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 - Page 329: 8



Work Step by Step

$u=2x^2$, therefore $du=4xdx$. Let's rewrite the integral! $\int (xsin(2x^2))dx$ We do the substitution! $\int (xsin(2x^2))dx=\int (sinu)\frac{du}{4}=\frac{1}{4}\int (sinu)du=\frac{1}{4}(-cosu)+c$ We re-do the substitution! $-\frac{1}{4}(cosu)+c=-\frac{1}{4}(cos(2x^2))+c$
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.