Answer
$$\int^\pi_0\sin^5\frac{x}{2}dx=\frac{16}{15}$$
Work Step by Step
$$A=\int^\pi_0\sin^5\frac{x}{2}dx$$ $$A=2\int^{\pi/2}_0\sin^5\frac{x}{2}d\Big(\frac{x}{2}\Big)$$
We set $a=\frac{x}{2}$.
$$A=2\int^{\pi/2}_0\sin^5ada$$
Following the integral form $\int\sin^mx\cos^nxdx$, we would apply Case 1 here, where $m=5$ and $n=0$.
$$A=2\int^{\pi/2}_0(\sin^2a)^2\sin ada$$ $$A=-2\int^{\cos(\pi/2)}_{\cos0}(1-\cos^2a)^2d(\cos a)$$
We set $u=\cos a$.
Limits of integration: $\cos(\pi/2)=0$ and $\cos0=1$
$$A=-2\int^0_1(1-u^2)^2du$$ $$A=-2\int^0_1(1-2u^2+u^4)du$$ $$A=-2\Big(u-\frac{2u^3}{3}+\frac{u^5}{5}\Big)\Big]^0_1$$ $$A=-2\Big(0-\Big(1-\frac{2}{3}+\frac{1}{5}\Big)\Big)$$ $$A=(-2)\Big(-\frac{8}{15}\Big)=\frac{16}{15}$$
