrac{\cosh x}{\sinh x} dx$ Now, consider $\sinh x=t \implies dt=\cosh x dx$ Thus, $\int^{\ln 4}_{\ln 2} \dfrac{\cosh x}{\sinh x} dx=\int^{15/8}_{3/ 4} (\dfrac{dt}{t}) =[\ln |t|]^{15/8}_{3/ 4}=\ln (\dfrac{15}{8})-\ln (\dfrac{3}{4})=\ln (\dfrac{15}{8} \cdot \dfrac{4}{3})$ Hence, $\int^{\ln 4}_{\ln 2} coth x dx=\ln (\dfrac{5}{2})$
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