Evaluating $\int \frac {\sqrt{\tan \theta}} {\sin 2\theta} \ d \theta$

Question

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.

Answer 1

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, $$ \mathrm{d}\theta=\frac{\mathrm{d}u}{1+u^2} $$

Answer 2

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.

Answer 3

Let $ \theta=\arctan x$ and use the fact that $\sin(2 \arctan x)=\displaystyle\frac{2x}{x^2+1}$ $$\int \frac {\sqrt{\tan \theta}} {\sin 2\theta} \ d \theta=\int \frac {\sqrt{x}} {\displaystyle\frac{2x}{x^2+1}} \cdot \frac{1}{x^2+1} \ dx=\sqrt x +C=\sqrt {\tan \theta} +C $$