## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 8: Techniques of Integration - Section 8.3 - Trigonometric Integrals - Exercises 8.3 - Page 462: 3

#### Answer

\begin{aligned} \int \cos ^{3} x \sin x d x =-\frac{1}{4} \cos ^{4} x+C \end{aligned}

#### Work Step by Step

Given $$$$\int \cos ^{3} x \sin x d x$$$$ So, we have \begin{aligned} I&= \int \cos ^{3} x \sin x d x\\ &=-\int \cos ^{3} x(-\sin x) d x\\ &=-\frac{1}{4} \cos ^{4} x+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.