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Theoretical Integration

  1. Feb 7, 2006 #1
    I'm stuck on explaining this. Does anybody know how to help?

    (a) By writing [tex]\cos^nx = cos^{n-1}xcosx [/tex] use integration by parts to show that

    [tex] \int \cos^nxdx = \cos^{n-1}xsinx + (n-1) \int \sin^2xcos^{n-2}xdx. [/tex]

    (b) Using the result of part (a) derive the reduction formula

    [tex] n\int \cos^nxdx = \cos^{n-1}x\sinx + (n-1) \int \cos^{n-2}xdx. [/tex]​

    My Working:

    (a) All i got so far is

    u = cosx dv/dx =cos^{n-1}x

    du/dx = -\sinx v = \int \cos^{n-1}x
    Last edited: Feb 7, 2006
  2. jcsd
  3. Feb 7, 2006 #2


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    You usually don't want to pick a function you can't immediately integrate as dv/dx, because then you have a nested integral, which is no fun. Try switching your choices of u and dv/dx.
  4. Feb 7, 2006 #3
    ok i swithced u and v and got:

    [tex] u = \cos^{n-1} \mbox{ and so } \frac{du}{dx} = \frac{-sin(n-1)x}{(n-1)^2} [/tex]

    is this right so far?
  5. Feb 7, 2006 #4


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    I don't think so, although I don't really understand what you've written for du/dx. Just use the chain rule to differentiate (cos(x))^(n-1).
  6. Feb 7, 2006 #5
    ok i got the first bit. using the chain rule works, thanks. but how do you do (b)? how can multiplying by n cancel out the sin^2x in the integration?
  7. Feb 7, 2006 #6


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    Use sin2(x)=1-cos2(x). Also, I think there's a mistake in the formula you posted. There should be a sin(x) in the first term on the right side.
    Last edited: Feb 7, 2006
  8. Feb 7, 2006 #7
    hmm, i can't see why using sin^2x = cos^2x - 1 helps, sorry. Could you explain it further please?
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