Answer
$$\int x\,cos4x\,dx=\frac{x\,sin4x}{4}+ \frac{cos4x}{16}$$
Work Step by Step
$${(\frac{x\,sin4x}{4})}'=\frac{sin4x}{4}+x\,cos4x$$
$$\int x\,cos4x\,dx=\frac{x\,sin4x}{4}-\int \frac{sin4x}{4}dx$$
$$=\frac{x\,sin4x}{4}+ \frac{cos4x}{16}$$
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 - 8.1 Integration by Parts - Exercises - Page 395: 4
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