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: 5


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

Work Step by Step

Given $$\begin{equation} \int \sin ^{3} x d x \end{equation}$$ So, we have \begin{aligned} I&= \int \sin ^{3} x d x\\ &=\int \sin ^{2} x \sin x d x\\ &=\int\left(1-\cos ^{2} x\right) \sin x d x\\ &=\int \sin x d x-\int \cos ^{2} x \sin x d x\\ &=-\cos x+\frac{1}{3} \cos ^{3} x+C \end{aligned}
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