Answer
$$\frac{1}{2}\cos e^{-2x}+c.$$
Work Step by Step
Let $ u=e^{-2x}$, then $ du=-2e^{-2x}dx $.
Now, we have
$$\int e^{-2x}\sin e^{-2x}dx =-\frac{1}{2}\int \sin udu
=\frac{1}{2}\cos u+c=\frac{1}{2}\cos e^{-2x}+c.$$
Calculus (3rd Edition)
by Rogawski, Jon; Adams, Colin
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
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