integral: square root of tan x


by Jeremy
Tags: integral, root, square
Jeremy
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#1
May13-06, 05:11 PM
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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
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Emieno
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#2
May13-06, 06:03 PM
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why don't you volunteer to get acreditted ?
Orion1
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#3
May13-06, 11:05 PM
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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
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#4
May14-06, 06:14 AM
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integral: square root of tan x


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
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#5
May14-06, 06:32 AM
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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
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May14-06, 07:15 AM
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Hmm..because it is wrong perhaps?
(Differentiate your last expression and see if you get your integrand)
arunbg
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#7
May14-06, 07:30 AM
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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
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#8
May14-06, 07:35 AM
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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
dextercioby
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May16-06, 05:22 AM
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It's much more interesting to consider

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

Daniel.
Orion1
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#10
May16-06, 07:11 AM
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[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
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May16-06, 12:37 PM
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Quote Quote by dextercioby
It's much more interesting to consider

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

Daniel.
As helpful as usual...
arildno
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May16-06, 01:13 PM
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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]

Orion1
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#13
May16-06, 10:19 PM
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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
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May17-06, 12:36 AM
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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
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May17-06, 01:20 AM
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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
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May17-06, 01:26 AM
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[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
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May17-06, 05:13 AM
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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
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#18
May17-06, 06:27 AM
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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


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