Cyclic Integration

So I had a revelation recently on how to compute integration by parts that appear to be cyclic. Consider the function: \[ I = \int e^x \sin x \ \mathrm{d}x \] Integrating this by parts is trivial using the standard method: \[ \int u \frac{\mathrm{d}v}{\mathrm{d}x} \ \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \ \mathrm{d}x \] Hence for the function I: \[ \begin{array}{rlc} u = e^x & \frac{\mathrm{d}v}{\mathrm{d}x} = \sin x \\ \frac{\mathrm{d}u}{\mathrm{d}x} = e^x & v = -\cos x \end{array} \] So: \[ I = -e^x \cos x + \int e^x \cos x \ \mathrm{d}x \] The end term must be integrated by parts again: \[ \begin{array}{rlc} u = e^x & \frac{\mathrm{d}v}{\mathrm{d}x} = \cos x \\ \frac{\mathrm{d}u}{\mathrm{d}x} = e^x & v = \sin x \end{array} \] So this is: \[ I = -e^x \cos x +e^x \sin x - \int e^x \sin x \ \mathrm{d}x \] But the final integral is the same as the original integral so this can be expressed as: \[ I = -e^x \cos x +e^x \sin x - I \] This is: \[ 2I = -e^x \cos x +e^x \sin x \] which means: \[I = \frac{1}{2}e^x(\sin x - \cos x) \]