Evaluating $\int \frac {\sqrt{\tan \theta}} {\sin 2\theta} \ d \theta$ 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.
 A: Let $ \theta=\arctan x$ and use the fact that $\sin(2 \arctan x)=\displaystyle\frac{2x}{x^2+1}$
$$\int \frac {\sqrt{\tan \theta}} {\sin 2\theta} \ d \theta=\int \frac {\sqrt{x}} {\displaystyle\frac{2x}{x^2+1}} \cdot \frac{1}{x^2+1} \ dx=\sqrt x +C=\sqrt {\tan \theta} +C $$
A: $$\begin{aligned}\sin 2\theta &= 2\sin\theta\cos\theta\\&=2\tan\theta\cos^2\theta\end{aligned} $$
$$\begin{aligned}\int\frac{\sqrt{\tan\theta}}{\sin2\theta}\,\mathrm{d}\theta&=\int\frac{\sqrt{\tan\theta}}{2\tan\theta}\sec^2\theta\,\mathrm{d}\theta\\&=\int\frac{1}{2\sqrt{\tan\theta}}\sec^2\theta\,\mathrm{d}\theta\end{aligned} $$
Now, set $u = \tan\theta$ and $\mathrm{d}u=\sec^2\theta\,\mathrm{d}\theta$ to get
$$\int\frac{\sqrt{\tan\theta}}{\sin2\theta}\,\mathrm{d}\theta = \sqrt{\tan\theta}+C $$
A: Substituting $u=\tan(\theta)$ yields
$$
\begin{align}
\int\frac{\sqrt{\tan(\theta)}}{\sin(2\theta)}\,\mathrm{d}\theta
&=\int\frac{\sqrt{u}}{\frac{2u}{1+u^2}}\frac{\mathrm{d}u}{1+u^2}\\
&=\int\frac1{2\sqrt{u}}\,\mathrm{d}u\\
&=\sqrt{u}+C\\
&=\sqrt{\tan(\theta)}+C
\end{align}
$$
The Substitution
if $u=\tan(\theta)$, then
$$
\sin(2\theta)=2\sin(\theta)\cos(\theta)=\frac{2\tan(\theta)}{\sec^2(\theta)}=\frac{2u}{1+u^2}
$$
Furthermore,
$$
\mathrm{d}u=\sec^2(\theta)\,\mathrm{d}\theta=(1+u^2)\,\mathrm{d}\theta
$$
Therefore,
$$
\mathrm{d}\theta=\frac{\mathrm{d}u}{1+u^2}
$$
A: Just for noting a point here describing why @robjohn's substitution works.

Let you want to solve $\int R\big(\sin(x),\cos(x)\big)dx$ and you know that $$R\big(-\sin(x),-\cos(x)\big)\equiv R\big(\sin(x),\cos(x)\big)$$ (Check the integrand for that); then you can always take $\tan(x)=t$ for a good substitution.

