# Integral: square root of tan x

1. ### Jeremy

24
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. ### Emieno

109
why don't you volunteer to get acreditted ?

990
4. ### arildno

12,015
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. ### Hootenanny

9,681
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. ### arildno

12,015
Hmm..because it is wrong perhaps?

7. ### arunbg

600
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. ### Hootenanny

9,681
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. ### dextercioby

12,303
It's much more interesting to consider

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

Daniel.

10. ### Orion1

990
Last edited: May 17, 2006

3,048

12. ### arildno

12,015
$$\int\frac{2u^{2}du}{\sqrt{1-u^{4}}}$$

13. ### Orion1

990

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

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

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

2,000
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. ### Curious3141

2,970
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. ### Orion1

990
$$F(z|m) = \text{EllipticF}[z,m] = \int_0^z \frac{1}{\sqrt{1 - m \sin^2 t}} dt$$

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

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

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

12,015
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. ### Orion1

990

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. ### arildno

12,015
The argument of the real log function should ALWAYS be written within absolute value signs.

20. ### Orion1

990

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