Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.3 Integration By Substitution - Exercises Set 4.3 - Page 286: 29



Work Step by Step

Let $u=2-\sin4\theta$. Then $du=-4\cos 4\theta d\theta$, so $\cos 4\theta d\theta=-\frac{1}{4}du$. $\int\cos 4\theta\sqrt{2-\sin 4\theta}d\theta$ $=\int\sqrt{2-\sin 4\theta}\cos 4\theta d\theta$ $=\int\sqrt{u}*(-\frac{1}{4})du$ $=-\frac{1}{4}\int u^{\frac{1}{2}}du$ $=-\frac{1}{4}*\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C$ $=-\frac{1}{6}u^{\frac{3}{2}}+C$ $=-\frac{1}{6}(2-\sin4\theta)^{\frac{3}{2}}+C$
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