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: 12


$$\int^{\pi/6}_0(1-\cos3t)\sin3tdt=\frac{1}{6}$$ $$\int^{\pi/3}_{\pi/6}(1-\cos3t)\sin3tdt=\frac{1}{2}$$

Work Step by Step

Part a) $$A=\int^{\pi/6}_0(1-\cos3t)\sin3tdt$$ We set $u=1-\cos3t$, which means $$du=-3(-\sin3t)dt=3\sin3tdt$$ $$\sin3tdt=\frac{1}{3}du$$ For $t=0$, we have $u=1-\cos0=1-1=0$ For $t=\pi/6$, we have $u=1-\cos(\pi/2)=1-0=1$ Therefore, $$A=\frac{1}{3}\int^1_0udu=\frac{1}{3}\times\frac{u^2}{2}\Big]^1_0=\frac{1}{6}\times u^2\Big]^1_0$$ $$A=\frac{1}{6}(1-0)=\frac{1}{6}$$ Part b) $$A=\int^{\pi/3}_{\pi/6}(1-\cos3t)\sin3tdt$$ We set up the same substitution as in part a). For $t=\pi/3$, we have $u=1-\cos\pi=1-(-1)=2$ For $t=\pi/6$, we have $u=1-\cos(\pi/2)=1-0=1$ Therefore, $$A=\frac{1}{3}\int^2_1udu=\frac{1}{6}\times u^2\Big]^2_1$$ $$A=\frac{1}{6}(2^2-1^2)=\frac{1}{6}\times(3)=\frac{1}{2}$$
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