University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 109



Work Step by Step

$$A=\int^{1}_{1/\sqrt3}\frac{dy}{y\sqrt{4y^2-1}}=\int^{1}_{1/\sqrt3}\frac{dy}{y\sqrt{(2y)^2-1^2}}$$ Set $u=2y$, which means $$du=2dy$$ $$dy=\frac{1}{2}du$$ Also, we would have $y=1/2u$ For $y=1$, we have $u=2$ and for $y=1/\sqrt3$, we have $u=2/\sqrt3$ Therefore, $$A=\int^{2}_{2/\sqrt3}\frac{\frac{1}{2}du}{\frac{1}{2}u\sqrt{u^2-1^2}}=\int^{2}_{2/\sqrt3}\frac{du}{u\sqrt{u^2-1^2}}$$ We have $$\int\frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}\sec^{-1}\Big|\Big(\frac{x}{a}\Big)\Big|+C$$ Therefore, $$A=\frac{1}{1}\sec^{-1}|u|\Big]^{2}_{2/\sqrt3}$$ $$A=\sec^{-1}2-\sec^{-1}\frac{2}{\sqrt3}$$ $$A=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$$
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