University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Section 8.5 - Integral Tables and Computer Algebra Systems - Exercises - Page 451: 41


$$\int \sin^52xdx=-\frac{\sin^42x\cos2x}{10}-\frac{2\sin^22x\cos2x}{15}-\frac{4}{15}\cos2x+C$$

Work Step by Step

$$A=\int \sin^52xdx$$ Use Reduction Formula 67, which states that $$\int \sin^nax=-\frac{\sin^{n-1}ax\cos ax}{na}+\frac{n-1}{n}\int\sin^{n-2}axdx$$ for $a=2$ and $n=5$ $$A=-\frac{\sin^42x\cos2x}{10}+\frac{4}{5}\int\sin^32xdx$$ Use Reduction Formula 67 one more time, this time for $a=2$ and $n=3$ $$A=-\frac{\sin^42x\cos2x}{10}+\frac{4}{5}\Big(-\frac{\sin^22x\cos 2x}{6}+\frac{2}{3}\int\sin2xdx\Big)$$ $$A=-\frac{\sin^42x\cos2x}{10}+\frac{4}{5}\Big(-\frac{\sin^22x\cos 2x}{6}+\frac{2}{3}\times\Big(-\frac{1}{2}\Big)\cos2x\Big)+C$$ $$A=-\frac{\sin^42x\cos2x}{10}+\frac{4}{5}\Big(-\frac{\sin^22x\cos 2x}{6}-\frac{1}{3}\cos2x\Big)+C$$ $$A=-\frac{\sin^42x\cos2x}{10}-\frac{2\sin^22x\cos2x}{15}-\frac{4}{15}\cos2x+C$$
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