## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - 8.2 Trigonometric Integrals - Exercises - Page 404: 67

#### Answer

$$\frac{1}{16} \left( 2x- \sin 2 x\cos 2x \right) +c$$

#### Work Step by Step

\begin{aligned} \int \sin ^{2} x \cos ^{2} x d x &=\int \frac{1}{2}(1-\cos 2 x) \cdot \frac{1}{2}(1+\cos 2 x) d x \\ &=\frac{1}{4} \int\left(1-\cos ^{2} 2 x\right) d x \\ &=\frac{1}{4} \int\left(1-\frac{1}{2}(1+\cos 4 x)\right) d x \\ &=\frac{1}{4} \int\left( \frac{1}{2}- \frac{1}{2}\cos 4 x \right) d x\\ &=\frac{1}{4} \left( \frac{1}{2}x- \frac{1}{8}\sin 4 x \right) +c \\ &=\frac{1}{16} \left( 2x- \sin 2 x\cos 2x \right) +c \end{aligned}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.