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Math Help - Integration of ∫e^(-1/2)t cos t dt

  1. #1
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    Integration of ∫e^(-1/2)t cos t dt

    How would you integrate \int e^{(-1/2)t}(\cos t)\,dt?

    I tried integrating by parts, but the problem I discovered is that it doesn't work because when you integrate either function, it just yields another function. For example, if I choose u=\cos t and dv=e^{(-1/2)t} dv, cos just turns into sin x, which turns into -cos x, etc and e^(-1/2)t never goes away either.

    Please help!
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  2. #2
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    let k = -\frac{1}{2}

    \int e^{kt} \cos{t} \, dt<br />

    u = \cos{t} ... dv = e^{kt} \, dt

    du = -\sin{t} \, dt ... v = \frac{1}{k}e^{kt}

    \int e^{kt} \cos{t} \, dt = \frac{1}{k}e^{kt} \cos{t} + \frac{1}{k} \int e^{kt} \sin{t} \, dt

    parts again w/ the last integral term ...

    u = \sin{t} ... dv = e^{kt} \, dt<br />

    du = \cos{t} \, dt ... v = \frac{1}{k}e^{kt}

    \int e^{kt} \cos{t} \, dt = \frac{1}{k}e^{kt} \cos{t} + \frac{1}{k}\left[\frac{1}{k}e^{kt} \sin{t} - \frac{1}{k} \int e^{kt} \cos{t} \, dt \right]

    \int e^{kt} \cos{t} \, dt = \frac{1}{k}e^{kt} \cos{t} + \frac{1}{k^2}e^{kt} \sin{t} - \frac{1}{k^2} \int e^{kt} \cos{t} \, dt

    \int e^{kt} \cos{t} \, dt + \frac{1}{k^2} \int e^{kt} \cos{t} \, dt = \frac{1}{k}e^{kt} \cos{t} + \frac{1}{k^2}e^{kt} \sin{t}

    back substitute k = -\frac{1}{2} ...

    \int e^{-\frac{1}{2}t} \cos{t} \, dt + 4\int e^{-\frac{1}{2}t} \cos{t} \, dt = -2e^{-\frac{1}{2}t} \cos{t} + 4e^{-\frac{1}{2}t} \sin{t}

    5\int e^{-\frac{1}{2}t} \cos{t} \, dt = -2e^{-\frac{1}{2}t} \cos{t} + 4e^{-\frac{1}{2}t} \sin{t}

    \int e^{-\frac{1}{2}t} \cos{t} \, dt = -\frac{2}{5}e^{-\frac{1}{2}t} \cos{t} + \frac{4}{5}e^{-\frac{1}{2}t} \sin{t} + C
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  3. #3
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    Thanks a bunch skeeter I would have never figured that one out =\
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