Answer
Divergent.
Work Step by Step
$$\int_{0}^{\infty }sin^{2}\alpha d\alpha=\lim_{t\rightarrow \infty}\int_{0}^{t}sin^{2}\alpha d\alpha$$
$$=\lim_{t\rightarrow \infty}\int_{0}^{t}\frac{1-cos2\alpha}{2}d\alpha$$
$$=\lim_{t\rightarrow \infty} \left|\frac{\alpha}{2}-\frac{sin2\alpha}{4}\right |_{0}^{t}$$
$$= \lim_{t\rightarrow \infty} (\frac{t}{2}-\frac{sin2t}{4})$$
$$=\infty,\ divergent$$
