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Math Help - integral of e^2x * sinx

  1. #1
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    integral of e^2x * sinx

    i gotta integrate the following:

    e^2x * sinx

    i got this as my answer:

    (4/5)(-e^2x*cosx + .5e^2x * sinx)

    is it right?
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  2. #2
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    Is done by parts twice.


    Wolfram Mathematica Online Integrator
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  3. #3
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    An alternative method is to take the derivative twice...

    \frac{d}{dx}(e^{2x}\sin{x}) = e^{2x}\cos{x} + 2e^{2x}\sin{x}.

    Therefore \int{e^{2x}\cos{x} + 2e^{2x}\sin{x}\,dx} = e^{2x}\sin{x}

    \int{e^{2x}\cos{x}\,dx} + 2\int{e^{2x}\sin{x}\,dx} = e^{2x}\sin{x}

    \int{e^{2x}\sin{x}\,dx} = \frac{1}{2}e^{2x}\sin{x} - \frac{1}{2}\int{e^{2x}\,\cos{x}\,dx}.

    Call this equation 1. Notice that now we have another integral to find... so take the derivative again...


    \frac{d}{dx}(e^{2x}\cos{x}) = 2e^{2x}\cos{x} - e^{2x}\sin{x}.

    Therefore \int{2e^{2x}\cos{x} - e^{2x}\sin{x}\,dx} = e^{2x}\cos{x}

    2\int{e^{2x}\cos{x}\,dx} - \int{e^{2x}\sin{x}\,dx} = e^{2x}\cos{x}

    \int{e^{2x}\cos{x}\,dx} = \frac{1}{2}e^{2x}\cos{x} + \frac{1}{2}\int{e^{2x}\sin{x}\,dx}.

    Substitute this back into equation 1:


    \int{e^{2x}\sin{x}\,dx} = \frac{1}{2}e^{2x}\sin{x} - \frac{1}{2}\left(\frac{1}{2}e^{2x}\cos{x} + \frac{1}{2}\int{e^{2x}\sin{x}\,dx}\right)

    \int{e^{2x}\sin{x}\,dx} = \frac{1}{2}e^{2x}\sin{x} - \frac{1}{4}e^{2x}\cos{x} - \frac{1}{4}\int{e^{2x}\sin{x}\,dx}

    \frac{5}{4}\int{e^{2x}\sin{x}\,dx} = \frac{1}{2}e^{2x}\sin{x} - \frac{1}{4}e^{2x}\cos{x}

    \int{e^{2x}\sin{x}\,dx} = \frac{2}{5}e^{2x}\sin{x} - \frac{1}{5}e^{2x}\cos{x} + C.
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