TITLE: Evaluating $\int \frac {\sqrt{\tan \theta}} {\sin 2\theta} \ d \theta$
QUESTION [16 upvotes]: 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.

REPLY [2 votes]: $$\begin{aligned}\sin 2\theta &= 2\sin\theta\cos\theta\\&=2\tan\theta\cos^2\theta\end{aligned} $$
$$\begin{aligned}\int\frac{\sqrt{\tan\theta}}{\sin2\theta}\,\mathrm{d}\theta&=\int\frac{\sqrt{\tan\theta}}{2\tan\theta}\sec^2\theta\,\mathrm{d}\theta\\&=\int\frac{1}{2\sqrt{\tan\theta}}\sec^2\theta\,\mathrm{d}\theta\end{aligned} $$
Now, set $u = \tan\theta$ and $\mathrm{d}u=\sec^2\theta\,\mathrm{d}\theta$ to get
$$\int\frac{\sqrt{\tan\theta}}{\sin2\theta}\,\mathrm{d}\theta = \sqrt{\tan\theta}+C $$