$\int \sqrt{\frac{\cos x-{\cos }^{3}x}{1-{\cos }^{3}x}}dx$ is equal to 

$$\begin{gathered} \int \sqrt{\frac{\cos \mathrm{x}-{\cos }^3\mathrm{x}}{1-{\cos }^3\mathrm{x}}} \\ =\int \frac{\sqrt{\cos \mathrm{x}(1-{\cos }^2\mathrm{x})}}{{(1-{\cos }^3\mathrm{x})}^{1/2}}\mathrm{dx} \\ =\int \frac{\sqrt{\cos \mathrm{x}} \sin \mathrm{x}}{{(1-{\cos }^3\mathrm{x})}^{1/2}}\mathrm{dx} \\ \begin{array}{c}=\frac{-2}{3}\int \frac{\mathrm{dt}}{\sqrt{1-{\mathrm{t}}^2}}\end{array}\overline{)\begin{array}{c}\mathrm{Put} {\cos }^{3/2}\mathrm{x}=\mathrm{t}\\ \frac{3}{2}{\cos }^{1/2}\mathrm{x}(-\sin \mathrm{x})\mathrm{dx}=\mathrm{dt}\\ \sqrt{\cos \mathrm{x}} \sin \mathrm{x} \mathrm{dx}\\ =\frac{-2}{3}\mathrm{dt}\end{array} } \\ =\frac{2}{3} {\cos }^{-1}\left(\mathrm{t}\right)+\mathrm{c} \left(\therefore -\int \frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}={\cos }^{-1}\mathrm{x}+\mathrm{c}\right) \\ =\frac{2}{3}{\cos }^{-1}\left({\cos }^{3/2}\mathrm{x}\right)+\mathrm{c} \end{gathered}$$