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: 47


$\tanh (x-\dfrac{1}{2}) +C$

Work Step by Step

Given: $\int sech^2(x-\dfrac{1}{2}) dx$ Plug in $x-\dfrac{1}{2}=t$ and $dx =dt$ Thus, $\int sech^2(x-\dfrac{1}{2}) dx=\int sech^2 t dt=\tanh t +C$ or, $= \tanh (x-\dfrac{1}{2}) +C$
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