University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 338: 43



Work Step by Step

$$A=\int^{2}_{\sqrt2}\frac{\sec^2(\sec^{-1}x)dx}{x\sqrt{x^2-1}}$$ We set $u=\sec^{-1}x$, which means $$du=\frac{1}{x\sqrt{x^2-1}}dx$$ - For $x=\sqrt2$, we have $u=\sec^{-1}\sqrt2=\cos^{-1}\frac{\sqrt2}{2}=\frac{\pi}{4}$ - For $x=2$, we have $u=\sec^{-1}2=\cos^{-1}\frac{1}{2}=\frac{\pi}{3}$ Therefore, $$A=\int^{\pi/3}_{\pi/4}\sec^2udu$$ $$A=\tan u\Big]^{\pi/3}_{\pi/4}$$ $$A=\tan\frac{\pi}{3}-\tan\frac{\pi}{4}$$ $$A=\sqrt3-1$$
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