University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 40

Answer

$$\int x^2\sin{x^3}dx=-\frac{\cos x^3}{3}+C$$

Work Step by Step

$$A=\int x^2\sin{x^3}dx$$ We set $a=x^3$. We then have $$da=3x^2dx$$ $$x^2dx=\frac{1}{3}da$$ Therefore, $$A=\frac{1}{3}\int \sin ada$$ $$A=-\frac{\cos a}{3}+C$$ $$A=-\frac{\cos x^3}{3}+C$$
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