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: 11


\begin{aligned} \int \sin ^{3} x \cos ^{3} x d x =\frac{1}{4} \sin ^{4} x-\frac{1}{6} \sin ^{6} x+C \end{aligned}

Work Step by Step

Given $$ \int \sin ^{3} x \cos ^{3} x d x $$ So, we have \begin{aligned} I&= \int \sin ^{3} x \cos ^{3} x d x\\ &=\int \sin ^{3} x \cos ^{2} x \cos x d x\\ &=\int \sin ^{3} x\left(1-\sin ^{2} x\right) \cos x d x\\ &=\int \sin ^{3} x \cos x d x-\int \sin ^{5} x \cos x d x\\ &=\frac{1}{4} \sin ^{4} x-\frac{1}{6} \sin ^{6} x+C \end{aligned}
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