Ideas for this integral: $\int \frac{\sqrt{\tan{x}}}{\sin{x}} dx$ $$\int \frac{\sqrt{\tan{x}}}{\sin{x}} \mathrm{d}x$$
So I was wondering if this correctly by converting $\sqrt{\tan{x}}$ into $\frac{\sqrt{\sin{x}}}{\sqrt{\cos{x}}}$ therefore I can divide it with $\sin{x}$ and that will give me $\frac{\sqrt{\cos{x}}}{\sqrt{\sin{x}}}$ and that will be $\sqrt{\cot{x}}$ so the integral is formed into $\displaystyle \int \sqrt{\cot x}\mathrm{d}x $. Is this so far correctly done?

Hint:
$$\int \frac{\sqrt{\tan{x}}}{\sin{x}} dx=\int\frac{1}{\sqrt{\cos\left(x\right)\sin\left(x\right)}} dx=\int\frac{\sqrt{2}}{\sqrt{2}\sqrt{\cos\left(x\right)\sin\left(x\right)}}dx=\int\frac{\sqrt{2}}{\sqrt{\sin\left(2x\right)}}dx$$
Substituting $t=2x$ you have $dx=\frac 12dt$:
$$=\frac{1}{2}\int\frac{dt}{\sqrt{\sin\left(t\right)}}$$
where the last integral can be reduced to a complete elliptical integral of first species.
