Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers

up vote 9 down vote accepted

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

share|improve this answer
    
Makes sense now. Thanks! –  Joe Oct 10 '12 at 5:06
add comment

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

share|improve this answer
    
Ah, clever solution (+1). Thanks! –  Joe Oct 21 '12 at 5:08
add comment

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.

share|improve this answer
1  
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
1  
The magic $t$ substitution! I've ran into Weierstrass substitution a few times, thanks for the info. –  Joe Oct 10 '12 at 21:13
    
Nice, Babak!...hoping to see you before I turn in for the night...;-) –  amWhy Mar 26 '13 at 3:27
    
@amWhy: Thanks Amy, I hope so Angel. ;-) –  B. S. Mar 26 '13 at 6:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.