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: 4


$$\int x\,cos4x\,dx=\frac{x\,sin4x}{4}+ \frac{cos4x}{16}$$

Work Step by Step

$${(\frac{x\,sin4x}{4})}'=\frac{sin4x}{4}+x\,cos4x$$ $$\int x\,cos4x\,dx=\frac{x\,sin4x}{4}-\int \frac{sin4x}{4}dx$$ $$=\frac{x\,sin4x}{4}+ \frac{cos4x}{16}$$
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