t = (\sin x)^{3\over 2} \\
dt = \frac{-3\cos x \sqrt{\sin x}}{2} dx \\
\sqrt{\sin x} dx = -\frac{2 dt}{3\cos x} \\
\frac{\sqrt{\sin x} dx}{\cos x} = -\frac{2 dt}{3(cos x)^2} \\
= -\frac{1}{3}\frac{2 dt}{1 - (sin x)^2} \\
= -\frac{2}{3}\frac{dt}{1 - t^{4\over 3}} \\
\int {\sqrt{\sin x} \over \cos{x}} dx = -\frac{2}{3} \int {\frac{dt}{1 - t^{4\over 3}}}