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Integration Question

  1. Feb 25, 2007 #1
    1. The problem statement, all variables and given/known data
    integral cos^5(9*t) dt


    2. Relevant equations
    half sets?


    3. The attempt at a solution
    integral cos(9t)^5 dt

    integral cos(9t)^2 * cos(9t)^2 * cos(9t)

    cos(9t)^2 = (1/2)*[ 1 + cos(18t) ]

    integral (1/2)*[ 1 + cos(18t) ] * (1/2)*[ 1 + cos(18t) ] * cos(9t)

    Am I on the right path?
     
  2. jcsd
  3. Feb 25, 2007 #2
    Hint:
    cos^2 u = 1-sin^2 u
     
  4. Feb 25, 2007 #3
    u = cos(9t)
    du = -9sin(9t)

    [ 1 - sin(9t)^2 ] * [ 1 - sin(9t)^2 ] * u du?
     
  5. Feb 26, 2007 #4
    [tex]\cos^5{9t}[/tex] is the same as [tex]\cos{9t}(1-\sin^2{9t})^2[/tex]

    Expand it and integrate
     
  6. Feb 26, 2007 #5
    Then how do you integrate that beat o_O.

    cos(9t) * (1 - sin(9t)^2)^2 * dt

    u = 1-sin(9t)^2
    du = -18sin(9t) * cos(9t) * dt

    du / -18sin(9t) = cos(9t) * dt

    1 / -18sin(9t) <-integral-> u^2 * du

    1/-18sin(9t) * u^3

    1/-18sin(9t) * (1-sin(9t)^2)^3 ?
     
  7. Feb 26, 2007 #6
    try again, try u = something else.
     
  8. Feb 26, 2007 #7
    You need to expand it. Then you will be able to use [tex]\frac{d}{dx}\sin{x} = \cos{x}[/tex]
     
  9. Feb 26, 2007 #8

    Gib Z

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    Homework Helper

    So basically when we have to integrate something with only cosine in it, and cosine is odd powered, we take the most even powers out and transform those into the sines as mentioned, expand and use substitution u=sin x to do the rest.
     
  10. Feb 26, 2007 #9
    I don't understand... can you give me an example?

    cos^2 * cos^2 * cos(x) = (1 - sin^2)^2 * cos(x).

    I don't understand :(
     
  11. Feb 27, 2007 #10

    Gib Z

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    Homework Helper

    Do you know the Identity [itex]\sin^2 x + \cos^2 x=1[/itex]?
     
  12. Feb 27, 2007 #11

    HallsofIvy

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    Science Advisor

    What if you let u= sin(9t) instead?
     
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