Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Practice Exercises - Page 517: 8


$\dfrac{-e^{-2x}}{13}[3 \cos 3x+2 \sin 3x]$

Work Step by Step

Here, $I_1=\int e^{-2x} \sin 3x dx=\dfrac{-1}{3}(e^{-2x})( \cos 3x) - \dfrac{2}{3} \int (e^{-2x}) (\cos 3x) dx$ $\implies I_1=\dfrac{9}{4}[(-1/3)(e^{-2x})( \cos 3x) - (2/3) \int e^{-2x} (\cos 3x) dx]$ Next, let us consider that $I_2=\int e^{-2x} \sin 3x dx=-\dfrac{1}{2} (e^{-2x}) (\sin 3x) - \dfrac{3}{2} \int e^{-2x} \cos 3x dx$ Now, $I=I_1+I_2$ $I=(\dfrac{9}{4}+1) \int e^{-2x} \sin 3x dx$ $\implies I=(\dfrac{-3}{4})e^{-2x} \cos 3x-(\dfrac{1}{2})e^{-2x} \sin 3x$ Thus, $\implies I=-e^{-2x}[(3/4) \cos 3x+(1/2)\sin 3x]=\dfrac{-e^{-2x}}{13}[3 \cos 3x+2 \sin 3x]$
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