I am trying to evaluate
$\int \frac {\sqrt{\tan \theta}} {\sin 2\theta} \ d \theta$
I tried rewriting it as $\int {\sqrt{\tan \theta}} \cdot \csc(2\theta) \ d\theta$
Supposedly letting $u = \sqrt{\tan \theta}$ cleans up the integral to just $1$, but I don't see how. $du = \frac{\sec^2 \theta}{2\sqrt{\tan(\theta)}} d\theta \implies 2u \ du = \sec^2 \theta \ d\theta$
Here's where I'm not sure how expressing $\csc(2\theta)$ as a double angle and in terms of $u$ cleans it up so nicely.
Note: a subtle hint or nudge in the right direction is preferred rather than a full solution.