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Square Root Indefinite Integral

  1. Mar 11, 2013 #1

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    Hello everyone..
    1. The problem statement, all variables and given/known data
    ∫√((1+(e^-x))^2)dx




    2. The attempt at a solution
    I first tried to do a u sub and then attempt a trig sub however I cant do anything with the e^-x left in the u sub. Does anyone have another way I can integrate this thing??

    Thank you for any suggestions/help!
     
    Last edited: Mar 11, 2013
  2. jcsd
  3. Mar 11, 2013 #2

    Ray Vickson

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    Before doing anything, try computing the square root.
     
  4. Mar 11, 2013 #3

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    The integral is within an arc length problem if it matters. However when i integrate (1+(e^-x)^2)^1/2 from 0 to 2 i get roughly 2.2214187 and when i integrate 1+e^-x from 0 to 2 i get 2.8647.
    Am I allowed to square root 1+(e^-x)^2 and receive 1+e^-x?
     
  5. Mar 11, 2013 #4

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    I'm sorry I forgot the parenthesis around the (e^-x)^2 it should be square root of ((1+((e^-x)^2)))
     
  6. Mar 11, 2013 #5

    SammyS

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    Well that changes everything.


    Then, of course, [itex]\displaystyle \ \sqrt{1+\left(e^{-x}\right)^2}\ne 1+e^{-x}\ .\ [/itex] You wouldn't expect the two integrals to give the same result.
     
  7. Mar 11, 2013 #6

    Ray Vickson

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    Using too many brackets is almost as confusing as not using enough. Do you mean (a): ##\sqrt{1+(e^{-x})^2}## or (b): ##\sqrt{(1 + e^{-x})^2}##? If you mean (a) you can write it as sqrt(1 + (e^{-x})^2); if you mean (b) you can write it as sqrt((1 + e^{-x})^2). You can even write e^-x without the {} bracket, as you have done, but some people would argue with that.

    I thought in your OP you meant (b), and that is why I suggested you do the square root first. In (a) you can still simplify it usefully before tackling the integral.
     
    Last edited: Mar 11, 2013
  8. Mar 11, 2013 #7

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    your right, so is there a specific technique of integration I should use for this integral?
     
  9. Mar 11, 2013 #8

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    A is what the integral i must find looks like. I truly wish it had looked like B though...
     
  10. Mar 11, 2013 #9

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    Anyone have an idea?
     
  11. Mar 11, 2013 #10

    SammyS

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    Looking at the results from WolframAlpha, the indefinite integral is messy, but not impossible.
     
  12. Mar 11, 2013 #11

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    I'm still attempting to solve it but haven't gotten anywhere yet.
     
  13. Mar 11, 2013 #12

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    If I substituted u= e^x in any integral and the e^x didn't cross out can I just let it equal u and multiply 1/u into the original integral?? Or is that illegal?
     
  14. Mar 11, 2013 #13

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    After three substitutions and a partial fraction I have the answer and it matches wolfram alpha. Thanks for the help
     
  15. Mar 11, 2013 #14
    So, just to clarify, the integral is

    ##\int \sqrt{1+e^{-2x}}\;dx##?
     
  16. Mar 11, 2013 #15

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    Yes that's correct
     
  17. Mar 11, 2013 #16
    Alright, well here's what I did:

    ##\displaystyle\int\sqrt{1+e^{-2x}}\;dx\\
    \displaystyle\int\sqrt{1+e^{-2x}}\cdot\dfrac{\sqrt{1+e^{-2x}}}{\sqrt{1+e^{-2x}}}\;dx\\
    \displaystyle\int \dfrac{1+e^{-2x}}{\sqrt{1+e^{-2x}}}\;dx\\
    \displaystyle\int \dfrac{1+e^{-2x}}{\sqrt{1+e^{-2x}}}\cdot\dfrac{\sqrt{1-e^{-2x}}}{\sqrt{1-e^{-2x}}}\;dx\\
    \displaystyle\int \dfrac{\left(1+e^{-2x}\right)\sqrt{1-e^{-2x}}}{1-e^{-2x}}\;dx##

    As a substitution, let ##u=e^{-x},\;du=-e^{-x}dx\Rightarrow dx=-\dfrac{du}{u}##.

    ##\displaystyle-\int \dfrac{\left(1+u^2\right)\sqrt{1-u^2}}{1-u^2}\;\dfrac{du}{u}##

    Then, a trig substitution with ##u = \sin t,\;du=\cos t\;dt##.

    ##\displaystyle-\int \dfrac{\left(1+\sin^2t\right)\sqrt{1-\sin^2t}}{\sin t\left(1-\sin^2t\right)}\cos t\;dt##

    Something tells me there may have been a slightly shorter route in the first set of rewriting, though...
     
  18. Mar 11, 2013 #17
    Actually, I take it back. After looking over the first few lines, it doesn't look like stopping after one of the earlier steps gives you an easy integral to work with.
     
  19. Mar 12, 2013 #18

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    Another way you could possibly do this would be a u= e^-x then a trig sub (tan)
     
  20. Mar 12, 2013 #19

    Ray Vickson

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    For ##\int \sqrt{1+e^{-2x}}\, dx,## substitute ##e^{-2x} = u.##
     
  21. Mar 12, 2013 #20

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    I would leave the e^-x squared that way the trig sub would work after u=e^-x
     
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