Integral: square root of tan x

In summary, the conversation was about integrating the square root of tan x and various methods and approaches were suggested. It was also mentioned that a similar integral involving the square root of sin x was more interesting to consider. The importance of verifying the integrability of a function before attempting to find its integral was also emphasized. A specific domain for the function was given and a discussion about equations and expressions ensued. Finally, a Calculus I student was encouraged to practice integrating the given equation.
  • #1
Jeremy
28
0
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
 
Physics news on Phys.org
  • #2
why don't you volunteer to get acreditted ?
 
  • #4
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
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
 
  • #6
Hmm..because it is wrong perhaps?
(Differentiate your last expression and see if you get your integrand)
 
  • #7
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
 
Last edited:
  • #8
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
 
  • #9
It's much more interesting to consider

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

Daniel.
 
  • #11
dextercioby said:
It's much more interesting to consider

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

Daniel.
As helpful as usual...
 
  • #12
dextercioby said:
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:
 
  • #13
arildo said:
[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]?
 
Last edited:
  • #14
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]
 
  • #15
dx said:
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
 
  • #16
[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/EllipticIntegrals/EllipticF/02/
 
Last edited:
  • #17
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.
 
Last edited:
  • #18

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
https://www.physicsforums.com/showpost.php?p=990017&postcount=10
 
Last edited:
  • #19
The argument of the real log function should ALWAYS be written within absolute value signs.
 
  • #20

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

Is this valid, correct?
 
Last edited:
  • #21
Yes, that is correct.
 
  • #22

Any Calculus I students interested in integrating this equation?
[tex]\int \frac{dx}{\sqrt{\tan x}}[/tex]
 
  • #23
Orion1 said:

Any Calculus I students interested in integrating this equation?
[tex]\int \frac{dx}{\sqrt{\tan x}}[/tex]
Not really. Oh, I forgot, I'm not in CalcI. :frown:
 
  • #24
also, that's not an equation :(
 
  • #25

[tex]\int \frac{1}{\sqrt{\tan x}} dx = F(x) + C[/tex]
 
Last edited:
  • #26
Orion1 said:

Data, why is that 'not an equation', please elaborate and clarify your statement.

It is an expression (more specifically, an integrand), not an equation. An equation symbolises the equal relationship between two expressions.
 
  • #27

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

Then what are your equations for this specific integrand expression?
 
Last edited:
  • #28
Orion1 said:

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

Then what are your equations for this specific integrand expression?

You mean, how would I evaluate that integral? Like Arildno said, I'll leave it to someone in CalcI. :biggrin:

No, seriously, I think it's better that a student does these, they stand to gain from the practice.
 
  • #29
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 :smile:
 
  • #30

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

Data Denominator Theorem:
[tex]\int \frac{1}{u(x)}} dx = F(x) + C[/tex]
 
Last edited:
  • #31
At any rate, it is a fairly easy integral. Two subtitutions and a partial fraction decomposition will give the answer. It's messy though, much neater to do a definite integral working out the substitutions along the way.
 
  • #32
Orion1 said:

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

Data Denominator Theorem:
[tex]\int \frac{1}{u(x)}} dx = F(x) + C[/tex]
"Data Denominator Theorem" ?! Did you invent that? It makes no sense to me at all.
 
  • #33
TD said:
"Data Denominator Theorem" ?! Did you invent that? It makes no sense to me at all.
Of course he did. :biggrin: :approve:
 
  • #34
Jeremy said:
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

it is not easy n also very lenthy n time taking
 
  • #35
http://math.ucsd.edu/~wgarner/math10b/int_sqrt_tan.htm [Broken]
 
Last edited by a moderator:
<h2>1. What is the integral of the square root of tangent x?</h2><p>The integral of the square root of tangent x is equal to <code>2/3 * tan^(3/2) x + C</code>, where C is a constant of integration.</p><h2>2. How do you solve an integral with a square root of tangent x?</h2><p>To solve an integral with a square root of tangent x, you can use the substitution method. Let <code>u = tan x</code> and <code>du = sec^2 x dx</code>. Then the integral becomes <code>∫√(tan x) dx = ∫√u * du = 2/3 * u^(3/2) + C = 2/3 * tan^(3/2) x + C</code>.</p><h2>3. Can the integral of the square root of tangent x be evaluated using basic integration rules?</h2><p>No, the integral of the square root of tangent x cannot be evaluated using basic integration rules. It requires the use of substitution or other advanced integration techniques.</p><h2>4. Is there a specific range of values for x that the integral of the square root of tangent x is defined?</h2><p>Yes, the integral of the square root of tangent x is only defined for values of x where <code>tan x > 0</code>. This means that x must be in the interval <code>(-π/2, π/2)</code> or any interval that is <code>2π</code> units away from it.</p><h2>5. Can the integral of the square root of tangent x be expressed in terms of elementary functions?</h2><p>Yes, the integral of the square root of tangent x can be expressed in terms of elementary functions. It can be written as <code>2/3 * tan^(3/2) x + C</code>, which is a combination of trigonometric and power functions.</p>

1. What is the integral of the square root of tangent x?

The integral of the square root of tangent x is equal to 2/3 * tan^(3/2) x + C, where C is a constant of integration.

2. How do you solve an integral with a square root of tangent x?

To solve an integral with a square root of tangent x, you can use the substitution method. Let u = tan x and du = sec^2 x dx. Then the integral becomes ∫√(tan x) dx = ∫√u * du = 2/3 * u^(3/2) + C = 2/3 * tan^(3/2) x + C.

3. Can the integral of the square root of tangent x be evaluated using basic integration rules?

No, the integral of the square root of tangent x cannot be evaluated using basic integration rules. It requires the use of substitution or other advanced integration techniques.

4. Is there a specific range of values for x that the integral of the square root of tangent x is defined?

Yes, the integral of the square root of tangent x is only defined for values of x where tan x > 0. This means that x must be in the interval (-π/2, π/2) or any interval that is units away from it.

5. Can the integral of the square root of tangent x be expressed in terms of elementary functions?

Yes, the integral of the square root of tangent x can be expressed in terms of elementary functions. It can be written as 2/3 * tan^(3/2) x + C, which is a combination of trigonometric and power functions.

Similar threads

  • Calculus
Replies
3
Views
3K
Replies
3
Views
900
Replies
20
Views
2K
Replies
3
Views
1K
  • Calculus
Replies
15
Views
4K
  • Calculus
Replies
6
Views
1K
Replies
1
Views
793
  • Introductory Physics Homework Help
Replies
14
Views
505
Replies
0
Views
508
Replies
1
Views
1K
Back
Top