Answer
$$\int x\,sin(3-x)dx=x\,cos(3-x)-sin(3-x)+C$$
Work Step by Step
$${[x\,cos(3-x)]}'= cos(3-x)+x\,sin(3-x)$$
$$\int x\,sin(3-x)dx=x\,cos(3-x)-\int cos(3-x)dx$$
$$=x\,cos(3-x)-sin(3-x)+C$$
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: 12
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