Answer
$\tanh (x-\dfrac{1}{2}) +C$
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, $= \tanh (x-\dfrac{1}{2}) +C$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 7 - Section 7.3 - Hyperbolic Functions - Exercises - Page 418: 47
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