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

Published by Pearson

# Chapter 8 - Section 8.5 - Integral Tables and Computer Algebra Systems - Exercises - Page 451: 25

#### Answer

$$\int\cos\frac{\theta}{3}\cos\frac{\theta}{4}d\theta=6\sin\frac{\theta}{12}+\frac{6}{7}\sin\frac{7\theta}{12}+C$$

#### Work Step by Step

$$A=\int\cos\frac{\theta}{3}\cos\frac{\theta}{4}d\theta$$ Use Formula 69c, which states that $$\int \cos ax\cos bx dx=\frac{\sin(a-b)x}{2(a-b)}+\frac{\sin(a+b)x}{2(a+b)}+C$$ for $a=1/3$ and $b=1/4$: $a-b=1/3-1/4=1/12$ while $a+b=1/3+1/4=7/12$ $$A=\frac{\sin\frac{1}{12}\theta}{2\times\frac{1}{12}}+\frac{\sin\frac{7}{12}\theta}{2\times\frac{7}{12}}+C$$ $$A=\frac{\sin\frac{\theta}{12}}{\frac{1}{6}}+\frac{\sin\frac{7\theta}{12}}{\frac{7}{6}}+C$$ $$A=6\sin\frac{\theta}{12}+\frac{6}{7}\sin\frac{7\theta}{12}+C$$

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