Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

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


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

Work Step by Step

Given $$\begin{equation} \int \cos ^{3} x \sin x d x \end{equation}$$ 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}
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