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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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4 Answers 4

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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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  • $\begingroup$ Makes sense now. Thanks! $\endgroup$
    – Joe
    Commented Oct 10, 2012 at 5:06
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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.

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    $\begingroup$ Also, in many cases of rational trigonometric functions (and more), the $z$-substitution (of which this is a thinly veiled example) is very useful. $\endgroup$
    – robjohn
    Commented Oct 10, 2012 at 20:03
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    $\begingroup$ The magic $t$ substitution! I've ran into Weierstrass substitution a few times, thanks for the info. $\endgroup$
    – Joe
    Commented Oct 10, 2012 at 21:13
  • $\begingroup$ Nice, Babak!...hoping to see you before I turn in for the night...;-) $\endgroup$
    – amWhy
    Commented Mar 26, 2013 at 3:27
  • $\begingroup$ @amWhy: Thanks Amy, I hope so Angel. ;-) $\endgroup$
    – Mikasa
    Commented Mar 26, 2013 at 6:56
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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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  • $\begingroup$ Ah, clever solution (+1). Thanks! $\endgroup$
    – Joe
    Commented Oct 21, 2012 at 5:08
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$$\begin{aligned}\sin 2\theta &= 2\sin\theta\cos\theta\\&=2\tan\theta\cos^2\theta\end{aligned} $$ $$\begin{aligned}\int\frac{\sqrt{\tan\theta}}{\sin2\theta}\,\mathrm{d}\theta&=\int\frac{\sqrt{\tan\theta}}{2\tan\theta}\sec^2\theta\,\mathrm{d}\theta\\&=\int\frac{1}{2\sqrt{\tan\theta}}\sec^2\theta\,\mathrm{d}\theta\end{aligned} $$ Now, set $u = \tan\theta$ and $\mathrm{d}u=\sec^2\theta\,\mathrm{d}\theta$ to get $$\int\frac{\sqrt{\tan\theta}}{\sin2\theta}\,\mathrm{d}\theta = \sqrt{\tan\theta}+C $$

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