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    I want to integrate e^x.cos(nx) (where n is number)

    I use the integration by parts method where uv - Int(v.du/dx)

    u = cos(nx) dv/dx = e^x
    du/dx = -nsin(nx) v = e^x

    Therefore

    = cos(nx).e^x - Int(-nsin(nx).e^x)
    = cos(nx).e^x + n.Int(sin(nx).e^x)

    I have to repeat this same method again for "Int(sin(nx).e^x)"

    u = sin(nx) dv/dx = e^x
    du/dx = ncos(nx) v = e^x

    So now:
    cos(nx).e^x + n [sin(nx).e^x - n.Int(cos(nx).e^x)]
    Using I = Int[e^x.cos(nx)]
    cos(nx).e^x + n [sin(nx).e^x - n.I]
    cos(nx).e^x + nsin(nx).e^x - n^2.I
    I.n^2 = cos(nx).e^x + nsin(nx).e^x
    I = [e^x (cos(nx) + nsin(nx))]/n^2


    Im sorry if thats hard to understand
    but ive also attached the handwritten version


    I checked my answer on Wolphram alpha,
    they have the same thing except in the denominator they have
    n^2 +1 .. where as I have n^2 only

    If someone could point out the mistake I've made - I would really appreciate it!
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    (Original post by Sifr)
    I want to integrate e^x.cos(nx) (where n is number)

    I use the integration by parts method where uv - Int(v.du/dx)

    u = cos(nx) dv/dx = e^x
    du/dx = -nsin(nx) v = e^x

    Therefore

    = cos(nx).e^x - Int(-nsin(nx).e^x)
    = cos(nx).e^x + n.Int(sin(nx).e^x)

    I have to repeat this same method again for "Int(sin(nx).e^x)"

    u = sin(nx) dv/dx = e^x
    du/dx = ncos(nx) v = e^x

    So now:
    cos(nx).e^x + n [sin(nx).e^x - n.Int(cos(nx).e^x)]
    Using I = Int[e^x.cos(nx)]
    cos(nx).e^x + n [sin(nx).e^x - n.I]
    cos(nx).e^x + nsin(nx).e^x - n^2.I
    I.n^2 = cos(nx).e^x + nsin(nx).e^x
    I = [e^x (cos(nx) + nsin(nx))]/n^2


    Im sorry if thats hard to understand
    but ive also attached the handwritten version


    I checked my answer on Wolphram alpha,
    they have the same thing except in the denominator they have
    n^2 +1 .. where as I have n^2 only

    If someone could point out the mistake I've made - I would really appreciate it!
    You have I=[\mathrm{stuff}] - n^2I, adding n^2I to both sides gives I+n^2I = [\mathrm{stuff}] where the LHS factorises to (1+n^2)I; not In^2 i.e. your mistake is two lines from the bottom of your written working.

    Unrelated, but your handwriting is excellent!
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    Wow, agreed, your handwriting is very nice

    On the last line, did you draw the fraction with a ruler or by hand 0.o
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    (Original post by Farhan.Hanif93)
    You have I=[\mathrm{stuff}] - n^2I, adding n^2I to both sides gives I+n^2I = [\mathrm{stuff}] where the LHS factorises to (1+n^2)I; not In^2 i.e. your mistake is two lines from the bottom of your written working.

    Unrelated, but your handwriting is excellent!
    The fact that I have good handwriting just proves why I am a terrible mathematician!
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    (Original post by Sifr)
    The fact that I have good handwriting just proves why I am a terrible mathematician!
    It also means you'll never become a doctor
 
 
 
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