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 329: 19



Work Step by Step

$$A=\int\theta\sqrt[4]{1-\theta^2}d\theta$$ We set $u=1-\theta^2$. Then $$du=-2\theta d\theta$$ That means, $$\theta d\theta=-\frac{1}{2}du$$ Therefore, $$A=-\frac{1}{2}\int\sqrt[4]udu=-\frac{1}{2}\int u^{1/4}du$$ $$A=-\frac{1}{2}\times\frac{u^{5/4}}{\frac{5}{4}}+C=-\frac{u^{5/4}}{\frac{5}{2}}+C=-\frac{2u^{5/4}}{5}+C$$ $$A=-\frac{2(1-\theta^2)^{5/4}}{5}+C$$
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