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Homework Help: Should be easy integration

  1. Jan 31, 2010 #1
    1. The problem statement, all variables and given/known data
    F(x) = sin ([tex]\int[/tex]0 to x sin ( [tex]\int[/tex] 0 to y sin^3(u) du) dy )

    find F'(x)
    2. Relevant equations
    FTC


    3. The attempt at a solution

    Is the answer

    cos ([tex]\int[/tex]0 to x sin ( [tex]\int[/tex] 0 to y sin^3(u) du) dt ) sin( [tex]\int[/tex] 0 to x sin^3(u) du) dy ) sin^3(x)?

    because when you differentiate the integral, the function you get becomes with respect to the boundary variable and so the last terms become with respect to x and I have to differentiate them as well?

    ([tex]\int[/tex]0 to x sin ( [tex]\int[/tex] 0 to y sin^3(u) du) dy ) this becomes sin ( tex]\int[/tex] 0 to x sin^3(u) du), then since the integral in the argument is with respect to x, I must differentiate that as well to get sin^3(x)

    thanks
     
    Last edited: Jan 31, 2010
  2. jcsd
  3. Jan 31, 2010 #2

    LCKurtz

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    I doubt anyone can make sense of that. Use a little tex to make it readable. For example, you do an integral like this:

    [tex] \int_{lower limit here}^{upper limit here} integrand\ dx [/itex]

    To see what I did just click on it. You can copy and paste it and use it to post your question so we can read it.
     
  4. Jan 31, 2010 #3
    thanks, I didn't know how to do that

    anyway, I'm looking for the derivative of:

    [tex]
    F(x) = sin(\int_{0}^{x} sin(\int_{0}^{y} sin^3(u)\ du)\ dy ) [/tex]


    I'm pretty sure that it comes out to be:


    [tex]
    F'(x) = [cos(\int_{0}^{x} sin(\int_{0}^{y} sin^3(u)\ du)\ dy ) \ dx] sin(\int_{0}^{x} sin^3(u)\ du) sin^3(x) [/tex]

    But I'm not 100% clear about the steps. Why do I have to keep differentiating after I differentiate [tex] (\int_{0}^{x} sin(\int_{0}^{y} sin^3(u)\ du)\ dy ) [/tex] (from the chain rule)

    Is it because when I differentiate the expression, [tex] sin(\int_{0}^{y} sin^3(u)\ du)\ dy ) [/tex], it is evaluated at x and so I must differentiate that expression? (since I am differentiating with respect to x)

    thank you
     
  5. Jan 31, 2010 #4

    LCKurtz

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    You have an extra dx at the end of your F(x) that shouldn't be there.

    Call

    [tex]

    g(x)= \int_{0}^{x} sin(\int_{0}^{y} sin^3(u)\ du)\ dy
    [/tex]

    so your F(x) = sin(g(x)) and F'(x) = cos(g(x)) g'(x) by the chain rule.

    Now by the fundamental theorem of calculus to get g'(x) you simply write the integrand in g(x) with the dummy variable y replaced by x:

    [tex]g'(x) = sin(\int_{0}^{x} sin^3(u)\ du)[/tex]

    Put those together; I think it is a bit different than what you wrote. It's late and I'm a little sleepy so check it. :zzz:
     
  6. Jan 31, 2010 #5
    thanks, I noticed the extra dx. I left it there by accident after copying and pasting the integral latex code

    by the chain rule it should be F'(x) = cos(g(x) )g'(x), but how come the answer has sin^3(x) at the end? is it wrong? (I got this answer from someone's notes, refer to post #3)
    thanks
     
  7. Feb 1, 2010 #6
    forgive me for bumping.. but bump
     
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