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anonymous

  • one year ago

Evaluate the integral..

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    is arctan possible here

  3. anonymous
    • one year ago
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    \[\int\limits _{-\infty }^{\infty }\:\left(\frac{dx}{e^x+e^{-x}}\right)\]

  4. Zale101
    • one year ago
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    Yes, it is possible

  5. freckles
    • one year ago
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    I would make it more obvious that is so by multiplying e^x on top and bottom

  6. anonymous
    • one year ago
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    it is in fact arctan in one step right?

  7. anonymous
    • one year ago
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    The problem is I don't know how first time seeing such problem\[\int\limits \frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C \] took this from my formula ntbk

  8. freckles
    • one year ago
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    \[\int\limits_{}^{}\frac{ du}{u^2+1} \text{ where } u=e^x \]

  9. freckles
    • one year ago
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    the lower limit of integration changes since as x goes to -inf then u goes to 0

  10. anonymous
    • one year ago
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    @freckles that is clever, i would have taking \[\lim_{t\to -\infty}\tan^{-1}(e^t)\]

  11. anonymous
    • one year ago
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    *taken

  12. anonymous
    • one year ago
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    what about e^-x do I have to bring that up?

  13. freckles
    • one year ago
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    who cares we aren't English snobs we are math snobs @satellite73

  14. anonymous
    • one year ago
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    \[\frac{e^x}{e^x}\]

  15. freckles
    • one year ago
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    \[\int\limits_{-\infty}^{\infty} \frac{1}{e^{x}+e^{-x} } dx \\ \text{ multiply \top and bottom by } e^{x} \\ \int\limits_{-\infty}^{\infty} \frac{e^{x}}{e^{2x}+1} dx \\ \text{ \let } u=e^{x} \\ du=e^{x} dx \\ \\ \int\limits_0^\infty \frac{du }{u^2+1}\]

  16. freckles
    • one year ago
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    @arvnoodle that have written out does it make more sense or still no sensE?

  17. anonymous
    • one year ago
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    Wait I'll analyze it

  18. anonymous
    • one year ago
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    Why do you have to multiply e^x? Is that algebra or what, you do that if you want to bring something up to numerator?

  19. Zale101
    • one year ago
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    |dw:1438745594477:dw|

  20. freckles
    • one year ago
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    \[\frac{e^x}{e^{x}}=1 \\ \text{ I multiplied by this fancy 1 to put it in the obvious } \arctan \text{ integral form }\]

  21. freckles
    • one year ago
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    yes that is algebra

  22. anonymous
    • one year ago
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    I noticed the lower limit changed to 0 why is that

  23. freckles
    • one year ago
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    as x approaches -inf e^x approaches 0

  24. anonymous
    • one year ago
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    Oh okay tried it in my calcu it reaches 0 so that make sense

  25. freckles
    • one year ago
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    \[u=e^x \\ x \rightarrow \infty \text{ then } u=e^{x} \rightarrow \infty \\ x \rightarrow -\infty \text{ then } u=e^{x} \rightarrow 0\]

  26. freckles
    • one year ago
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    |dw:1438745936722:dw| notice as x gets infinitely negative large the y values are getting close to 0

  27. anonymous
    • one year ago
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    \[\int\limits _0^{\infty }\:\frac{du}{u^2+1}\] sooo

  28. freckles
    • one year ago
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    you can integrate that correct? you actually gave a formula for that earlier and other people had mentioned something about the pretty arctan( ) thingy

  29. anonymous
    • one year ago
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    \[\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right)+C\]

  30. anonymous
    • one year ago
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    is this the one? is my formula right?

  31. freckles
    • one year ago
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    but a is 1 and we don't need the C because we have a definite integral

  32. freckles
    • one year ago
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    \[\int\limits_0^\infty \frac{1}{u^2+1} du=\lim_{z \rightarrow \infty}[\arctan(u)|_0^{z}]\]

  33. anonymous
    • one year ago
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    Oh okay so arctan(e^x) ? with limits of 0 and inf

  34. freckles
    • one year ago
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    well if you change it back to arctan(e^(x)) then you need to go back to the original limits

  35. anonymous
    • one year ago
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    So will the final answer be infinity?

  36. freckles
    • one year ago
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    no

  37. freckles
    • one year ago
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    \[\int\limits\limits_0^\infty \frac{1}{u^2+1} du=\lim_{z \rightarrow \infty}[\arctan(u)|_0^{z}] \\ =\lim_{z \rightarrow \infty}[\arctan(z)-\arctan(0)] \\ =\lim_{z \rightarrow \infty}\arctan(z)\] y=tan(x) on (-pi/2,pi/2) looks like: |dw:1438746457408:dw| so y=arctan(x) on (-inf,inf) looks like: |dw:1438746492037:dw|

  38. freckles
    • one year ago
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    so looking at this y=arctan(x) on (-inf,inf) in your graphing calculator you should see as x gets super large then the y values are approaching y=pi/2

  39. freckles
    • one year ago
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    anyways can you tell me the following limit: \[\lim_{z \rightarrow \infty}\arctan(z)\] it should be obvious from the graph and what I just said that this is ....

  40. anonymous
    • one year ago
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    okay let me get this cllear for me

  41. anonymous
    • one year ago
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    \[\left[\tan^{-1}\left(\frac{e^x}{1}\right)\right]\]

  42. anonymous
    • one year ago
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    that's right? ^

  43. freckles
    • one year ago
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    if we put it back in terms of x then we have to go back to the original limits

  44. anonymous
    • one year ago
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    okay -inf to +inf

  45. freckles
    • one year ago
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    which is fine

  46. freckles
    • one year ago
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    but you will have to break that integral up

  47. anonymous
    • one year ago
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    the -inf turns to 0

  48. anonymous
    • one year ago
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    Hmmm break?

  49. anonymous
    • one year ago
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    oh my this problem kind of hard lol sorry for not getting it quickly

  50. freckles
    • one year ago
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    \[\int\limits_{-\infty}^{\infty} \frac{1}{e^x+e^{-x} } dx \\ =\int\limits_{-\infty}^0 \frac{1}{e^{x}+e^{-x} } dx+\int\limits_0^\infty \frac{1}{e^{x}+e^{-x} } dx \\ \text{ check to see both integrals converge } \] which we know they will because we already seen the answer another way so the first integral will converge to the sum of the other two integrals in this case

  51. freckles
    • one year ago
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    \[= \lim_{u \rightarrow -\infty}[\arctan(e^{x})|_u^0]+\lim_{z \rightarrow \infty}[\arctan(e^{x})|_0^z]\]

  52. anonymous
    • one year ago
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    the answer is pi/2

  53. freckles
    • one year ago
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    yes

  54. anonymous
    • one year ago
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    I'm unsatisfied because by my logic the e^x reaches inf it means the answer is inf

  55. anonymous
    • one year ago
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    it doesn't make sense pi/2 to me but ohwell

  56. freckles
    • one year ago
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    as x gets large then e^x gets large then arctan(getting super large)=?

  57. freckles
    • one year ago
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    |dw:1438747311900:dw| look at the graph of y=arctan(x) as x gets supper large y approaches pi/2

  58. freckles
    • one year ago
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    super!

  59. anonymous
    • one year ago
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    oh I get by graph!!!

  60. anonymous
    • one year ago
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    oh my does it mean I need to graph when I'm answering this kind of question, kind of tedious lol

  61. anonymous
    • one year ago
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    but okay thankyou tho

  62. freckles
    • one year ago
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    well you eventually won't need to graph every thing you will just remember the end behavior of most graphs

  63. freckles
    • one year ago
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    the graph was more of me trying to convince you that it is pi/2 it is not needed in the answer

  64. UnkleRhaukus
    • one year ago
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    as the opposite side of a right angled triangle approaches infinitely greater in length than the adjacent side, the considered angle approaches 90° = π/2

  65. anonymous
    • one year ago
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    I got it now I got now!! by chain rule

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