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Integrating with Change of variables -method

  1. Oct 9, 2007 #1
    [SOLVED] Integrating with Change of variables -method

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

    Solve by "Change of variables"-method ∫(dx)/(eˆx + eˆ(-x))

    2. Relevant equations

    ∫f(x)dx=∫f(x(t))*x'(t)dt

    3. The attempt at a solution

    ∫(dx)/(eˆx + eˆ(-x))=∫(eˆx)/(1 + eˆ2x) t=eˆx x=ln (t) (dt)/(dx)=t => dx=1/t dt

    ∫t/(tˆ2 + 1) * 1/t => ∫1/(tˆ2 + 1) = WRONG!

    Where am I making a mistake? How do I know which term I should take as "t"? I have tried to read more in internet of this subject but couldn't find anything helpful. I read one topic here about this subject. The helper had used chain-method to solve this sort of problem but I couldn't understand it. Could someone help me to understand it? I'm just started my studies at university and my knowledge is very very limited. (Sorry my english, I'm finnish...)
     
    Last edited: Oct 9, 2007
  2. jcsd
  3. Oct 9, 2007 #2

    Dick

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    You are doing fine. Why do you say integral(1/(t^2+1)) is wrong? What is that integral?
     
  4. Oct 9, 2007 #3
    I'm not entirely sure if your algebra and methods are right, but I think you're trying it the hard way.

    Think about it a second: the derivative of the e^x is e^x, and the integral of the same function is also itself (plus a constant). With these facts, the integral should be easy.
     
  5. Oct 9, 2007 #4

    Dick

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    It's not the integral of e^x+e^(-x). It's the integral of 1/(e^x+e^(-x)). I overlooked the '/' on first reading as well.
     
  6. Oct 10, 2007 #5
    Oh wow, can't believe I missed that...

    I looked onto the integral again and his method doesn't seem wrong (not really an expert here).
     
    Last edited: Oct 10, 2007
  7. Oct 10, 2007 #6
    ∫(dx)/(eˆx + eˆ(-x))=∫(eˆx)/(1 + eˆ2x) t=eˆx x=ln (t) (dt)/(dx)=t => dx=1/t dt

    ∫t/(tˆ2 + 1) * 1/t dt => ∫1/(tˆ2 + 1) dt

    Yeah! I solved it. You're right Dick, it isn't wrong at all. I just misthough there.

    The next phase is using a formula ∫1/(xˆ2 + 1) dx = arctan x + C

    So the answer is arctan eˆx + C

    Thank you for help!

    How can I add [solved] to the title?
     
    Last edited: Oct 10, 2007
  8. Oct 10, 2007 #7

    Dick

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    I don't think you can put the solved mark on it - you have to have special powers, I'll try and do it.
     
  9. Oct 16, 2007 #8
    We had yesterday class where we discussed about this problem and this was a solution. The teacher called it "trick" mathematics. He said you just have to invent what formula do you use. The second option could be using trigonometrical functions or something like that in change of variables. But I can get nowhere using them...
     
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