Answer
$$s =\frac{1}{8} \left (\alpha -\frac{1}{2\alpha}\sin 2\alpha\right) +c$$
Work Step by Step
$$
\frac{d s}{d \alpha}=\sin ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2}
$$
Since $\sin2 \alpha=2\sin\alpha\cos \alpha$, then
\begin{align*}
ds&= \sin ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2}d\alpha\\
s&= \int \left(\sin \frac{\alpha}{2} \cos \frac{\alpha}{2}\right)^2d\alpha\\
&=\frac{1}{4} \int \sin^2\alpha d\alpha\\
&=\frac{1}{8} \int (1-\cos 2\alpha) d\alpha\\
&=\frac{1}{8} \left (\alpha -\frac{1}{2\alpha}\sin 2\alpha\right) +c
\end{align*}
