University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises: 7

Answer

$-\frac{1}{3}(cos(3x))+c$

Work Step by Step

$u=3x$, therefore $du=3dx$. Let's rewrite the integral! $\int (sin3x)dx$ We do the substitution! $\int (sin3x)dx=\int (sinu)\frac{du}{3}=\frac{1}{3}\int (sinu)du=\frac{1}{3}(-cosu)+c$ We re-do the substitution! $-\frac{1}{3}(cosu)+c=-\frac{1}{3}(cos(3x))+c$
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.