## Thomas' Calculus 13th Edition

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