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: 63


$$\int \frac{3dr}{\sqrt{1-4(r-1)^2}}=\frac{3}{2}\sin^{-1}\Big(2(r-1)\Big)+C$$

Work Step by Step

$$A=\int \frac{3dr}{\sqrt{1-4(r-1)^2}}$$ $$A=3\int\frac{dr}{\sqrt{1-(2(r-1))^2}}$$ We set $a=2(r-1)$, which means $$da=2dr$$ $$dr=\frac{1}{2}da$$ Therefore, $$A=\frac{3}{2}\int\frac{da}{\sqrt{1-a^2}}$$ $$A=\frac{3}{2}\sin^{-1}a+C$$ $$A=\frac{3}{2}\sin^{-1}\Big(2(r-1)\Big)+C$$
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