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 330: 29


$$\int x^{1/2}\sin(x^{3/2}+1)dx=-\frac{2}{3}\cos(x^{3/2}+1)+C$$

Work Step by Step

$$A=\int x^{1/2}\sin(x^{3/2}+1)dx$$ We set $u=x^{3/2}+1$ Then $$du=\frac{3}{2}x^{1/2}dx$$ That means $$x^{1/2}dx=\frac{2}{3}du$$ Therefore, $$A=\frac{2}{3}\int\sin udu=-\frac{2}{3}\cos u+C$$ $$A=-\frac{2}{3}\cos(x^{3/2}+1)+C$$
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