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: 7


$\dfrac{e^x}{5}[2 \sin 2x+\cos 2x]$

Work Step by Step

Here, $I_1=\int e^x \cos 2x dx=\dfrac{1}{2}e^x \sin 2x - \dfrac{1}{2}\int e^x (\sin 2x) dx$ Now, do multiplication by $4$, we have $\implies I_1=(\dfrac{4}{2})e^x \sin 2x - (\dfrac{4}{2})\int e^x \sin 2x dx=2(e^x) (\sin 2x) - 2 )\int (e^x) (\sin 2x) dx$ Let us consider that $I_2=\int (e^x) (\cos 2x) dx=(e^x) (\cos 2x)+2 \int (e^x) (\sin 2x) dx$ Now, $I=I_1+I_2$ or, $I=2(e^x) (\sin 2x)+(e^x)( \cos 2x)$ Thus, $I=\dfrac{e^x}{5}[2 \sin 2x+\cos 2x]$
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