Thomas' Calculus 13th Edition

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


$-\frac{1}{3}cos 3x +C$

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
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.