# Integral: square root of tan x

1. May 13, 2006

### Jeremy

My class, teacher included, cannot seem to figure out the integral of the square root of tan x. Maybe someone here can help?

thanks,
jeremy

2. May 13, 2006

### Emieno

why don't you volunteer to get acreditted ?

3. May 13, 2006

### Orion1

4. May 14, 2006

### arildno

The simplest way is to set u=sqrt(tan(x)); you'll end up with a rational integrand that you may decompose with partial fractions.
(Remember that sec^2(x)=tan^2(x)+1=u^4+1)

5. May 14, 2006

### Hootenanny

Staff Emeritus
This may be a bit simplistic but why can't you simply do;

$$\int \sqrt{\tan x} \;\; dx = \int \tan^{\frac{1}{2}}x \;\; dx$$

$$= \frac{3}{2}\tan^{\frac{3}{2}} x = \frac{3}{2}\tan x \sqrt{\tan x}$$

~H

6. May 14, 2006

### arildno

Hmm..because it is wrong perhaps?

7. May 14, 2006

### arunbg

Well Hoot, what you have done is considered tan(x)=u and integrated
u^1/2 du .But you haven't changed dx to du.You can do this as
$$u=tan(x)^{\frac{1}{2}}$$

$$\frac{du}{dx}=\frac{sec^2(x)}{2\sqrt{tan(x)}}$$
and then find du and so the integrand changes.

We had the exact same question for our final board exams in India.
It took me 10 mins of precious time and two pages of trial to finally get to the answer( a very big one mind you).And to think you did it all for 3 marks in a 100 mark paper.Phew!

PS:Something wrong with latex? I just can't seem to edit them.

PPSoot, even if you are integrating u^(1/2) it would be 2/3u^3/2

Last edited: May 14, 2006
8. May 14, 2006

### Hootenanny

Staff Emeritus
Ahh, dammit! I knew it was too simple. It's a repeat of my xmas exams when I did a very similar thing with secant! Sorry guys!

~H

9. May 16, 2006

### dextercioby

It's much more interesting to consider

$\int \sqrt{\sin x} \ dx$

Daniel.

10. May 16, 2006

### Orion1

Last edited: May 17, 2006
11. May 16, 2006

### nrqed

12. May 16, 2006

### arildno

$$\int\frac{2u^{2}du}{\sqrt{1-u^{4}}}$$

13. May 16, 2006

### Orion1

$$\int\frac{2u^{2}du}{\sqrt{1-u^{4}}} = \frac{-2\,{\sqrt{1 - u^2}}\,{\sqrt{1 + u^2}}\,\left( -\text{EllipticE}(\sin^{-1} u,-1) + \text{EllipticF}(\sin^{-1} u,-1) \right) }{{\sqrt{1 - u^4}}}$$

Arildno, what are you suggesting for $$u$$?

Last edited: May 17, 2006
14. May 17, 2006

### dx

while were on the topic of integrating expressions that contain square roots of trigonometric functions, I was having a hard time a while ago evaluating this
$$\int{\frac{1}{\sqrt{sin x}}dx$$

15. May 17, 2006

### Curious3141

It is always a good idea when confronted with an unfamiliar integral, to verify that it can be done before expending effort to figure out how. Mathematica is a good tool, or you use the free WebMathematica equivalent at http://integrals.wolfram.com/index.jsp

16. May 17, 2006

### Orion1

$$F(z|m) = \text{EllipticF}[z,m] = \int_0^z \frac{1}{\sqrt{1 - m \sin^2 t}} dt$$

$$\int{\frac{1}{\sqrt{\sin x}}dx = \int_0^{\frac{1}{2} \left( \frac{\pi}{2} - x \right)} \frac{1}{\sqrt{1 - 2 \sin^2 t}} dt = -2\text{EllipticF} \left[ \frac{1}{2} \left( \frac{\pi}{2} - x \right), 2 \right]$$

Reference:
http://functions.wolfram.com/EllipticIntegrals/EllipticF/02/

Last edited: May 17, 2006
17. May 17, 2006

### arildno

Given a function f with domain D, the function
$$G(x)=\int_{x_{0}}^{x}f(y)dy, x_{0}, y, x\in{D}$$
is seen to have no larger domain than f. Since the definite integral can't generate any singularities on its own (integration is a "smoothing" process), it is seen that G doesn't have a less domain than f.
Thus, G has the same domain as f.

Last edited: May 17, 2006
18. May 17, 2006

### Orion1

Given that $\sqrt{\tan x}$ is valid in Quadrants I,III then the specific domains for this function are:

$$D: \left[ 0, \frac{\pi}{2} \right) \; \; \; I$$
$$D: \left[ \pi, \frac{3 \pi}{2} \right) \; \; \; III$$

The third equation component in post #10 is:
$$\ln ( - \tan x + \sqrt{2} \sqrt{\tan x} - 1 )$$

Placing the component in a point within its own domain produces:
$$\ln \left( - \tan \frac{\pi}{4} + \sqrt{2} \sqrt{\tan \frac{\pi}{4}} - 1 \right) = \ln ( - 1 + \sqrt{2} \sqrt{1} - 1) = \ln ( \sqrt{2} - 2)$$

Taking the 'sign' of internal component $\ln [sgn(\sqrt{2} - 2)]$ yields:
$$\ln (-1)$$

Reference:
http://mathworld.wolfram.com/Singularity.html
https://www.physicsforums.com/showpost.php?p=990017&postcount=10

Last edited: May 17, 2006
19. May 17, 2006

### arildno

The argument of the real log function should ALWAYS be written within absolute value signs.

20. May 17, 2006

### Orion1

Then the third equation component in post #10 should actually be:
$$\ln (| -\tan x + \sqrt{2} \sqrt{\tan x} - 1 |)$$

Is this valid, correct?

Last edited: May 17, 2006