University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.2 - Trigonometric Integrals - Exercises - Page 434: 11



Work Step by Step

$$A=\int\sin^3x\cos^3xdx$$ Following the integral form $\int\sin^mx\cos^nxdx$, we would apply Case 1 here, where $m=3$ and $n=3$. $$A=\int\sin^3x\cos^2x(\cos xdx)$$ $$A=\int\sin^3x(1-\sin^2x)d(\sin x)$$ We set $u=\sin x$. $$A=\int u^3(1-u^2)du$$ $$A=\int(u^3-u^5)du$$ $$A=\frac{u^4}{4}-\frac{u^6}{6}+C$$ $$A=\frac{\sin^4x}{4}-\frac{\sin^6x}{6}+C$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.