Answer
$-\frac{\cos(4x)}{4}+C$
Work Step by Step
$\int \sin(4x)dx$
let
$u=4x$
$4dx=du$
$dx=\frac{1}{4}du$
$\int \sin(4x)dx$
$=\frac{1}{4} \int \sin(u)du$
$=\frac{1}{4}-\cos(u)+C$
$=-\frac{\cos(4x)}{4}+C$
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: 34
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