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Integrating trig(sqr)

  1. Jan 27, 2005 #1


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    I need to know how to integrate this function:
    I did this:
    u = sqrt(x)
    du/dx = 1/(2sqrt(x))

    S(sin(sqrt(x))dx) = S(sin(u)*dx*du/dx/(2sqrt(x)) = S(Sin(u)/u du)

    But then I got stuck: integration by parts won't work, trig substitution is out...

    The one thing I did come up with was:
    (u/v)' = (u'v - uv')/v^2
    S((u/v)') = uv = S((u'v - uv')/v^2)

    So I changed
    S(sin(u)*u/u^2 + cos(u)/u^2 - cos(u)/u^2)
    = S((sin(u)*u + cos(u))/u^2) - S(cos(u)/u^2)
    = cos(u)/u - S(cos(u)/u^2)

    But I get the feeling I'm barking up the wrong tree. If I do integration by parts on that integral I'm going to end up with sin(u)/u again!! :grumpy:
  2. jcsd
  3. Jan 27, 2005 #2


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    Are you sure you need an elementary formula for the integral? Getting, say, a Taylor series for an antiderivative would be fairly straightforward, as would estimating a definite integral with numerical approximation.
  4. Jan 27, 2005 #3


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    [tex] Si(x)=:\int_{0}^{x} \frac{\sin t}{t} dt [/tex]

    is called SINE INTEGRAL and is not an "elementary" function.

  5. Jan 28, 2005 #4
    I apologize, I have no desire to write out the latex code...it's much too late for that :yuck:

    Attached Files:

  6. Jan 28, 2005 #5


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    Well,Alkatran,your initial substitution was good,unfortunately you got lost on the way.

    [tex] \int \sin\sqrt{x} \ dx =...?? [/tex]

    Make the substitution:
    [tex] \sqrt{x}=u [/tex] equivalently:
    [tex] x=u^{2} [/tex]

    Can u take it from here?

  7. Jan 28, 2005 #6


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    I don't see how [tex] x=u^{2} [/tex] helps? There are no x's in the equation. However...

    [tex] \int \sin\sqrt{x} \ dx = \int {\sin\sqrt{u^{2}}}/u \ du [/tex]
    [tex] \sin\sqrt(u)^{2} = {1 - \cos{2*\sqrt{u}}}/2 [/tex]

    But I get the feeling I'm headed for circles again.

    We haven't covered integral approximation or taylor series yet so I would have no idea how to do that. However, the initial problem was slightly simpler: It had upper and lower limits (and I believe the answer involved pi).

    On another note:
    A small question: I was given a question once with the hint: this is the area of a common shape...

    Well I realized it was a circle, and did the question without even using integration, but I was just wondering how I would integrate a circle's function (well, top half of a circle's function)?
  8. Jan 28, 2005 #7


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    [tex] x=u^{2} \Rightarrow dx=2u \ du [/tex]

    Write the integral in terms of "u".

  9. Jan 29, 2005 #8
    Maybe what is missing is the idea of Intergation by Parts:

    [tex]\int{ysin(y)dy}=-ycos(y) + \int{cos(y)dy}[/tex]

    To integrate the circle function (X^2+Y^2 =R^2 in the first quadrant, we solve for Y= [tex]\sqrt{R^2-X^2}[/tex] and then to solve for X=0 to 1, we will require a trigonometric substitution. (This occurs because pi is a transcendental function and can not be obtained from a finite polynomial.)
    Last edited: Jan 29, 2005
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