Answer
$\frac{1}{8}\sin{8x}+C$
Work Step by Step
$\int\cos{8x}dx$
Let $u=8x\hspace{8mm}du=8dx\hspace{8mm}dx=\frac{du}{8}$
$\int\cos{8x}dx=\int\cos{u}\frac{du}{8}=\frac{1}{8}\int\cos{u}du$
$\frac{1}{8}\sin{u}+C=\frac{1}{8}\sin{8x}+C$
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 4 - Integration - 4.5 Exercises - Page 301: 35
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