Answer
$$\sqrt{\frac{3}{2}}-\frac{2}{3}$$
Work Step by Step
We integrate as follows:
\begin{align*}
\int_{\pi / 3}^{\pi / 2} \frac{\sin ^{2} x}{\sqrt{1-\cos x}} d x&=\int_{\pi / 3}^{\pi / 2} \frac{\sin ^{2} x}{\sqrt{1-\cos x}} \frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}} d x\\
&=\int_{\pi / 3}^{\pi / 2} \frac{\sin ^{2} x \sqrt{1+\cos x}}{\sqrt{1-\cos ^{2} x}} d x\\
&=\int_{\pi / 3}^{\pi / 2} \frac{\sin ^{2} x \sqrt{1+\cos x}}{\sqrt{\sin ^{2} x}} d x\\
&=\int_{\pi / 3}^{\pi / 2} \sin x \sqrt{1+\cos x} d x\\
&=\left[-\frac{2}{3}(1+\cos x)^{3 / 2}\right]_{\pi / 3}^{\pi / 2}\\
&=-\frac{2}{3}\left(1+\cos \left(\frac{\pi}{2}\right)\right)^{3 / 2}+\frac{2}{3}\left(1+\cos \left(\frac{\pi}{3}\right)\right)^{3 / 2}\\
&=-\frac{2}{3}+\frac{2}{3}\left(\frac{3}{2}\right)^{3 / 2}\\
&=\sqrt{\frac{3}{2}}-\frac{2}{3}
\end{align*}
