Integral: square root of tan x

  1. 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. jcsd
  3. why don't you volunteer to get acreditted ?
     
  4. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

    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

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    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. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

    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: May 14, 2006
  8. Hootenanny

    Hootenanny 9,681
    Staff Emeritus
    Science Advisor
    Gold Member

    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. dextercioby

    dextercioby 12,309
    Science Advisor
    Homework Helper

    It's much more interesting to consider

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

    Daniel.
     
  10. Last edited: May 17, 2006
  11. nrqed

    nrqed 3,048
    Science Advisor
    Homework Helper

    As helpful as usual...
     
  12. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

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

    :confused:
     

  13. [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: May 17, 2006
  14. dx

    dx 2,004
    Homework Helper
    Gold Member

    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. Curious3141

    Curious3141 2,970
    Homework Helper

    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: May 17, 2006
  17. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

    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: May 17, 2006

  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: May 17, 2006
  19. arildno

    arildno 12,015
    Science Advisor
    Homework Helper
    Gold Member

    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: May 17, 2006
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?