Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.2 - Trigonometric Integrals - 7.2 Exercises: 12

Answer

$$\int_{0}^{\frac{\pi}{2}}(2-sin\theta)^{2}d\theta=\frac{9}{4}\pi - 4$$

Work Step by Step

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