Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises - Page 524: 12



Work Step by Step

$$\int_{0}^{\pi/2}(2-sin\theta)^{2}d\theta=\int_{0}^{\pi/2}(4-4sin\theta+sin^{2}\theta)d\theta$$ $$=\int_{0}^{\pi/2}(4-4sin\theta+\frac{1-cos2\theta}{2})d\theta$$ $$=\left |\frac{9}{2}\theta+4cos\theta-\frac{sin2\theta}{4} \right |_{0}^{\pi/2}$$ $$=\frac{9}{4}\pi+4*0-\frac{sin\pi}{4}-4*1+\frac{sin0}{4}$$ $$=\frac{9}{4}\pi-4$$
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