Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 7 - Exponential Functions - Chapter Review Exercises - Page 387: 69


$$\frac{1}{2}\cos e^{-2x}+c.$$

Work Step by Step

Let $ u=e^{-2x}$, then $ du=-2e^{-2x}dx $. Now, we have $$\int e^{-2x}\sin e^{-2x}dx =-\frac{1}{2}\int \sin udu =\frac{1}{2}\cos u+c=\frac{1}{2}\cos e^{-2x}+c.$$
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