Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises 5.5 - Page 295: 10


$\int(1 - \cos\frac{t}{2})^2 \sin\frac{t}{2}dt = \frac{2}{3}(1-\cos\frac{t}{2})^3 + C$

Work Step by Step

$\int(1 - \cos\frac{t}{2})^2 \sin\frac{t}{2}dt\space$ and $\space u = 1-\cos\frac{t}{2}$ $du = \frac{1}{2}\sin\frac{t}{2}\cdot dt$ Doing the substitution $\space u = 1-\cos\frac{t}{2}$ $\int(u)^2 \cdot2du \space\Rightarrow\space2\int u^2 du$ Applying the integrative rules $2\int u^2 du = \frac{2}{3}u^3+C$ Backing to $t$ $\int(1 - \cos\frac{t}{2})^2 \sin\frac{t}{2}dt = \frac{2}{3}(1-\cos\frac{t}{2})^3 + C$
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