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Integration - Special Case

  1. Jan 15, 2014 #1
    Hello,

    I have been going round in circles on a problem, but I am aware of a special case regarding this integral, however I don't know how to use or really understand it.

    1. The problem statement, all variables and given/known data

    Integrate the following function with respect to x.

    [itex]e^{x} \sin 2x[/itex]


    2. Relevant equations

    [itex]2 \int e^{x} \sin x dx = e^{x} (\sin x - \cos x) + c[/itex]

    I believe this is the relevant equation I need to use...

    3. The attempt at a solution

    [itex]\int e^{x} \sin 2x = uv - \int u \frac{dv}{dx} dx [/itex]

    [itex] v = e^{x}[/itex]

    [itex] \frac{dv}{dx} = e^{x}[/itex]

    [itex] \frac{du}{dx} = \sin 2x[/itex]

    [itex] u = -\frac{1}{2} \cos 2x[/itex]

    So,

    [itex]\int e^{x} \sin 2x = -\frac{1}{2} e^{x} \cos 2x - \int -\frac{1}{2} e^{x} \cos 2x[/itex]

    Consider

    [itex] \int -\frac{1}{2} e^{x} \cos 2x = uv - \int u \frac{dv}{dx} dx [/itex]

    [itex] v = e^{x} [/itex]

    [itex] \frac{dv}{dx} = e^{x}[/itex]

    [itex] \frac{du}{dx} = -\frac{1}{2} \cos 2x[/itex]

    [itex] u = -\frac{1}{4} \sin 2x [/itex]

    so,

    [itex] \int -\frac{1}{2} e^{x} \cos 2x = -\frac{1}{4} e^{x} \sin 2x - \int -\frac{1}{4} e^{x} \sin 2x[/itex]

    Bringing that together gives,

    [itex]\int e^{x} \sin 2x = -\frac{1}{2} e^{x} \cos 2x + \frac{1}{4} e^{x} \sin 2x - \int -\frac{1}{4} e^{x} \sin 2x[/itex]


    This doesn't seem to be getting me anywhere as i'm sure repeating this process will leave me with another integral on the end...

    I was trying to follow the process my book uses to arrive at the 'special case' I gave in the relevant equations section, but I don't know how to proceed.

    Thanks!
     
  2. jcsd
  3. Jan 15, 2014 #2

    Office_Shredder

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    Subtracting[itex] \frac{1}{4} \int e^{x} \sin 2x dx [/itex] to both sides

    [tex] \frac{3}{4} \int e^{x} \sin 2x dx = -\frac{1}{2} e^{x} \cos 2x + \frac{1}{4} e^{x} \sin 2x+C[/tex]

    However I think you have a minus sign issue in getting to that last equation you should go back and look for. The final step will be exactly the same idea though.
     
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