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Trig substitution

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    Use a trigonometric substitution to evaluate the integral.


    2. Relevant equations
    [itex]\int e^x\,dx [/itex] [itex]/\sqrt{1-e^2x}[/itex]


    3. The attempt at a solution
    e^x = sin∅
    x=lnsin∅
    dx=cos∅/sin∅


    [itex]\frac{sin∅*cos∅}{sin∅*\sqrt{1-(sin∅)^2}}[/itex]





    [itex]\int sin∅cos∅
    /
    sin∅(cos∅)\,d∅ [/itex]



    [itex]\int \,d∅ = ∅[/itex]

    e^x = sin∅
    ∅ = arcsin(e^x)

    answer:
    ∅ + C
    arcsin(e^x) + c
     
    Last edited: Mar 8, 2013
  2. jcsd
  3. Mar 8, 2013 #2

    tiny-tim

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    hi whatlifeforme! :smile:
    nooo :redface:

    you forgot the sqrt :wink:

    (btw it's easier to say ex = sinθ, so exdx = cosθdθ)
     
  4. Mar 8, 2013 #3
    so it should be

    [itex]\int sin∅cos∅
    /
    sin∅(cos∅)\,dx[/itex]
     
  5. Mar 8, 2013 #4

    thus,


    [itex]\int 1\,d∅[/itex]

    e^x = sin∅
    ∅ = arcsin(e^x)

    answer: arcsin(e^x) + c
     
    Last edited: Mar 8, 2013
  6. Mar 8, 2013 #5

    tiny-tim

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    yup! :biggrin:

    (except of course that integral should have been ∫ 1 dθ :wink:)
     
  7. Mar 8, 2013 #6
    ooopss. sry fixed.
     
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