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: 26


$\frac{\sin^6 2t}{12}+C$

Work Step by Step

Let $u=\sin 2t$. Then $du=2\cos 2tdt$, so $\cos 2tdt=\frac{1}{2}du$. $\int\cos 2t\sin^5 2tdt$ $=\int\sin^5 2t\cos 2tdt$ $=\int u^5*\frac{1}{2}du$ $=\frac{1}{2}*\frac{u^6}{6}+C$ $=\frac{u^6}{12}+C$ $=\frac{\sin^6 2t}{12}+C$
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