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The integral of cot^3(x)

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data
    Integrate
    ∫[itex]cot^{3}(x)[/itex]


    2. Relevant equations

    u*v-∫vdu

    3. The attempt at a solution

    I used the integration by parts formula and I got:

    [itex]cot^{3}(x)[/itex](ln|sin(x)|)-∫ln|sin(x)|-x-cot(x)dx

    I don't know how to integrate the integrand.
     
  2. jcsd
  3. Feb 14, 2012 #2

    SammyS

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    Please ... let us know what you used for u & v. We could guess, but why make us guess ?
     
  4. Feb 14, 2012 #3

    LCKurtz

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    Try writing ##\cot^3(x) = \cot^2x\cot x = (\csc^2(x)-1)\cot x##. Then a couple of appropriate u-substitutions might work for you.
     
  5. Feb 14, 2012 #4
    [itex]cot^{2}x[/itex] as u, [itex]-csc^{2}x[/itex] as du
    cot(x) as dv, ln|sin(x)| as v



    That was what I tried to do first. I made u=cot(x), du= -[itex]csc^{2}x[/itex]
    when i integrate i get [itex]u^{3}[/itex]/3 - u
     
  6. Feb 14, 2012 #5

    LCKurtz

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    The u sub only works for the first term. Try writing ##\cot x = \frac{\cos x}{\sin x}## for the second term.
     
  7. Feb 14, 2012 #6
    Okay, now i got:

    ∫(1-csc(x))*[itex]\frac{cos(x)}{sin(x)}[/itex]

    ∫(1-[itex]\frac{1}{sin(x)}[/itex] * [itex]\frac{cos(x)}{sin(x)}[/itex]

    u=sin(x), du=cos(x)

    ∫(1-[itex]\frac{1}{u}[/itex])

    u-lnu

    sin(x)-ln|sin(x)|
     
  8. Feb 14, 2012 #7

    Dick

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    No, you don't have it yet. But you are getting closer. Write cot(x)^3=cos(x)^3/sin(x)^3. Now try u=sin(x).
     
  9. Feb 14, 2012 #8
    I just ended up with ln|sin(x)| I don't know what I'm doing wrong.
     
  10. Feb 14, 2012 #9

    Dick

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    It would be really hard to say what you are doing wrong if you don't show what you are doing. Now wouldn't it??
     
  11. Feb 14, 2012 #10
    ∫ cos(x)^3/sin(x)^3.

    u=sin(x), du=cos(x)

    ∫1/u

    ln(u)

    ln|sin(x)|
     
  12. Feb 14, 2012 #11

    Dick

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    No. What happened the power 3? That would be ok, if it was cos(x)/sin(x). It's not. It's cos(x)^3/sin(x)^3.
     
  13. Feb 14, 2012 #12
    ∫[itex]\frac{1}{u^3}[/itex]

    [itex]\frac{2}{sin^2}[/itex]

    Is this right?
     
  14. Feb 14, 2012 #13

    Dick

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    Not even a little. If you substitute u=sin(x) du=cos(x) dx into [itex]\int \frac{cos^3(x)}{sin^3(x)} dx[/itex] you get [itex]\int \frac{cos^2(x)}{u^3} du[/itex]. Now you just need to express cos(x)^2 in terms of u.
     
    Last edited: Feb 14, 2012
  15. Feb 14, 2012 #14
    ∫[itex]\frac{u^2}{u^3}[/itex]

    [itex]\frac{2u^3}{u^2}[/itex]

    [itex]\frac{2cos^3}{cos^2}[/itex]



    I don't think I'm doing it right. Trigonometric Integrals confuse me.
     
  16. Feb 14, 2012 #15

    Dick

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    Yes, they do confuse you. cos(x)^2=1-sin(x)^2. Express that in terms of u.
     
  17. Feb 14, 2012 #16

    [itex]\frac{1-sin^2(x)}{sin^3}[/itex]
     
  18. Feb 14, 2012 #17

    Dick

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    This is not going well. cos(x)^2=1-sin(x)^2=1-u^2. You might be too tired right now to think straight. I know I am. Gotta go now.
     
  19. Feb 14, 2012 #18
    That is not in terms of u. :p
    u = sin x, so:
    [itex]∫\frac{1-u^2}{u^3}du[/itex]
     
  20. Feb 14, 2012 #19
    I think it may be the fact that I'm very bad at math. Thanks for the help though, I appreciate it.

    do i integrate that?
     
  21. Feb 14, 2012 #20
    Also, just another way to solve, if you want some practice :devil::

    Mod note: Removed complete solution.

    Yes, split the faction into partial fractions. All you have is 1/u^3 - u^2/u^3.
     
    Last edited by a moderator: Feb 14, 2012
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