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: 8

Answer

$-\frac{1}{4}(cos(2x^2))+c$

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! $-\frac{1}{4}(cosu)+c=-\frac{1}{4}(cos(2x^2))+c$
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