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