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How do you take integral of cot ?

  1. Mar 24, 2005 #1
    integral of (cot2x)^3 dx

    are you supposet to convert to cos and sin?
     
  2. jcsd
  3. Mar 24, 2005 #2
    Try using the identity [itex] \csc^2{x} - \cot^2{x} = 1[/itex] and later making the substitution [itex]u = \csc{2x}[/itex] (similar to integrals of powers of sines and cosines, or secants and tangents. Of course, you'll have to deduce the differential of u! :smile:).
     
    Last edited: Mar 24, 2005
  4. Mar 24, 2005 #3

    xanthym

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    The answer to your question for this case: YES (for easy solution)
    Then sub {w = sin(2*x) ::AND:: dw = 2*cos(2*x)*dx ::OR:: (dw/2) = cos(2*x)*dx}:

    ∫ {cot(2*x)}^3 dx = ∫ {cos(2*x)/sin(2*x)}^3 dx =
    = ∫ {cos(2*x)}^2/{sin(2*x)}^3 {cos(2*x)}dx =

    = ∫ (1 - w^2)/w^3 {dw/2} = ::: Sub "w" using {cos^2() = 1 - sin^2()} in num
    = (1/2)*∫ {w^(-3) - w^(-1)} dw =
    ::: (integrating) :::

    = (1/2)*{(-1/2)*w^(-2) - Loge(|w|)} + C =
    = (-1/4)*{sin(2*x)}^(-2) - (1/2)*Loge{|sin(2*x)|} + C =

    = (-1/4)*{csc(2*x)}^2 - (1/4)*Loge{{sin(2*x)}^2} + C


    ~~
     
    Last edited: Mar 25, 2005
  5. Mar 24, 2005 #4
    Here's the solution using my substitution:

    Note that [itex] u = \csc(2x)[/itex] and thus [itex] du = -2\csc(2x)\cot(2x)dx[/itex]

    [tex] \int \cot^3(2x) \ dx = \int (\csc^2(2x) - 1)\cot(2x) \ dx[/tex]
    [tex] = \frac{-1}{2}\int \frac{u^2 - 1}{u} \ du = \frac{-1}{2}\left( \frac{u^2}{2} - \ln{|u|} \right) + C [/tex]

    [tex]= \frac{1}{4}\ln \left(\csc^2(2x)\right) - \frac{1}{4}\csc^2(2x) + C[/tex]
     
    Last edited: Mar 24, 2005
  6. Mar 24, 2005 #5
    wow, 2 different solutions! thanks a lot guys, this helps a lot.
     
  7. Mar 24, 2005 #6

    xanthym

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    The 2 solutions presented in this thread are equivalent in final results. Note that:
    csc^2() = sin^(-2)()
    ::: ⇒ Loge{csc^2()} = (-1)*Loge{sin^2()}

    Thus, we have:

    [tex] :(1): \ \ \ \ \frac{1}{4}\ln \left(\csc^2(2x)\right) \ - \ \frac{1}{4}\csc^2(2x) \ + \ C \ \ = \ \ \frac{-1}{4}\csc^2(2x) \ + \ \color{red} (-1)*\color{black}\frac{1}{4}\ln \left(sin^2(2x)\right) \ + \ C [/tex]

    which is equivalent to that presented in the EASY (:wink: :wink:) solution in msg #3 involving only sin() & cos() functions, identities, & derivatives (except for the very last equation)!!


    ~~
     
    Last edited: Mar 25, 2005
  8. Mar 24, 2005 #7
    There's no error. I just squared the csc inside the logarithm and as a result pulled out an extra [tex]\frac{1}{2}[/tex]. Doing it that way gets rid of the absolute value signs for one thing. Note that in your original solution the argument to [itex]\ln[/itex] was [itex] \sin{2x}[/itex], not [itex]\sin^2{2x}[/itex] :smile:
     
  9. Mar 24, 2005 #8

    xanthym

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    Indeed, there was no error. And the squared Loge argument is a good idea. After adjusting orig msgs, both solutions are now equiv in final result.


    ~~
     
    Last edited: Mar 25, 2005
  10. Mar 24, 2005 #9
    Well, we actually did both make a mistake. Pesky constants of integration :wink:
     
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