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Simple Integration by Parts Question

  1. Jul 11, 2012 #1
    1. The problem statement, all variables and given/known data
    integral of e^-xsinxdx


    2. Relevant equations
    uv-/vdu


    3. The attempt at a solution
    u=e^-x
    du = -e^-xdx
    v=sinx
    dv=cosxdx

    e^-xsinx-/(-e^-x)sinx
    =e^-x(sinx+cosx)

    Wolfram alpha is telling me that the indefinite integral is actually "-1/2e^-x(sinx+cosx)" where did the -1/2 come from? :(
     
  2. jcsd
  3. Jul 11, 2012 #2

    LCKurtz

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    dv = sin(x)
    v = - cos(x) dx

    I don't understand what you did in those last two lines.You should have, after that integration by parts,$$
    \int e^{-x}\sin x\, dx = -e^{-x}\cos x -\int e^{-x}\cos x\,dx $$
    You have to do the second integral on the right and solve for your original integral. Then you will see where the 1/2 comes from.
     
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