Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - Chapter Review Exercises - Page 278: 64



Work Step by Step

Let $u=\cos\theta$, then $du=-\sin\theta d\theta$; we have \begin{align*} \int\sin\theta\sqrt{4-\cos\theta}d\theta &= -\int \sqrt{4- u}du \\ &=\frac{1}{3/2}(4-u)^{3/2}+c\\ &=\frac{2}{3}(4-\cos\theta)^{3/2}+c. \end{align*}
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