University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.6 - Definite Integral Substitutions and the Area Between Curves - Exercises - Page 337: 20



Work Step by Step

$$A=\int^{\pi/4}_0(1-\sin2t)^{3/2}\cos2tdt$$ We set $u=1-\sin2t$, which means $$du=-2\cos2tdt$$ $$\cos2tdt=-\frac{1}{2}du$$ For $t=\pi/4$, we have $u=1-\sin\pi/2=1-1=0$ For $t=0$, we have $u=1-\sin0=1-0=1$ Therefore, $$A=-\frac{1}{2}\int^0_1u^{3/2}du$$ $$A=-\frac{1}{2}\times\frac{2u^{5/2}}{5}\Big]^0_1=-\frac{1}{5}u^{5/2}\Big]^0_1$$ $$A=-\frac{1}{5}(0-1)$$ $$A=\frac{1}{5}$$
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