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U substitution on ∫sin(x^5)dx

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data

    Use Part 2 of the Fundamental Theorem of Calculus to find the derivative.

    [tex] \int_3^x sin(t^{5}) \, dt [/tex]


    2. Relevant equations



    3. The attempt at a solution

    I know the general idea of what I'm supposed to do as far as evaluate the indefinate integral and then do a subtraction of the upper limit and lower limits...but I can't even get to the point of finding the indefinate integral. (maybe it's that I just "think" I know what i'm supposed to do..)

    [tex] \int_3^x sin(t^{5}) \, dt [/tex]

    I'm letting u = [itex]t^{5}[/itex]

    so du = [itex]5t^{4}[/itex]

    then it looks like dt can be replaced by [itex]\frac{1}{5t^{4}}[/itex]

    so that [tex] \int_3^x \frac{1}{5t^4} sin(u) \, du [/tex]

    However, our professor has instructed that mixing variables within the integral is not allowed because it can't be evaluated.

    So how to I do u-substitution on this integral?
     
  2. jcsd
  3. Sep 3, 2012 #2

    vela

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    You don't. Note that the problem says to use the fundamental theorem of calculus. Don't ignore that bit of info.
     
  4. Sep 3, 2012 #3

    SammyS

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    What vela said, plus ... you're to find the derivative.
     
  5. Sep 3, 2012 #4
    oh, it says find the derivative...lol. I just saw the integral sign and began trying to integrate. I didn't even notice that there is a d/dt notation to the right of the integral clear as day. And so I figured the fundamental theorem being referred to was the property that is actually FTC1. Anyway, Thanks :) sin(x^5) duh. I turned a 30 second problem into a 20 minute headache. Thanks Vela and Sammy.
     
  6. Sep 3, 2012 #5

    SammyS

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    The limits od the integral are 3 and x, so the integral is a function of x, not t.

    The derivative of this integral is with respect to x.

    I.e.

    Find
    [itex]\displaystyle \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)\ .[/itex]​
     
  7. Sep 4, 2012 #6

    Thanks, I noticed that right after I submitted the post and editted with the correct notation right afterwards.
     
  8. Sep 6, 2012 #7
    Is this just [itex] 5x^4 sin(x^{5}) [/itex] ?
     
  9. Sep 6, 2012 #8

    SammyS

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    No.

    The anti-derivative of [itex] 5x^4 \sin(x^{5}) [/itex] is [itex] -\cos(x^{5}) +C[/itex],

    not [itex]\displaystyle \int \sin(x^{5}) dx\ .[/itex]
     
  10. Sep 6, 2012 #9
    So this should be [itex] sin(x^{5}) [/itex], correct?
     
  11. Sep 7, 2012 #10

    HallsofIvy

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    No, it isn't. What is the derivative of sin(u(x)) with respect to x. What is the derivative of cos(u(x)) with respect to x?
     
  12. Sep 7, 2012 #11
    cos(u(x)) u'(x) and -sin(u(x)) u'(x).
     
  13. Sep 7, 2012 #12
    This result looks correct to me. There is no x under the integral sign. I don't understand what Halls of Ivy is saying.

    Chet
     
  14. Sep 7, 2012 #13

    SammyS

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    If you mean:
    Is [itex]\displaystyle
    \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)=\sin(x^5)\ ?[/itex]​

    Then I agree with Chestermiller that, "Yes it is."

    Here's how I arrive at that:

    Let F(t) be an anti-derivative of sin(t5). In other words, let [itex]\displaystyle F(t)=\int sin(t^{5}) \, dt\ .[/itex]

    Then [itex]\displaystyle \int_3^x sin(t^{5}) \, dt=F(x)-F(3)\ .[/itex]

    Therefore, [itex]\displaystyle
    \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)=\frac{d}{dx}\left(F(x)-F(3)\right)=\sin(x^5)-0\ .[/itex]
     
  15. Sep 7, 2012 #14

    Mark44

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    "When all else fails, read (or reread) the instructions."
     
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