How Do You Integrate the Square Root of Tan x?

[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

 

[Emieno]

why don't you volunteer to get acreditted ?

 

[Orion1]

Examine the given references listed in the archive.

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

 

[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)

 

[Hootenanny]

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

 

[arildno]

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

 

[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.
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.

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

 

[Hootenanny]

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]

It's much more interesting to consider

\int \sqrt{\sin x} \ dx

Daniel.

 

[Orion1]

\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 ) + ...

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

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

 

[nrqed]

It's much more interesting to consider

\int \sqrt{\sin x} \ dx

Daniel.

As helpful as usual...

 

[arildno]

It's much more interesting to consider

\int \sqrt{\sin x} \ dx

Daniel.

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

 

[Orion1]

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

\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?
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[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

 

[Curious3141]

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

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]

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/

 

[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.

 

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

 

[arildno]

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

 

[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?
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[arildno]

Yes, that is correct.

 

[Orion1]

Any Calculus I students interested in integrating this equation?
\int \frac{dx}{\sqrt{\tan x}}
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[arildno]

Any Calculus I students interested in integrating this equation?
\int \frac{dx}{\sqrt{\tan x}}
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Not really. Oh, I forgot, I'm not in CalcI.

 

[Data]

also, that's not an equation :(

 

[Orion1]

\int \frac{1}{\sqrt{\tan x}} dx = F(x) + C
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[Curious3141]

Data, why is that 'not an equation', please elaborate and clarify your statement.
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It is an expression (more specifically, an integrand), not an equation. An equation symbolises the equal relationship between two expressions.

 

[Orion1]

I understand, I posted a short-hand integrand expression and called it an equation.

Then what are your equations for this specific integrand expression?
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[Curious3141]

I understand, I posted a short-hand integrand expression and called it an equation.

Then what are your equations for this specific integrand expression?
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You mean, how would I evaluate that integral? Like Arildno said, I'll leave it to someone in CalcI.

No, seriously, I think it's better that a student does these, they stand to gain from the practice.

 

[Data]

Integrating negative powers of tangent isn't very different from integrating positive powers of tangent, because of the way it's defined. So I think a calcI student should be able to handle that one pretty well, given the discussion already in this thread

 

[Orion1]

I understand, a CalcI student should use the Data Denominator Theorem.

Data Denominator Theorem:
\int \frac{1}{u(x)}} dx = F(x) + C
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