Physics Forums

Physics Forums (http://www.physicsforums.com/index.php)
-   Calculus (http://www.physicsforums.com/forumdisplay.php?f=109)
-   -   Integral: square root of tan x (http://www.physicsforums.com/showthread.php?t=120727)

Jeremy May13-06 05:11 PM

integral: square root of tan x
 
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

Emieno May13-06 06:03 PM

why don't you volunteer to get acreditted ?

Orion1 May13-06 11:05 PM



Examine the given references listed in the archive.


Reference:
http://www.physicsforums.com/showthread.php?t=91866
http://www.physicsforums.com/showthread.php?t=83012

arildno May14-06 06:14 AM

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)

Hootenanny May14-06 06:32 AM

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

[tex]\int \sqrt{\tan x} \;\; dx = \int \tan^{\frac{1}{2}}x \;\; dx[/tex]

[tex] = \frac{3}{2}\tan^{\frac{3}{2}} x = \frac{3}{2}\tan x \sqrt{\tan x} [/tex]

~H

arildno May14-06 07:15 AM

Hmm..because it is wrong perhaps?
(Differentiate your last expression and see if you get your integrand)

arunbg May14-06 07:30 AM

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
[tex]u=tan(x)^{\frac{1}{2}}[/tex]

[tex]\frac{du}{dx}=\frac{sec^2(x)}{2\sqrt{tan(x)}}[/tex]
and then find du and so the integrand changes.
Just follow Orion's thread to see how it is done.

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.

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

Hootenanny May14-06 07:35 AM

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! :frown: Sorry guys!

~H

dextercioby May16-06 05:22 AM

It's much more interesting to consider

[itex] \int \sqrt{\sin x} \ dx [/itex]

Daniel.

Orion1 May16-06 07:11 AM

[itex]\int \sqrt{\tan x} \;\; dx = \frac{1}{2 \sqrt{2}} [ 2 \tan^{-1} (1 - \sqrt{2} \sqrt{\tan x} ) + 2 \tan^{-1} ( \sqrt{2} \sqrt{\tan x} + 1 ) + ...[/itex]

[itex]\ln (| - \tan (x) + \sqrt{2} \sqrt{\tan x} - 1 |) - \ln (| \tan x + \sqrt{2} \sqrt{\tan x} + 1 |)][/itex]

Reference:
http://www.physicsforums.com/showthread.php?t=91866
http://www.physicsforums.com/showthread.php?t=83012

nrqed May16-06 12:37 PM

Quote:

Quote by dextercioby
It's much more interesting to consider

[itex] \int \sqrt{\sin x} \ dx [/itex]

Daniel.

As helpful as usual...

arildno May16-06 01:13 PM

Quote:

Quote by dextercioby
It's much more interesting to consider

[itex] \int \sqrt{\sin x} \ dx [/itex]

Daniel.

What's interesting about the integral:
[tex]\int\frac{2u^{2}du}{\sqrt{1-u^{4}}}[/tex]

:confused:

Orion1 May16-06 10:19 PM

Quote:

Quote by arildo
[tex]\int\frac{2u^{2}du}{\sqrt{1-u^{4}}}[/tex]


[tex]\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}}}[/tex]

Arildno, what are you suggesting for [tex]u[/tex]?

dx May17-06 12:36 AM

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
[tex]\int{\frac{1}{\sqrt{sin x}}dx[/tex]

Curious3141 May17-06 01:20 AM

Quote:

Quote by 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
[tex]\int{\frac{1}{\sqrt{sin x}}dx[/tex]

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

Orion1 May17-06 01:26 AM

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

[tex]\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][/tex]

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

arildno May17-06 05:13 AM

Given a function f with domain D, the function
[tex]G(x)=\int_{x_{0}}^{x}f(y)dy, x_{0}, y, x\in{D}[/tex]
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.

Orion1 May17-06 06:27 AM


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

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

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

Placing the component in a point within its own domain produces:
[tex]\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)[/tex]

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

Reference:
http://mathworld.wolfram.com/Singularity.html
http://www.physicsforums.com/showpos...7&postcount=10


All times are GMT -5. The time now is 10:37 PM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums