University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Practice Exercises - Page 342: 41


$arcsec (x)+\dfrac{2\pi}{3}$

Work Step by Step

We need to integrate: $y=\int \dfrac{dx}{x\sqrt{x^2-1}}$ We use the differential trigonometric formula: $\int \dfrac{dx}{x\sqrt{x^2-a^2}}=arcsec (x/a)+C$ $y=arcsec (x)+c$ Take initial conditions: $y(2)=\pi$ Then $y(2)=arcsec (2)+c =\pi \implies c=\dfrac{2\pi}{3}$ Thus, we have $y=arcsec (x)+\dfrac{2\pi}{3}$
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