University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Practice Exercises - Page 343: 49


$$\int\sqrt t\sin(2t^{3/2})dt=-\frac{1}{3}\cos(2t^{3/2})+C$$

Work Step by Step

$$A=\int\sqrt t\sin(2t^{3/2})dt$$ We set $a=2t^{3/2}$, which means $$da=2\times\frac{3}{2}t^{1/2}dt=3\sqrt tdt$$ $$\sqrt tdt=\frac{1}{3}da$$ Therefore, $$A=\frac{1}{3}\int\sin ada$$ $$A=-\frac{1}{3}\cos a+C$$ $$A=-\frac{1}{3}\cos(2t^{3/2})+C$$
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