## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 25

#### Answer

$$\frac{-1}{4}$$

#### Work Step by Step

\begin{aligned} \int_{0}^{\pi/4} \sin 3 x \cos 5 x d x &=\int_{0}^{\pi/4} \frac{1}{2}(\sin 8 x+\sin -2 x) d x \\ &=\int_{0}^{\pi/4} \frac{1}{2}(\sin 8 x-\sin 2 x) d x \\ &=\frac{1}{2} \int_{0}^{\pi/4} \sin 8 x d x-\frac{1}{2} \int_{0}^{\pi/4} \sin 2 x d x \\ &=\frac{1}{2} \frac{-\cos 8 x}{8}-\frac{1}{2} \frac{-\cos 2 x}{2}\bigg|_{0}^{\pi/4}\\ &= \frac{-1}{4} \end{aligned}

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