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A bit tricky integral

  1. May 7, 2010 #1
    a bit tricky integral!!

    1. The problem statement, all variables and given/known data

    Integral of (sqrt(cot[x])dx)

    2. Relevant equations

    I just need a hint :)

    3. The attempt at a solution
     
  2. jcsd
  3. May 7, 2010 #2

    Mark44

    Staff: Mentor

    Re: a bit tricky integral!!

    I believe that an ordinary substitution will work: u = cot(x). At least, that's where I would start.

    An identity that might come in handy is cot2(x) + 1 = csc2(x).
     
  4. May 7, 2010 #3
    Re: a bit tricky integral!!

    I think that won't work ,, bcoz i'll get Integral of (Sqrt(u)/(1+u^2))du

    then what the next step?!!

    i also tried integration by parts after the step u mentioned ,, but it just gets more complicated :confused:

    More Help Is Appreciated :)
     
  5. May 7, 2010 #4

    Mark44

    Staff: Mentor

    Re: a bit tricky integral!!

    I think that I would go with a trig substitution next, tan w = u/1.

    I haven't worked this through, so caveat emptor!
     
  6. May 7, 2010 #5
    Re: a bit tricky integral!!

    what I would do is convert cot x to cos x/sin x. Let u = sin x and go from there.
     
  7. May 7, 2010 #6
    Re: a bit tricky integral!!

    So... √(du/u) ???
     
  8. May 7, 2010 #7
    Re: a bit tricky integral!!

    That won't work out, will it? The square root poses a problem then.
     
  9. May 7, 2010 #8
    Re: a bit tricky integral!!

    Umm ,,, i tried what you said ,, i think that will turn the integral back to itself :S

    except i'll get Integral of Sqrt(tan(x))dx

    I'm really stuck now :biggrin:

    Thanks by the way
    I appreciate it :D

    and if you get any ideas i'll be glad to try some :)
     
  10. May 7, 2010 #9

    Cyosis

    User Avatar
    Homework Helper

    Re: a bit tricky integral!!

    This is quite a cumbersome integral. Start out with the substitution given by Mark in post #2 then make a substitution [itex]y=\sqrt{u}[/itex]. Enjoy the partial fractions.
     
  11. May 7, 2010 #10

    Mark44

    Staff: Mentor

    Re: a bit tricky integral!!

    Here's another tack: Let u = sqrt(cot(x)), du = -csc^2(x)dx/2sqrt(cot(x))

    This turns the original integral into -2 int{u^2/(u^4 + 1) * du}

    I think that one is amenable to a trig substitution, tan w = u^2

    This is a quite messy integral!
     
  12. May 7, 2010 #11
    Re: a bit tricky integral!!

    are you sure about the partial fractions? ,, cuz i didn't enjoy any !!

    i got the Integral of (2(y^2) / (1 + y^4))dy !!
     
  13. May 7, 2010 #12
    Re: a bit tricky integral!!

    it tried it ,, it would also give me : (constant) times int{Sqrt(tan w) dw}
     
  14. May 7, 2010 #13

    Cyosis

    User Avatar
    Homework Helper

    Re: a bit tricky integral!!

    At this step you need to integrate a rational function. To do that use partial fractions.
     
  15. May 7, 2010 #14
    Re: a bit tricky integral!!

    at what step exactly?!

    and how can i do that ?! would you please show me how :) :blushing:
     
  16. May 7, 2010 #15
    Re: a bit tricky integral!!

    hey guys ,, i managed to make the integral look like this [2][int{1/(1+(u^4)) du}]

    would that be any good !!
     
  17. May 7, 2010 #16

    Cyosis

    User Avatar
    Homework Helper

    Re: a bit tricky integral!!

    You need to use partial fractions to continue from:

    [tex]
    -2\int \frac{y^2}{y^4+1}
    [/tex]
     
  18. May 7, 2010 #17
    Re: a bit tricky integral!!

    First, make a substitution:

    [tex]
    u = \sqrt{\cot(x)}.
    [/tex]

    Then you will get an integral of a rational function. The denominator is [itex]u^{4} + 1[/itex]. In order to factorize it, you will need all the complex fourth roots of [itex]-1[/itex]. These are:

    [tex]

    \begin{array}{rcl}

    e^{\iota \frac{\pi}{4}} & = & \frac{1 + \iota}{\sqrt{2}} \\

    e^{\iota \frac{3 \pi}{4}} & = & \frac{-1 + \iota}{\sqrt{2}} \\

    e^{\iotai \frac{5 \pi}{4}} & = & \frac{-1 - \iota}{\sqrt{2}} \\

    e^{\iota \frac{7 \pi}{4}} & = & \frac{1 - \iota}{\sqrt{2}}
    \end{array}
    [/tex]

    i.e. you have 2 pairs of complex conjugate roots. Therefore:

    [tex]
    u^{4} + 1 = (u^{2} + \sqrt{2}u + 1)(u^{2} - \sqrt{2}u + 1)
    [/tex]

    From here, you should be able to proceed in the standard fashion.
     
  19. May 7, 2010 #18
    Re: a bit tricky integral!!

    But i can't factor (x^4)+1

    i mean it's (x^4)+1 ,, not (x^4)-1

    i don't know how to use partial fractions with that !!
     
  20. May 7, 2010 #19
    Re: a bit tricky integral!!

    Ahaaaa ,, that's entirely awesome :D

    thanks soooooo much ,, to all of you guys :)

    now ,, i can go to rest :)
     
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