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Math Help - Toughie Integral.

  1. #1
    Super Member Showcase_22's Avatar
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    Toughie Integral.

    I(x):=\frac{1}{6}\int_{2x^3}^{3x^2} \frac{sin(t)}{t}dt for 0<x<\frac{3}{2}.
    I know that Si(x)=\int_0^{x}\frac{sin(t)}{t}dt. I can't see how this helps!!

    Integration by parts isn't useful either as it makes an infinitely long sequence.

    Macluarin expansions also don't help. They make something very messy.

    gah!!
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  2. #2
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    No elementary primitive.
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  3. #3
    Super Member Showcase_22's Avatar
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    What do you mean by that?

    Is there a special trick to integrate it?

    It seems like there is something to it because it has it's own definition for it's integral: Si(x).
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  4. #4
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    No trick, you can't find a primitive.

    Perhaps this is a question involving the FTC.
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  5. #5
    Super Member Showcase_22's Avatar
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    What does FTC stand for?

    I tried googling it but it didn't work =S

    Is a primitive the same as an antiderivative?
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  6. #6
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    The answer for your last line is yes.

    And, FTC means Fundamental Theorem of Calculus; but dunno what you want to do with your problem.
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  7. #7
    Super Member Showcase_22's Avatar
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    umm, I want to find an expression for the integral.

    I have the answer, but I can't work out how to get there.

    I get stuck on having sin(3x^2) and sin(2x^3). Is there a way to change these into just sin(x)? (my question gives the answer in terms of just sin(x)).

    Double angle formulae don't work for this part either...
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  8. #8
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    Please, give me the answer, I'm pretty sure what is this.
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  9. #9
    Super Member Showcase_22's Avatar
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    Okay then, here's the rest of the question:

    Show that \frac{dI}{dx}=(1-x)sin(x) and deduce that I(x)=1-cos(x)+xcos(x)-sin(x)
    .
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  10. #10
    Moo
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    Quote Originally Posted by Showcase_22 View Post
    Okay then, here's the rest of the question:

    .
    So differentiate I(x)=\frac 16 \int_{2x^3}^{3x^2} \frac{\sin(t)}{t} ~dt

    Let F(t) be an antiderivative of \frac{\sin(t)}{t}

    Then I(x)=\frac 16 (F(3x^2)-F(2x^3))

    Now you know that F'(t)=\frac{\sin(t)}{t} by definition of an antiderivative.

    So \frac{dI}{dx}=x F'(3x^2)-x^2 F'(2x^3) by the chain rule of differentiation.

    \frac{dI}{dx}=x \cdot \frac{\sin(3x^2)}{3x^2}-x^2 \cdot \frac{\sin(2x^3)}{2x^3}=\frac{1}{x} \cdot \sin(3x^2)-\frac 1x \cdot \sin(2x^3)


    I'm sorry I have to go to eat... I'll think about the rest later
    Last edited by Moo; February 11th 2009 at 10:22 AM.
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  11. #11
    Super Member Showcase_22's Avatar
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    Well there's no need.

    I walked into todays lecture and he changed the question completely!

    I like your method Moo, i'll remember that for another time (hopefully for a similar question in the future!).

    If anyone can actually integrate this between the limits I would really like to see how it's done. It's really quite puzzling.
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