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Integration by parts

  1. Jun 8, 2013 #1
    1. The problem statement, all variables and given/known data


    Hi, I attached the question.Just integral trouble.

    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jun 8, 2013 #2

    SteamKing

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    I don't understand the expression you have written after "This is what I want". You have an integral in the denominator of an expression. I don't see how this is supposed to result in a recursion formula. Have you checked you attachment to make sure all expressions are correctly written?
     
  4. Jun 8, 2013 #3
    There isn't it is (n-1)/n (times) the integral It isn't supposed to be in the denominator.

    ∫〖sin〗^n x dx= (-〖sin〗^(n-1) xcosx)/n)+ ((n-1)/n)∫〖sin〗^(n-2) x dx

    thx
     
  5. Jun 9, 2013 #4

    haruspex

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    I find that hard to read. Do you mean ##\int \sin^n(x).dx = -\frac1n \sin^{n-1}(x)\cos(x) + \frac{n-1}n \int \sin^{n-2}(x).dx##? Looks right.
     
  6. Jun 9, 2013 #5
    Yeah that is what I mean.

    See I'm trying to figure out how to get to that. On my attachment there is a problem before that where you get

    \int \sin^n(x).dx = -\frac1n \sin^{n-1}(x)\cos(x) + \frac{n-1}n \int \sin^{n-2}(x)cos^{2}.dx
    They tell you to replace cosx^2 in the second integral and get to
    \int \sin^n(x).dx = -\frac1n \sin^{n-1}(x)\cos(x) + \frac{n-1}n \int \sin^{n-2}(x).dx

    I'm having trouble with that.
     
  7. Jun 9, 2013 #6
    How come it doesn't work then I copy and pasted your code for the equation? Anyways my first equation in the attachment I need to get to the one you have written above. Just look at my attachment it is all there/
     
  8. Jun 9, 2013 #7

    haruspex

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    For some reason, I can never read the .docx files you attach. I see the plain text, but equations and diagrams are blank.
    In the latex, you left out the double hash (##) at start and end.
    I don't understand how you got the first of those two equations. You should have
    ##\int \sin^n(x).dx = [-\sin^{n-1}(x)\cos(x)] + \int (n-1)\sin^{n-2}(x)\cos^2(x).dx##
    Replacing the cos2 in that gives:
    ##\int \sin^n(x).dx = [-\sin^{n-1}(x)\cos(x)] + (n-1)\int \sin^{n-2}(x)(1-\sin^2(x)).dx##
    ##\int \sin^n(x).dx = [-\sin^{n-1}(x)\cos(x)] + (n-1)\int \sin^{n-2}(x).dx - (n-1)\int \sin^{n}(x).dx##
    ##n\int \sin^n(x).dx = [-\sin^{n-1}(x)\cos(x)] + (n-1)\int \sin^{n-2}(x).dx ##
    etc.
     
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