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Math Help - Integration by Parts - Matching with Original Integral

  1. #1
    Member VitaX's Avatar
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    Integration by Parts - Matching with Original Integral

    \int e^{4x} sin(5x) dx

    I know I have to use integration by parts and match up with the original integral since neither function will get down to a coefficient.

    \int u dv = uv - \int v du

    Here u = e^{4x} dv = sin(5x) du=4e^{4x} v= - \frac{1}{5}cos(5x)

    \int e^{4x} sin(5x) dx = e^{4x} \left(-\frac{1}{5}cos(5x)\right) - \int -\frac{1}{5} cos(5x)(4e^{4x}) dx

    \int e^{4x} sin(5x) dx = e^{4x} \left(-\frac{1}{5}cos(5x)\right) + \frac{4}{5} \int e^{4x} cos(5x)dx

    \int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) +\frac{4}{5}\left[e^{4x}\left(\frac{1}{5}sin(5x)\right) - \int \frac{1}{5}sin(5x)(4e^{4x})dx\right]

    \int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) +\frac{4}{5}\left[\frac{1}{5}e^{4x}sin(5x) - \frac{4}{5} \int e^{4x}sin(5x)dx\right]

    \int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) + \frac{4}{25}e^{4x}sin(5x) - \frac{16}{25} \int e^{4x}sin(5x)dx

    At this point I'm stuck. I can see that on both sides of the equations I do have the matching integral but the coefficient in front of the integral on the right side is tripping me up. Or perhaps I screwed up somewhere.
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  2. #2
    Super Member General's Avatar
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    Quote Originally Posted by VitaX View Post
    \int e^{4x} sin(5x) dx

    I know I have to use integration by parts and match up with the original integral since neither function will get down to a coefficient.

    \int u dv = uv - \int v du

    Here u = e^{4x} dv = sin(5x) du=4e^{4x} v= - \frac{1}{5}cos(5x)

    \int e^{4x} sin(5x) dx = e^{4x} \left(-\frac{1}{5}cos(5x)\right) - \int -\frac{1}{5} cos(5x)(4e^{4x}) dx

    \int e^{4x} sin(5x) dx = e^{4x} \left(-\frac{1}{5}cos(5x)\right) + \frac{4}{5} \int e^{4x} cos(5x)dx

    \int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) +\frac{4}{5}\left[e^{4x}\left(\frac{1}{5}sin(5x)\right) - \int \frac{1}{5}sin(5x)(4e^{4x})dx\right]

    \int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) +\frac{4}{5}\left[\frac{1}{5}e^{4x}sin(5x) - \frac{4}{5} \int e^{4x}sin(5x)dx\right]

    \int e^{4x} sin(5x) dx = -\frac{1}{5}e^{4x}cos(5x) + \frac{4}{25}e^{4x}sin(5x) - \frac{16}{25} \int e^{4x}sin(5x)dx

    At this point I'm stuck. I can see that on both sides of the equations I do have the matching integral but the coefficient in front of the integral on the right side is tripping me up. Or perhaps I screwed up somewhere.
    Move - \frac{16}{25} \int e^{4x}sin(5x)dx to the left hand side, then take \int e^{4x}sin(5x)dx as common factor..
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  3. #3
    Member VitaX's Avatar
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    Quote Originally Posted by General View Post
    Move - \frac{16}{25} \int e^{4x}sin(5x)dx to the left hand side, then take \int e^{4x}sin(5x)dx as common factor..
    ok I didn't think about common factors at the time. by the way isn't there a short cut to integrating by parts if you need to do it more than once?
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  4. #4
    Super Member General's Avatar
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    Quote Originally Posted by VitaX View Post
    by the way isn't there a short cut to integrating by parts if you need to do it more than once?
    Yes. Reduction formula.
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