## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 435: 57

#### Answer

$$\int\sin^2\theta\cos3\theta d\theta=\frac{1}{6}\sin3\theta-\frac{1}{4}\sin\theta-\frac{1}{20}\sin5\theta+C$$

#### Work Step by Step

$$A=\int\sin^2\theta\cos3\theta d\theta$$ Use the identity: $$\sin^2\theta=\frac{1-\cos2\theta}{2}$$ $$A=\int\frac{1-\cos2\theta}{2}\cos3\theta d\theta$$ $$A=\frac{1}{2}\Big(\int\cos3\theta d\theta-\int\cos2\theta\cos3\theta d\theta\Big)$$ Now use the identity for the product of two cosine functions: $$\cos mx\cos nx=\frac{1}{2}[\cos(m-n)x+\cos(m+n)x]$$ Therefore, $$A=\frac{1}{2}\Big(\frac{1}{3}\sin3\theta-\frac{1}{2}\int[\cos(-\theta)+\cos5\theta]d\theta\Big)$$ $$A=\frac{1}{2}\Big(\frac{1}{3}\sin3\theta-\frac{1}{2}\int[\cos\theta+\cos5\theta]d\theta\Big)$$ $$A=\frac{1}{2}\Big[\frac{1}{3}\sin3\theta-\frac{1}{2}\Big(\sin\theta+\frac{1}{5}\sin5\theta\Big)\Big]+C$$ $$A=\frac{1}{6}\sin3\theta-\frac{1}{4}\sin\theta-\frac{1}{20}\sin5\theta+C$$

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