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 - Review Exercises - Page 579: 13

Answer

$-\frac{1}{4}(2x^{2}cos2x-2xsin2x-cos2x)+C$

Work Step by Step

Use integration by parts twice. $\int x^{2}sin2xdx$ $u=x^{2},dv=\int sin2xdx$ $u'=2xdx,v=-\frac{1}{2}cos2x$ $=-\frac{1}{2}x^{2}cos2x-\int (2x)(-\frac{1}{2}cos2x)dx$ $=-\frac{1}{2}x^{2}cos2x+\int xcos2xdx$ $u=x,dv=\int cos2x$ $u'=dx,v=\frac{1}{2}sin2x$ $=-\frac{1}{2}x^{2}cos2x+\frac{1}{2}xsin2x-\int\frac{1}{2}sin2xdx$ $=-\frac{1}{2}x^{2}cos2x+\frac{1}{2}xsin2x+\frac{1}{4}cos2x+C$ Take out $-\frac{1}{4}$ to get: $-\frac{1}{4}(2x^{2}cos2x-2xsin2x-cos2x)+C$
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