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: 14



Work Step by Step

$$A=\int^{\pi/2}_0\sin^2xdx$$ This is Case 3, so we will need to rewrite $$\sin^2x=\frac{1-\cos2x}{2}$$ Therefore, $$A=\int^{\pi/2}_0\frac{1-\cos2x}{2}dx$$ $$A=\int^{\pi/2}_0\frac{1}{2}dx-\frac{1}{2}\int^{\pi/2}_0\cos2xdx$$ $$A=\frac{x}{2}\Big]^{\pi/2}_0-\frac{1}{2}\Big(\frac{1}{2}\sin2x\Big)\Big]^{\pi/2}_0$$ $$A=\frac{\pi}{4}-\frac{1}{4}(\sin\pi-\sin0)$$ $$A=\frac{\pi}{4}-\frac{1}{4}(0-0)$$ $$A=\frac{\pi}{4}$$
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