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?

$$I=\int \frac{\sqrt{\tan x}}{\sin x} dx$$
Let $\tan x =t^2 \implies \sec^2 x dx=2t dt$
Then $$I=\int \frac{2}{\sqrt{1+t^4}} dt$$
This can be found interms of Elliptic functions.
