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SOme help with intergration by parts,

  1. Jan 15, 2008 #1
    SOme help with intergration by parts, plz

    1. Find: [tex] \int e^x cos (x) dx [/tex]

    3. The attempt at a solution

    I tried using integration by parts - what we are working on... all these 3 possibilities:

    [tex] u = cos x, u= e^x, u = e^x cos(x) [/tex]

    And the [tex] \int vdu [/tex] are, respectively:

    [tex]1. \int e^x sin(x) dx

    >> 2.\int sin(x) e^x dx

    >> 3.\int x e^x cos(x) dx [/tex]

    which won't work out very well... Please give me a suggestion.

    Thanks :)

    PS: I dont know how to start a new line in the coded body :S srry
  2. jcsd
  3. Jan 15, 2008 #2
    Either will be fine. The concept behind this problem is that you have to do Parts more than once. After the 2nd or 3rd time, you will notice that you get your original Integral back, bring it to the other side and then just divide by the constant and you're pretty much done.
  4. Jan 16, 2008 #3
    Use LAITE for future problems related to integration by parts.

    L - Logarithmic function
    A - Algebraic function
    I - Inverse function (like arctan)
    T - Trig function
    E - Exponential function

    So in your example, you have an exponential function and a trig function. According to the acronym, the trig function comes before Exponential function. So u= cos x.

    This method had worked for ALL the integration by parts I have done so far
    e.g. int(ln(x)) where you use 1 x ln(x). u= 1 in this case.
  5. Jan 16, 2008 #4

    Tom Mattson

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    You've misspelled the acronym: It's LIATE. You'd have a real hard time integrating [itex]x\arctan(x)[/itex] if you spelled it LAITE!
  6. Jan 16, 2008 #5
    What he (Tom Mattson) said.
  7. Jan 16, 2008 #6


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    If u= cos x, then du= -sin x dx and dv= exdx so v= ex. Then
    [tex]\int e^x cos x dx= -e^x cos x+ \int e^x sin x dx[/tex]
    Now, do it again, letting u= sin(x), dv= ex dx. Your result will be
    [tex]\int e^x cos(x)dx[/tex]= something involving that same integral. combine them and solve for [tex]\int e^x cos(x)dx[/tex].

    If u= ex then du= exdx and dv= cos x dx so v= sin x. Then
    [tex]\int e^x cos x dx= e^x sin x- \int e^x sin x dx[/tex]
    Again, repeat and do the same thing as in the first one

    if u= e^x cos x then du= e^x cos x- e^x sin x and dv= dx so v= x. Your result will be
    [tex]\int e^x cos x dx= x e^x cos x- \int xe^x cos x- xe^x sin x dx[/tex]
    which, I agree, doesn't seem to help. Use one of the first two methods.

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