Calculus 10th Edition

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: 9

Answer

$-\frac x 3 cos3x +\frac 1 9 sin3x+C$

Work Step by Step

$dv=sin3xdx$ $v=\int sin3xdx = -\frac 1 3 cos3x$ $u=x, du=dx$ $uv-\int vdu$ $x(-\frac 1 3 cos3x)-\int -\frac 1 3 cos3xdx$ $-\frac x 3 cos3x +\frac 1 9 sin3x+C$
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