Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 403: 43


$$\frac{-1}{8}\cos 4x+c$$

Work Step by Step

\begin{align*} \int \sin 2x\cos 2xdx&=\frac{1}{2}\int\sin 4xdx\\ &=\frac{-1}{8}\cos 4x+c \end{align*}
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