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: 7



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