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Integrating squares of sine & cosine

  1. Feb 23, 2012 #1
    1. The problem statement, all variables and given/known data
    find ∫4cosx*sin^2 x.dx


    2. Relevant equations



    3. The attempt at a solution

    ∫4cos x * 1/2 (1 - cos2x)

    ∫2cosx - 4cos^2 x.

    Then i dont know whereto go from here??
     
  2. jcsd
  3. Feb 23, 2012 #2
    your last step how did you reach to 4cos^2 x ?
     
  4. Feb 23, 2012 #3

    Mark44

    Staff: Mentor

    A simple substitution will work in this problem.
     
  5. Feb 23, 2012 #4
    ie,
    let u = cos x

    ∫ 2u - 4u^2

    = u^2 -4/3 (u)^3 + C

    This still does not lead me anywhere?

    Or did you mean a substitution not in terms of 'u' but in terms of another trig result?
    I feel lost.
     
  6. Feb 23, 2012 #5
    by expanding 4cosx * 1/2(1-cos2x)

    = 2cosx - 4cos^2 x
     
  7. Feb 23, 2012 #6

    Mark44

    Staff: Mentor

    1. Start from the original integral.
    2. That's almost the right substitution, but what is du? You are being very sloppy by omitting the differential, and this is exactly the type of problem where doing that will bite you in the butt.
     
    Last edited: Feb 23, 2012
  8. Feb 23, 2012 #7

    SammyS

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    Instead:

    Let u = sin(x) .
     
  9. Feb 23, 2012 #8

    Mark44

    Staff: Mentor

    No.
    cos(x) * cos(2x) [itex]\neq[/itex] cos2(x)
     
  10. Feb 23, 2012 #9
    Hmmm
    I think i need to stop rushing through my questions.
    Thanks for pointing out the obvious!!
     
  11. Feb 23, 2012 #10
    Yea I asked that too
     
  12. Feb 23, 2012 #11
    ∫4cosx * u^2 * du/cosx

    ∫4(u)^2

    4/3 (u)^3 + C

    = 4/3 (sin x)^3 +C

    Better?
     
  13. Feb 23, 2012 #12
    Now you can proceed , easily :)
     
  14. Feb 23, 2012 #13

    Mark44

    Staff: Mentor

    Yes, better, and that's the right answer, but some of the supporting work could be improved.

    ∫4cosx * u^2 * du/cosx

    =∫4(u)^2 du

    =4/3 (u)^3 + C

    = 4/3 (sin x)^3 +C

    In addition, it's usually better to make the substitution completely so that all the x's and dx's change to u's and du's, like this:

    ∫4cos(x) * sin2(x)dx [u = sin(x), du = cos(x)dx]
    = 4 ∫u2 du
    = 4 u3/3 + C
    = (4/3) sin3(x) + C
     
  15. Feb 23, 2012 #14
    okie
     
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