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: $$I = \frac{1}{2}e^x(\sin x - \cos x)$$