University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 24



Work Step by Step

$$A=\int^{\pi}_0\sqrt{1-\cos2x}dx$$ Use the identity $$\frac{1-\cos 2x}{2}=\sin^2x$$ $$1-\cos2x=2\sin^2x$$ That means $$A=\int^{\pi}_0\sqrt{2\sin^2x}dx$$ $$A=\sqrt2\int^{\pi}_0|\sin x|dx$$ As $\sin x\ge0$ on $[0,\pi]$, $|\sin x|=\sin x$ $$A=\sqrt2\int^{\pi}_0\sin xdx$$ $$A=-\sqrt2\cos x\Big]^{\pi}_0$$ $$A=-\sqrt2(\cos\pi-\cos0)$$ $$A=-\sqrt2(-1-1)=2\sqrt2$$
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