Answer
$$\int x\,cos2x\,dx=\frac{x}{2}sin2x+\frac{1}{4}cos\,2x+C$$
Work Step by Step
$${(\frac{x}{2}sin2x)}'=\frac{1}{2}sin2x+x\,cos2x$$
$$\int x\,cos2x\,dx=\frac{x}{2}sin2x-\frac{1}{2}\int sin2x\,dx$$
$$=\frac{x}{2}sin2x+\frac{1}{4}cos\,2x+C$$
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: 11
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
