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: 11

Answer

$-\frac{1}{13}(3e^{2x}cos3x-2e^{2x}sin3x)+C$

Work Step by Step

Use integration by parts twice. Set $A=\int e^{2x}sin3xdx$ $u=e^{2x},dv=\int sin3xdx$ $u'=2e^{2x}dx,v=-\frac{1}{3}cos3x$ $A=-\frac{1}{3}e^{2x}cos3x+\int (2e^{2x})(\frac{1}{3}cos3x)dx$ $u=2e^{2x},dv=\int \frac{1}{3}cos3xdx$ $u'=4e^{2x}dx,v=\frac{1}{9}sin3x$ $A=-\frac{1}{3}e^{2x}cos3x+\frac{2}{9}e^{2x}sin3x-\frac{4}{9}\int e^{2x}sin3xdx$ Cross out the $\frac{4}{9}\int e^{2x}sin3xdx$ and replace with $\frac{4}{9}A$ thus: $A=-\frac{1}{3}e^{2x}cos3x+\frac{2}{9}e^{2x}sin3x-\frac{4}{9}A$ Add $\frac{4}{9}A$ to other side. $\frac{13}{9}A=-\frac{1}{3}e^{2x}cos3x+\frac{2}{9}e^{2x}sin3x$ Multiply whole equation by $\frac{9}{13}$ to get: $A=-\frac{3}{13}e^{2x}cos3x+\frac{2}{13}e^{2x}sin3x$ Separate out $-\frac{1}{13}$ $A=-\frac{1}{13}(3e^{2x}cos3x-2e^{2x}sin3x)+C$
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