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: 69

Answer

$$\int\frac{(2r-1)\cos\sqrt{3(2r-1)^2+6}}{\sqrt{3(2r-1)^2+6}}dr=\frac{\sin\sqrt{3(2r-1)^2+6}}{6}+C$$

Work Step by Step

$$A=\int\frac{(2r-1)\cos\sqrt{3(2r-1)^2+6}}{\sqrt{3(2r-1)^2+6}}dr$$ Set $u=\sqrt{3(2r-1)^2+6}$, then $$du=\frac{\Big(3(2r-1)^2+6\Big)'}{2\sqrt{3(2r-1)^2+6}}dr=\frac{6(2r-1)(2r-1)'}{2\sqrt{3(2r-1)^2+6}}dr$$ $$du=\frac{6(2r-1)\times2}{2\sqrt{3(2r-1)^2+6}}dr=\frac{6(2r-1)}{\sqrt{3(2r-1)^2+6}}dr$$ That means, $$\frac{(2r-1)}{\sqrt{3(2r-1)^2+6}}dr=\frac{1}{6}du$$ Therefore, $$A=\frac{1}{6}\int\cos udu=\frac{\sin u}{6}+C$$ $$A=\frac{\sin\sqrt{3(2r-1)^2+6}}{6}+C$$
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