Answer
$$\int^{\pi/2}_{\pi/3}\frac{\sin^2x}{\sqrt{1-\cos x}}dx=\sqrt{\frac{3}{2}}-\frac{2}{3}$$
Work Step by Step
$$A=\int^{\pi/2}_{\pi/3}\frac{\sin^2x}{\sqrt{1-\cos x}}dx$$
Multiply both numerator and denominator with $\sqrt{1+\cos x}$
$$A=\int^{\pi/2}_{\pi/3}\frac{\sin^2x}{\sqrt{1-\cos x}}\frac{\sqrt{1+\cos x}}{\sqrt{1+\cos x}}dx$$ $$A=\int^{\pi/2}_{\pi/3}\frac{\sin^2x\sqrt{1+\cos x}}{\sqrt{1-\cos^2x}}dx$$ $$A=\int^{\pi/2}_{\pi/3}\frac{\sin^2x\sqrt{1+\cos x}}{\sqrt{\sin^2x}}dx$$ $$A=\int^{\pi/2}_{\pi/3}\frac{\sin^2x\sqrt{1+\cos x}}{\sin x}dx$$ (because on $[\pi/3,\pi,2]$, $\sin x\gt0$)
$$A=\int^{\pi/2}_{\pi/3}\sin x\sqrt{1+\cos x}dx$$ $$A=-\int^{\pi/2}_{\pi/3}\sqrt{1+\cos x}d(1+\cos x)$$
Set $u=1+\cos x$
For $x=\pi/2$, $u=1+0=1$ and for $x=\pi/3$, $u=1+1/2=3/2$
$$A=-\int^1_{3/2}\sqrt udu=\int^{3/2}_1u^{1/2}du$$ $$A=\frac{2}{3}u^{3/2}\Big]^{3/2}_1$$ $$A=\frac{2}{3}\Big[\Big(\frac{3}{2}\Big)^{3/2}-1\Big]=\frac{2}{3}\Big(\frac{3\sqrt3}{2\sqrt2}-1\Big)$$ $$A=\sqrt{\frac{3}{2}}-\frac{2}{3}$$
