# Finding $\int \frac{1}{\sqrt{\tan x}}dx$ [duplicate]

$$\int \frac{1}{\sqrt{\tan x}}dx = ?$$

I tried:

$$\int \frac{\sin^2 x + \cos^2x}{\sqrt{\tan x}}dx = \int \left(\frac{\sin^2x}{\sqrt{\tan x}}+\frac{\cos^2x}{\sqrt{\tan x}}\right)dx = \\ \int \sqrt{\sin^3x\cos x}dx + \int \sqrt{\frac{\cos^5x}{\sin x}}dx$$

Now solving the first one individually:

$$\int \sqrt{\sin^3x\cos x}dx = \int \sqrt{\sin x \cos x}*\sin x dx = \int \sqrt{\frac{\sin(2x)}{2}}*\sin x dx = \\ \frac{1}{\sqrt{2}}\int \sqrt{\sin(2x)}*\sin dx$$

I tried to integrate by parts but didn't get anywhere. Still, here it is:

$$\int \sqrt{\sin(2x)}*\sin dx$$

$$v = \sqrt{\sin(2x}$$

$$dv = \frac{\cos(2x)dx}{2\sqrt{\sin(2x)}}$$

$$du = \sin x dx$$

$$u = -\cos(x)$$

$$1/\sqrt{2} [\sqrt{\sin(2x}*-\cos (x) + 1/2 \int \frac{\cos(2x)}{\sqrt{\sin(2x)}}*\cos x dx]$$

I just keep going in circles. What do I do?