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