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Two questions regarding integrating items

  1. Oct 25, 2008 #1

    TFM

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    1. The problem statement, all variables and given/known data

    I have a couple of questions about some intriguing looking equations.

    Firstly, I have to integrate:

    [tex] \frac{sin\theta}{cos\theta} d\theta[/tex]

    Is there an answer to just dividing them? because sin/cos = tan, but that doesn't seem to help?

    Secondly, I have to integrate:

    [tex] 6e^{-x^2} [/tex]
    This One > [tex] 6e^{-x^2} [/tex] Should be displaying e to the minus x sqaured



    but I didn't think you could integrate a [tex] e^{x^2} [/tex]

    Again, this one V
    [tex] e^{x^2} [/tex] ?

    2. Relevant equations

    [tex]\frac{sin\theta}{cos\theta} = tan\theta[/tex]

    3. The attempt at a solution

    Given above

    Any help would be most appreciated,

    TFM
     
    Last edited: Oct 25, 2008
  2. jcsd
  3. Oct 25, 2008 #2

    rock.freak667

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    [tex]\int \frac{sin\theta}{cos\theta}d\theta[/tex]


    if you let [itex]u=cos\theta[/itex], what is [itex]\frac{du}{d\theta}[/itex]?

    For the 2nd one, I don't think you can express that in terms of elementary functions.
     
  4. Oct 25, 2008 #3

    TFM

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    [itex]\frac{du}{d\theta} = -sin\theta[/itex]

    Is there any way to integrate this, or should I leav it as it is (The question is solving differential equations)

    TFM
     
  5. Oct 25, 2008 #4

    rock.freak667

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    and so [itex]-du=sin\theta d\theta[/itex].

    Now just sub this into the integral and you can get it out.
     
  6. Oct 25, 2008 #5

    TFM

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    I have to admit making a small mistake...

    The equation to solve is:

    [tex] \frac{cos\theta}{tan\theta} [/tex]

    TFM
     
  7. Oct 25, 2008 #6

    rock.freak667

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    then you need to put tan as sin/cos and get

    [tex]\frac{cos^2 \theta}{sin\theta}[/tex]

    which in this case put u=sin
     
  8. Oct 25, 2008 #7

    TFM

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    So:

    [tex] u = sin\theta [/tex]

    [tex] \frac{du}{d\theta} = cos\theta [/tex]

    [tex] du = cos\theta d\theta [/tex]

    so sub-ing in:

    [tex] \frac{cos^2 \theta}{sin\theta}d\theta [/tex]

    [tex] \frac{du^2 \theta}{u}d\theta [/tex]

    Does this look right???

    Edit: No it doesn't. Should be

    [tex] \frac{du^2 }{u}[/tex]

    [tex] \frac{du^2 }{u}[/tex]



    Does this Look right?

    TFM
     
  9. Oct 25, 2008 #8

    Gib Z

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    I think what rock.freak667 wanted you to get was

    [tex] \frac{du}{u} \cdot \cos x = \frac{ \sqrt{1-u^2}}{u} du[/tex],

    which would warrant an integration by parts.

    An alternative method would be to write [tex]\cos^2 x = 1-\sin^2 x[/tex] and divide through.
     
  10. Oct 26, 2008 #9

    TFM

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    Could I not use double angle formula:

    [tex] cos^2(\theta) = \frac{1-cos(2\theta)}{2} [/tex]

    Giving

    [tex]\frac{1 - cos(2\theta)}{2sin(\theta)}[/tex]

    and then let [tex] u = cos\theta [/tex] again

    ??

    TFM
     
  11. Oct 26, 2008 #10

    HallsofIvy

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    How would that help you? What are you going to do with the [itex]cos(2\theta)[/itex] if [itex]u= cos(\theta)[/itex]?
     
  12. Oct 26, 2008 #11

    TFM

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    That's true, I just saw the cos squared and remebered that sometime that relation is used to break it down. So:

    [tex] \frac{cos^2 \theta}{sin\theta} [/tex]

    [tex] u = sin\theta [/tex]

    [tex] \frac{du}{d\theta} = cos\theta [/tex]

    [tex] du = cos\theta d\theta [/tex]

    This gives me

    [tex] \frac{cos\theta cos\theta}{sin\theta}d\theta [/tex]

    [tex] \frac{cos\theta }{u}du [/tex]


    [tex] cos\theta = \sqrt{1 - sin^2\theta} [/tex]

    [tex] cos\theta = \sqrt{1 - u^2} [/tex]

    so:


    [tex] \frac{\sqrt{1 - u^2}}{u}du [/tex]

    So integrating by parts:

    [tex]\int adb = ab - \int bda [/tex]

    [tex] db = u^{-1} ; u^{-2} [/tex]

    [tex] a = \sqrt{1 - u^2} ; da = [/tex]

    Does this look okay?

    TFM
     
  13. Oct 26, 2008 #12

    Gib Z

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    Well if you meant

    [tex]db = u^{-1}, b = \log_e u[/tex]

    Then that looks fine so far in terms of correctness, though it looks like it won't be fruitful. Perhaps rock.freak667 meant something different to what i Interpreted.

    Why didn't you like my suggestion though lol?
     
  14. Oct 26, 2008 #13

    TFM

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    I was actually going to do

    [tex]b = u^{-1}, db = u^{-2} [/tex]

    or is this going the wrong way?

    Was your suggestion


    An alternative method would be to write [tex] cos^2 x = 1-sin^2 x [/tex] and divide through.

    If so, should that not be

    [tex] cos^2(\theta) = \frac{1-cos(2\theta)}{2} [/tex] ???

    I'm wondering if i have made some mistakes in actually solving the Differential Equations, since out of four DEs I have three which seem to not be able to be completed?

    TFM
     
  15. Oct 26, 2008 #14

    TFM

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    If I may just check my calculations.

    First Question:

    [tex] x(t): x' = xt + 6te^{-t^{2}} [/tex]

    Subtract -xt from both sides:

    [tex] -xt \frac{dx}{dt} = 6te^{-t^{2}} [/tex]

    [tex] -x \frac{dx}{dt} = 6te^{-t^{2}}/t [/tex]

    [tex] -x dx = 6e^{-t^{2}} dt [/tex]

    Does this one look okay?

    TFM
     
  16. Oct 26, 2008 #15

    TFM

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    Does the above look correct?

    TFM
     
  17. Oct 26, 2008 #16

    rock.freak667

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    [tex]\frac{dx}{dt}=xt+6te^{-t^2}[/tex]

    [tex]\frac{dx}{dt}-xt=6te^{-t^2}[/tex]

    Do you know how to solve first order differential equations of the form (in your case)

    [tex]\frac{dx}{dt}+P(t)x=Q(t)[/tex]?
     
  18. Oct 27, 2008 #17

    TFM

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    That would be:

    [tex] = e^Fx \int^F q(t)dt + C [/tex]

    [tex] = e^Fx \int^F q(t)dt + C [/tex]

    [tex] = e^Fx \int^F q(t)dt + C [/tex]

    [tex] F = \int p(t)dt [/tex]

    ???

    TFM
     
  19. Oct 27, 2008 #18

    TFM

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    Sorry, Latex annoys me sometimes when editing:

    [tex] = e^Fx \int^F q(t)dt + C [/tex]

    [tex] F = \int p(t)dt [/tex]

    TFM
     
  20. Oct 27, 2008 #19

    TFM

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    Lets see. So:

    [tex] \frac{dx}{dt}-xt=6te^{-t^2} [/tex]

    [tex] \frac{dx}{dt}+P(t)x=Q(t) [/tex]

    solution:

    [tex] = e^Fx \int^F q(t)dt + C [/tex]

    [tex] F = \int p(t)dt [/tex]

    And:

    [tex] q(t) = 6te^{-t^2} [/tex]

    [tex] p(t) = -t [/tex]

    so:

    [tex] F = \int -t dt = \frac{-t^2}{2}[/tex]

    [tex] = e^{\frac{-t^2}{2}}x \int^{\frac{-t^2}{2}}(6te^{-t^2}) dt + C [/tex]

    We still seem to have to integrate the [tex] e^{-x^2} [/tex]? Which I thought ewas impossible?

    TFM
     
  21. Oct 27, 2008 #20

    rock.freak667

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    your integrating factor of exp(-t2/2) is correct.

    But your integrand should be


    [tex]e^\frac{t^2}{2} 6te^{t^2}[/tex]

    Which can be simplified. Then you can integrating using a u substitution.
     
    Last edited: Oct 27, 2008
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