Answer
$\frac 1 4 xsin(4x)+\frac 1 {16}cos(4x)+C$
Work Step by Step
$\int udv=uv-\int vdu$
$(x)(\frac 1 4 sin4x)-\int \frac 1 4 sin(4x)(1)dx$
$\frac 1 4 xsin(4x)+\frac 1 {16}cos(4x)+C$
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: 10
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