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Homework Help: Odd power of sine Integral

  1. May 6, 2007 #1
    1. The problem statement, all variables and given/known data

    Integral((sin(x))^2((cos(x))^2) dx


    2. Relevant equations



    3. The attempt at a solution

    Seperate Cos (x)^3

    sin(x)^2 (cos(x))(cos(x))^3

    Then:apply identities

    sin(x)^2(cos(x))(1-sin(x)^2)


    And now im lost!!!!:eek:



    Thanks Alot!!!
     
  2. jcsd
  3. May 6, 2007 #2
    Is that supposed to be (cosx)^4 or something (unless you divide by cos as well, you cannot have a cos^3 out of cos^2)?

    The way I would do this is to use the identity that sin^2 x is 1 - cos^2 x, and then you can integrate cos^n x with a reduction formula or double angle identities.

    Or you could start off with double angle identities and see where it leads you.
     
  4. May 6, 2007 #3
    Yea Sorry I screwed that up the original integral is
    Integral((sin(x))^2((cos(x))^3) dx


    Then
    Seperate Cos (x)^3

    sin(x)^2 (cos(x))(cos(x))^2

    Then:apply identities

    sin(x)^2(cos(x))(1-sin(x)^2)

    So now If I change the sin(x)^2 to 1-cos(x)^2 I wont have dx just as if I change it back????
     
  5. May 6, 2007 #4
    (sinx)^2 = 1-(cosx)^2
    so...

    Int((sin(x))^2((cos(x))^3) dx
    =Int[(cosx)^3(1-(cosx)^2)]dx
    =Int[(cosx)^3 - (cosx)^5]dx
    = -(cosx)^4/4sinx + (cosx)^6/6sinx

    Right?
     
    Last edited: May 6, 2007
  6. May 6, 2007 #5

    Office_Shredder

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    I would stick with changing cosine to sin. Then you get
    [tex]\int sin^2(x)cos(x)dx -\int sin^4(x)cos(x)dx[/tex]

    A discerning eye will note a great substitution there
     
  7. May 6, 2007 #6
    Perhaps you should try using

    [itex]\sin (2x) = 2 \sin (x) \cos (x)[/itex]

    edit: sorry, didn't see the new problem
     
  8. May 7, 2007 #7
    Ok heres my shot at this one at 6:40 AM :bugeye:

    so it becomes

    [tex]\int sin^2(x)cos(x)dx -\int sin^4(x)cos(x)dx[/tex]

    by using 1-sin(x)^2

    So I make u sin(x)
    du cos(x)

    Therefore I get (sin(x)^3)/3-(sin(x)^5)/5

    That sound about right?

    Thanks,
    Andy
     
  9. May 7, 2007 #8

    Office_Shredder

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    Looks good to me
     
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