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: 13


$$\int^{2\pi}_0\frac{\cos z}{\sqrt{4+3\sin z}}dz=\int^{\pi}_{-\pi}\frac{\cos z}{\sqrt{4+3\sin z}}dz=0$$

Work Step by Step

Part a) $$A=\int^{2\pi}_0\frac{\cos z}{\sqrt{4+3\sin z}}dz$$ We set $u=\sqrt{4+3\sin z}$, which means $$du=\frac{(4+3\sin z)'}{2\sqrt{4+3\sin z}}dz=\frac{3\cos z}{2\sqrt{4+3\sin z}}dz$$ $$\frac{\cos z}{\sqrt{4+3\sin z}}dz=\frac{2}{3}du$$ For $z=0$, we have $u=\sqrt{4+3\sin0}=\sqrt4=2$ For $z=2\pi$, we have $u=\sqrt{4+3\sin2\pi}=\sqrt4=2$ Therefore, $$A=\frac{2}{3}\int^2_2dz=0$$ Part b) $$A=\int^{\pi}_{-\pi}\frac{\cos z}{\sqrt{4+3\sin z}}dz$$ We set up the same substitution as in part a). For $z=\pi$, we have $u=\sqrt{4+3\sin\pi}=\sqrt{4}=2$ For $z=-\pi$, we have $u=\sqrt{4+3\sin(-\pi)}=\sqrt4=2$ Therefore, $$A=\frac{2}{3}\int^2_2dz=0$$
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