Answer
$$\frac{-1}{8}\cos 4x+c$$
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*}
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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