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: 11

Answer

$$\int_{0}^{\frac{\pi}{2}}sin^{2}x \, cos^{2}x \, dx=\frac{\pi}{16}$$

Work Step by Step

$$\int_{0}^{\frac{\pi}{2}}sin^{2}x \, cos^{2}x \, dx=\int_{0}^{\frac{\pi}{2}}(sinx \,cosx)^{2} \, dx$$ $$=\int_{0}^{\frac{\pi}{2}}(\frac{sin2x}{2})^{2} \, dx$$ $$=\int_{0}^{\frac{\pi}{2}}\frac{sin^{2}2x}{4} \, dx$$ $$=\int_{0}^{\frac{\pi}{2}}\frac{1-cos4x}{8} \, dx$$ $$=\left [\frac{1}{8}x -\frac{sin4x}{32} \right ]_{0}^{\frac{\pi}{2}}$$ $$=\frac{\pi}{16}$$
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