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Integration of a Sine function and a fraction

  1. Mar 24, 2009 #1
    1. The problem statement, all variables and given/known data
    It is a really long word problem, but I only need help with integrating one part of the proble.


    2. Relevant equations
    I need to know how to integrate 2+5sin((4*pi*x)/(25))


    3. The attempt at a solution
    I tried using the chain rule by first integrating the sin into negative cosine, and then integrating the inside, giving me: 2x-cos((4*pi*x)/(25))*(2*pi*x^2)/(25)

    Sorry, I don't know how to do all the symbols, or any of them really.
     
  2. jcsd
  3. Mar 24, 2009 #2

    Dick

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    No, not quite. Do a u substitution. u=4*pi*x/25, du=4*pi*dx/25.
     
  4. Mar 24, 2009 #3

    Mark44

    Staff: Mentor

    Here's your integral in latex tags, and split into two integrals.
    [tex]
    \int 2dx + 5sin(\frac{4 \pi x}{25})dx
    [/tex]

    The first integral gives you 2x + a constant.

    Your explanation of using the reverse chain rule and your work are incorrect. You don't integrate the sine part, and then integrate what's inside. This isn't how it works. If it were, you could differentiate your answer and get back to the integrand you started with.

    Use an ordinary substitution u = [itex]4/25 \pi x[/itex], so du = [itex]4/25 \pi dx[/itex]. Once you get this in place, your integral will be [itex]\int 5sin(u)du[/itex], which is pretty straightforward to do.
     
  5. Mar 24, 2009 #4
    Thanks for the explanation. I just want to confirm though, that the integral should be
    -5cos(du), right?
     
  6. Mar 24, 2009 #5

    Dick

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    The integral is -5*cos(u). Not du. But don't forget the factors that are coming from changing dx to du. Why don't you just do it and we'll check your answer?
     
  7. Mar 24, 2009 #6
    So would the answer then be -5cos((2*pi*x^2)/(25))?
     
  8. Mar 24, 2009 #7

    Dick

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    No again. Can you show us how you worked that out using the substitution u=4*pi*x/25?
     
  9. Mar 24, 2009 #8
    I tried to integrate the part after the cosine. I took out the "x" and got (x^2)/(2) and then multiplied it back into the (4*pi)/(25), and simplified it to (2*pi*x^2)/(25).

    Do I instead not bother with the inside of the cosine, and just leave it as is, as in (4*pi*x)/(25)?
     
  10. Mar 24, 2009 #9

    Dick

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    You just substitute. You have integral 5sin(4*pi*x/25)*dx. u=4*pi*x/25. du=(4*pi)dx/25. That gives you 5*sin(u)*dx. Now you just replace dx=25*du/(4*pi). Then you have an integral all in u. Integrate it. Now reverse the substitution to get an expression in x. Have you really never done this before?
     
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