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: 48


$\coth (5-x) +C$

Work Step by Step

Given: $\int csch^2(5-x) dx$ This can be re-written as: $\int csch^2(5-x) dx=\int csch^2(x-5) dx$ Plug in $x-5=t$ and $dx =dt$ Thus, $\int csch^2(5-x) dx=\int csch^2 t dt=-\coth t +C$ or, $= -\coth (x-5) +C$ or, $=\coth (5-x) +C$
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