Answer
$-\frac x 3 cos3x +\frac 1 9 sin3x+C$
Work Step by Step
$dv=sin3xdx$
$v=\int sin3xdx = -\frac 1 3 cos3x$
$u=x, du=dx$
$uv-\int vdu$
$x(-\frac 1 3 cos3x)-\int -\frac 1 3 cos3xdx$
$-\frac x 3 cos3x +\frac 1 9 sin3x+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises - Page 521: 9
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