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Math Help - Integration by parts

  1. #1
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    Integration by parts

    Hello!

    I am trying to solve the following indefinite integral by using integration by parts (as is recommended by the textbook): \int{(e^x\sin{x})dx}.

    However, I am stuck in the same circle over and over again.

    U = e^x and dU = (e^x)dx.
    dV = (\sin{x})dx and V = -\cos{x}.

    So: \int{(U)dV} = UV - \int{(V)dU} gives \int{(e^x\sin{x})dx} = -e^x\cos{x} - \int{(-e^x\cos{x})dx}.

    Resulting in the same problem when trying to find \int{(-e^x\cos{x})dx}.

    Even when I define U and dV the other way around, I am still stuck with e^x and either \sin{x} or \cos{x}.

    How do I do this? Please help!
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  2. #2
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    Re: Integration by parts

    \int e^x \sin{x} \, dx

    u = \sin{x} \, , \, dv = e^x \, dx

    du = \cos{x} \, dx \, , \, v = e^x

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

    parts again with the last integral expression ...

    u = \cos{x} \, , \, dv = e^x \, dx

    du = -\sin{x} \, dx \, , \, v = e^x

    \int e^x \sin{x} \, dx = e^x \sin{x} - \left[e^x \cos{x} - \int e^x (-\sin{x}) \. dx \right]

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

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

    \int e^x \sin{x} \, dx = \frac{e^x(\sin{x}-\cos{x})}{2} + C
    Thanks from Lotte1990
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