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: 10

Answer

$\frac{2(1-cos\frac{t}{2})^3}{3}+c$

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! $\frac{2u^3}{3}+c=\frac{2(1-cos\frac{t}{2})^3}{3}+c$
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