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Integral of sin^11x

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\int sin^{11}x.dx[/tex]


    2. Relevant equations



    3. The attempt at a solution
    [tex]\int (sin^{2}x)^{5}.sinx.dx[/tex]
    [tex]\int (1-cos^{2}x)^{5}.sinx.dx[/tex]

    let cosx be u, statement became

    [tex]- \int (1-u^{2})^{5}.du[/tex]

    and I'm stuck here. Any help is appreciated,
    thank you
     
  2. jcsd
  3. Nov 17, 2008 #2

    Dick

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    Why don't you just multiply the expression (1-u^2)^5 out to get just powers of u?
     
    Last edited: Nov 17, 2008
  4. Nov 17, 2008 #3
    This is potentially one of the 10 questions to be asked in tomorrow's 45 minutes exam, there should be a quicker way...

    Plus expanding the whole expression would be too long and impractical..

    Thanks anyways,
    any other suggestions?
     
  5. Nov 17, 2008 #4

    Dick

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    (1+a)^5=1+5a+10a^2+10a^3+5a^4+a^5. Put a=-u^2. It's pretty easy if you remember the binomial theorem. I can't think of anything easier.
     
  6. Nov 17, 2008 #5
    No quicker way,EXPAND it!
     
  7. Nov 17, 2008 #6
    [tex]
    \int {\sin ^k xdx} = \int {\sin ^{k - 1} x\sin xdx}
    [/tex]

    IBP:
    [tex]

    - \cos x\sin ^{k - 1} x - \int { - \cos x\left( {k - 1} \right)\sin ^{k - 2} x} \cos xdx

    [/tex]

    rewrite:


    [tex]

    - \cos x\sin ^{k - 1} x + \left( {k - 1} \right)\int {\cos ^2 x\sin ^{k - 2} x} dx

    [/tex]

    Use the identity:

    [tex]

    \cos ^2 x\sin ^{k - 2} x = \left( {1 - \sin ^2 x} \right)\sin ^{k - 2} x = \sin ^{k - 2} x - \sin ^k x

    [/tex]

    And so:

    [tex]

    - \cos x\sin ^{k - 1} x + \left( {k - 1} \right)\int {\sin ^{k - 2} x} dx - \left( {k - 1} \right)\int {\sin ^k x} dx

    [/tex]

    [tex]

    k\int {\sin ^k xdx} = - \cos x\sin ^{k - 1} x + \left( {k - 1} \right)\int {\sin ^{k - 2} x} dx

    [/tex]

    Conclusion:
    [tex]

    \int {\sin ^k xdx} = - \frac{{\cos x\sin ^{k - 1} x}}{k} + \frac{{k - 1}}{k}\int {\sin ^{k - 2} x} dx

    [/tex]
     
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