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Differentiation of Integral

  1. Oct 6, 2007 #1
    1. The problem statement, all variables and given/known data


    [tex]\frac{d}{dx}\int_{x^3}^{e^x}cost^2dt[/tex]



    3. The attempt at a solution

    [tex]\int cost^2dt=\frac{sint^2}{2t}+\int\frac{sint^2}{2t^2}dt[/tex]
    [tex]\int\frac{sint^2}{2t^2}dt=-\frac{sint^2}{2t}+\int cost^2dt[/tex]
    I came back to initial integral.
     
  2. jcsd
  3. Oct 6, 2007 #2

    Gib Z

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    That can happen with some of your choice for u and dv whilst doing integration by parts, the second time you apply it use different choices.
     
  4. Oct 6, 2007 #3

    arildno

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    Differentiate it, don't try to integrate it!
     
  5. Oct 6, 2007 #4

    Gib Z

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    I think he needs to evaluate the integral to be able to do that doesn't he >.<
     
  6. Oct 6, 2007 #5

    arildno

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    Nope.
    Here's how to do it properly:
    Let F(t) be an antiderivative of f, F'(t)=f(t).
    Thus, we have:
    [tex]\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)dt=\frac{d}{dx}(F(b(x))-F(a(x)))=F'(b(x))b'(x)-F'(a(x))a'(x)=f(b(x))b'(x)-f(a(x))a'(x)[/tex]

    As you can see, you do not need the explicit form of F, only the guarantee that some such F exists..:smile:
     
  7. Oct 6, 2007 #6

    D H

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    No. The integrand does not involve x. Simply apply the fundamental theorem of calculus.

    Hint:
    [tex]
    \frac{d}{dx}\int_{x^3}^{e^x}\cos t^2dt =
    \frac{d}{dx}\int_0^{e^x}\cos t^2dt \;\;-\;\;
    \frac{d}{dx}\int_0^{x^3}\cos t^2dt[/tex]
     
    Last edited: Oct 6, 2007
  8. Oct 6, 2007 #7
    So tha answer is
    [tex]2e^xsine^{2x}-6x^5sinx^6[/tex]
    yes?
     
  9. Oct 6, 2007 #8

    D H

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    No. Read arildno's post again. Sine is not involved.
     
  10. Oct 6, 2007 #9
    ok ok,my mistake
    [tex]2e^{2x}cos(e^{2x})-6x^5cos(x^6)[/tex]
     
    Last edited: Oct 6, 2007
  11. Oct 6, 2007 #10

    arildno

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    Eeh??
    Where do you get that 2-factor from??
     
  12. Oct 6, 2007 #11
    if we put [tex]e^x [/tex] to t shouldn't it be [tex]e^{2x}[/tex]
     
  13. Oct 6, 2007 #12

    arildno

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    I'm talking about the 2-factors in front of the cosine's, not the ones within the arguments.
     
  14. Oct 6, 2007 #13
    I typed wrongly instead of [tex]e^{2x}[/tex],I typed [tex]e^{x}[/tex]
    [tex]\frac{d}{dx}(e^{2x})=2xe^{2x}[/tex]
    This 2 are you asking ?
     
  15. Oct 6, 2007 #14

    arildno

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    What is a(x), and what is b(x); what are their derivatives?
     
  16. Oct 6, 2007 #15
    You say that answer is
    [tex]e^xcos(e^{2x})-3x^2cos(x^6)[/tex]?
     
  17. Oct 6, 2007 #16

    Gib Z

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    Today was not my best day obviously =] Yes I should have seen the proper method arildno and DH, maybe Ill have better luck tomorrow.
     
  18. Oct 6, 2007 #17

    HallsofIvy

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    No, no one has said that! Several people have asked you questions about this problem that you haven't answered.
     
  19. Oct 6, 2007 #18

    arildno

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    I didn't say that; but you said it correctly now! :smile:
     
  20. Oct 6, 2007 #19
    Ok.Thanks to everyone.
     
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