# 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.

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

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Makes sense now. Thanks! –  Joe Oct 10 '12 at 5:06

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

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Ah, clever solution (+1). Thanks! –  Joe Oct 21 '12 at 5:08
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.
Also, in many cases of rational trigonometric functions (and more), the $z$-substitution (of which this is a thinly veiled example) is very useful. –  robjohn Oct 10 '12 at 20:03
The magic $t$ substitution! I've ran into Weierstrass substitution a few times, thanks for the info. –  Joe Oct 10 '12 at 21:13