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Integration Problem

  1. Dec 12, 2007 #1
    1. The problem statement, all variables and given/known data

    f:=sin^3(x)cos(y)
    Integrate f, from x=y....pi/2)

    2. Relevant equations

    I can do it if the two letters were y... But i have no idea how to solve when they are combined..
     
    Last edited: Dec 12, 2007
  2. jcsd
  3. Dec 12, 2007 #2

    HallsofIvy

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    That is:
    [tex]\int_y^{\pi/2} sin^3(x)cos^3(y) dx[/tex]
    Find the anti-derivative (there's a standard method for odd powers of sin or cos) and evaluate between [itex]\pi/2[/itex] and y. Of course, the answer will not be number but will depend on y.
     
  4. Dec 12, 2007 #3

    cristo

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    Integrate f with respect to what? Is this a double integration, or a single integration? Please state the question exactly as it is given.
     
  5. Dec 12, 2007 #4
    f:=sin^3(x)*cos(y);

    Integrate f with respect to x.... from y to Pi/2
     
    Last edited: Dec 12, 2007
  6. Dec 13, 2007 #5
    In all fairness to jra_1574, I wouldn't say that the question was ambiguous at all. The variable of integration may not have been explicitly stated but the way it was written

    "from x=y...Pi/2"

    seemed fairly obvious to me, and is in fact the way that some CAS's denote a definite integral (the primary example in mind would be Maple).

    As for the actual question jra_1574, the integration doesn't depend on y, and so it's a constant and by linearity can be pulled out of the integral.
     
  7. Dec 13, 2007 #6

    HallsofIvy

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    Agree with everything you say except for the last line! y is the lower limit of integration and cannot be "pulled out of the integral".
     
  8. Dec 13, 2007 #7

    Office_Shredder

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    What if instead it was written as f=sin(x)^3cos(0) and was integrated from 0 to pi/2. Couldn't the cos(0) be pulled out of the integral? I'm not sure how this is supposed to be any different
     
  9. Dec 13, 2007 #8

    cristo

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    That may be true, but the fact that he said "I don't know what to do when there are two letters" implied that it could be a double integration. Besides, one should always state explicitly what one is doing.
    What you mean is that y is a constant when integrating wrt x, and so cos(y) can be treated as a constant.
     
  10. Dec 13, 2007 #9

    HallsofIvy

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    Since cos(y) is a constant (with respect to x) it can be "pulled out of the integral". That is different from saying y (which is also the lower limit of integration) can be "pulled out of the integral".

    [tex]\int_y^{\pi/2} sin^3(x)cos(y)dx= cos(y)\int_y^{\pi/2} sin^3(x) dx[/tex]
     
  11. Dec 13, 2007 #10

    Office_Shredder

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    I assumed that was what Kreizhn meant. Never mind then
     
  12. Dec 13, 2007 #11
    Yeah, sorry, meant the cos(y). It was late and I figured it was pretty obvious what I meant

    Edit: Hurrah for semantical trivialities. I know that I should've stated that better but come on HallsOfIvy, I would've imagined that you knew precisely what I was talking about and was nit-picking
     
    Last edited: Dec 13, 2007
  13. Dec 13, 2007 #12

    HallsofIvy

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    Actually, I was taken aback by it because I had focused on the "y" as a limit of the integral rather than the "cos(y)" while you were doing the opposite!

    Also I agree that that looks like "half" of a double integral!
     
  14. Dec 13, 2007 #13
    How would i type that in maple?

    f:=sin(x)^3*cos(y)?
    or f:=sin^3(x)*cos(y)?

    anyways i pulled the cos(y) out of the integral, then change sin^3(x) to sin(x)*(1-cos^2(x))... i think that led me to
    sin(x)-cos^2(x)*sin(x)

    so integrate sin(x)= cos(x) minus
    integration of cos^2(x)*sin(x) = i am not sure but is this g and g'? or i have to choose one to be u and do the du/dx?
     
  15. Dec 14, 2007 #14
    Don't forget that there's a [itex] \frac{1}{2} [/itex] in there as well.

    [tex] sin^2(x) = \displaystyle\frac{1}{2} \left( 1-cos(2x) \right) [/tex]

    then let [itex] u = cos(x) [/itex] so that [itex] du = - sin(x) dx[/itex] and proceed from there.

    In maple, you could write either

    f:=sin(x)^3*cos(y);

    or if you actually wanted to be able to evaluate f and certain points

    f:=(x,y) -> sin(x)^3*cos(y);
     
  16. Dec 14, 2007 #15
    OK thanks so much for your help!
     
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