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: 28


$-\frac{1}{24(1+2\sin 4\theta)^3}+C$

Work Step by Step

Let $u=1+2\sin4\theta$. Then $du=8\cos 4\theta d\theta$, so $\cos4\theta d\theta=\frac{1}{8}du$. $\int \frac{\cos 4\theta}{(1+2\sin4\theta)^4}d\theta$ $=\int \frac{\frac{1}{8}du}{u^4}$ $=\frac{1}{8}\int u^{-4}du$ $=\frac{1}{8}*\frac{u^{-3}}{-3}+C$ $=-\frac{1}{24u^3}+C$ $=-\frac{1}{24(1+2\sin 4\theta)^3}+C$
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