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