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: 91


$$\int^{\pi/2}_05(\sin x)^{3/2}\cos xdx=2$$

Work Step by Step

$$A=\int^{\pi/2}_05(\sin x)^{3/2}\cos xdx$$ We set $u=\sin x$, which means $$du=\cos xdx$$ For $x=\pi/2$, $u=1$ and for $x=0$, $u=0$ Therefore, $$A=5\int^1_0u^{3/2}du$$ $$A=\Big(\frac{5u^{5/2}}{\frac{5}{2}}\Big)\Big]^1_0=2u^{5/2}\Big]^1_0$$ $$A=2(1^{5/2}-0)$$ $$A=2\times1=2$$
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