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: 44



Work Step by Step

$$A=\int^2_{2/\sqrt3}\frac{\cos(\sec^{-1}x)dx}{x\sqrt{x^2-1}}$$ We set $u=\sec^{-1}x$, which means $$du=\frac{dx}{x\sqrt{x^2-1}}$$ - For $x=\frac{2}{\sqrt3}$, we have $u=\sec^{-1}\Big(\frac{2}{\sqrt3}\Big)=\cos^{-1}\Big(\frac{\sqrt3}{2}\Big)=\frac{\pi}{6}$ - For $x=2$, we have $u=\sec^{-1}2=\cos^{-1}\Big(\frac{1}{2}\Big)=\frac{\pi}{3}$ Therefore, $$A=\int^{\pi/3}_{\pi/6}\cos udu$$ $$A=\sin u\Big]^{\pi/3}_{\pi/6}$$ $$A=\sin\frac{\pi}{3}-\sin\frac{\pi}{6}$$ $$A=\frac{\sqrt3}{2}-\frac{1}{2}=\frac{\sqrt3-1}{2}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.