University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 59


$$\int \frac{5}{9+4r^2}dr=\frac{5}{6}\tan^{-1}\Big(\frac{2r}{3}\Big)+C$$

Work Step by Step

$$A=\int \frac{5}{9+4r^2}dr=5\int\frac{1}{3^2+(2r)^2}dr$$ We set $u=2r$. Then $$du=2dr$$ $$dr=\frac{1}{2}du$$ Therefore, $$A=\frac{5}{2}\int\frac{du}{3^2+u^2}$$ We have $$\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\Big(\frac{x}{a}\Big)+C$$ So $$A=\frac{5}{2}\times\frac{1}{3}\tan^{-1}\Big(\frac{u}{3}\Big)+C$$ $$A=\frac{5}{6}\tan^{-1}\Big(\frac{u}{3}\Big)+C$$ $$A=\frac{5}{6}\tan^{-1}\Big(\frac{2r}{3}\Big)+C$$
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