University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 329: 10



Work Step by Step

$u=1-cos\frac{t}{2}$, therefore $du=\frac{1}{2}-(-sin\frac{t}{2})dt=\frac{1}{2}sin\frac{t}{2}dt$. Let's rewrite the integral! $\int (1-cos\frac{t}{2})^2sin\frac{t}{2}dt$ We do the substitution!$\int (1-cos\frac{t}{2})^2sin\frac{t}{2}dt=\int u^2 2du=2\int u^2 du=2\frac{u^3}{3}+c=\frac{2u^3}{3}+c$ We re-do the substitution! $\frac{2u^3}{3}+c=\frac{2(1-cos\frac{t}{2})^3}{3}+c$
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.