Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 11


$$\int x\,cos2x\,dx=\frac{x}{2}sin2x+\frac{1}{4}cos\,2x+C$$

Work Step by Step

$${(\frac{x}{2}sin2x)}'=\frac{1}{2}sin2x+x\,cos2x$$ $$\int x\,cos2x\,dx=\frac{x}{2}sin2x-\frac{1}{2}\int sin2x\,dx$$ $$=\frac{x}{2}sin2x+\frac{1}{4}cos\,2x+C$$
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