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Math Help - How do I integrate a multiplication equation?

  1. #1
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    Question How do I integrate a multiplication equation?

    I always have problems integrating a multiplication equation. How do I integrate an equation like this one:

    <br />
\int sin2x . e^{3x}<br />

    I tried doing it by parts but still canot solve it.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    You could not use by parts?

    \int  e^{3x}\sin2x\ dx = \sin 2x. \dfrac{e^{3x}}{3} - \int \dfrac{e^{3x}}{3} .2\cos 2x\ dx

    Clean up;

    \int  e^{3x}\sin2x\ dx = \dfrac{e^{3x}}{3}\sin 2x  - \dfrac23 \int e^{3x}\cos 2x\ dx

    By parts for the second part;

    \int  e^{3x}\sin2x\ dx = \dfrac{e^{3x}}{3}\sin 2x  - \dfrac23\left[\cos2x. \dfrac{e^{3x}}{3} - \int  \dfrac{e^{3x}}{3}.-2\sin2x\ dx\right]

    Does the last part of the integral seem familiar?

    Clean up:

    \int  e^{3x}\sin2x\ dx = \dfrac{e^{3x}}{3}\sin 2x  - \dfrac23\left[\cos2x. \dfrac{e^{3x}}{3} + \dfrac23\int  e^{3x}\sin2x\ dx\right]

    Expand the brackets;

    \int  e^{3x}\sin2x\ dx = \dfrac{e^{3x}}{3}\sin 2x  - \dfrac{2 e^{3x}}{9}\cos2x - \dfrac49\int  e^{3x}\sin2x\ dx

    The trick now is to send the last part to the left.

    \int  e^{3x}\sin2x\ dx + \dfrac49\int  e^{3x}\sin2x\ dx= \dfrac{e^{3x}}{3}\sin 2x  - \dfrac{2 e^{3x}}{9}\cos2x

    Simplify;

    \dfrac{13}{9}\int  e^{3x}\sin2x\ dx = \dfrac{e^{3x}}{3}\sin 2x  - \dfrac{2 e^{3x}}{9}\cos2x

    Multiply by 9/13;

    \int  e^{3x}\sin2x\ dx = \dfrac{9}{13}\left[\dfrac{e^{3x}}{3}\sin 2x  - \dfrac{2 e^{3x}}{9}\cos2x\right]

    Expand and there you are!
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  3. #3
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    wow! That's a nice trick! Thanks man!
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  4. #4
    MHF Contributor Unknown008's Avatar
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    You're welcome
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