Answer
$-\frac{1}{3}cos 3x +C$
Work Step by Step
Put u = 3x.
Then $\frac{du}{dx}= 3$ or dx= du/3
Therefore $\int $sin 3x dx = $\int$ sin u $\frac{du}{3}$= $\frac{1}{3}\int$sin u du
= $\frac{1}{3}$(-cos u)+C= $-\frac{1}{3}$cos 3x +C
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 7
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