indefinite..

\hspace{-16}$Let$\bf{\;\; I = \int\sqrt{\frac{\sin x-\sin^3 x}{1-\sin^3 x}}\; dx = \int\sqrt{\frac{\sin x\cdot(1-\sin^2x)}{1-\sin^3 x}}\; dx}$\\\\\\ So $\bf{I = \int\sqrt{\frac{\sin x}{1-\sin^3 x}}\cdot \cos x\; dx = \int\frac{\sqrt{\sin x}}{\sqrt{1-\sin^3 x}}\cdot \cos x\; dx}$\\\\\\ Now Let $\bf{\sin x = t\;,}$ Then $\bf{\cos x\; dx = dt}$\\\\\\ So $\bf{I = \int\frac{\sqrt{t}}{\sqrt{1-\left(t^{\frac{3}{2}}\right)^2}}dt}$\\\\\\ Now Let $\bf{t^{\frac{3}{2}}=\sin \phi\;,}$ Then $\bf{\frac{3}{2}\sqrt{t}dt=\cos \phi\; d\phi\Rightarrow \sqrt{t}\;dt = \frac{2}{3}\cos \phi d\phi}$\\\\\\ So $\bf{I=\frac{2}{3}\int\frac{\cos \phi}{\cos \phi}\;d\phi = \frac{2}{3}\cdot \phi+\mathbb{C}=\frac{2}{3}\sin^{-1}\left(t^{\frac{3}{2}}\right)+\mathbb{C}}$\\\\\\ So $\bf{I = \int\sqrt{\frac{\sin x-\sin^3 x}{1-\sin^3 x}}\; dx=\frac{2}{3}\sin^{-1}\left(\sin^{\frac{3}{2}} x\right)+\mathbb{C}}$

Yes I am also getting this