Solve a returning integration-by-parts equation in two ways

Let $I=\int e^{2x}\cos4x\,dx$ and $J=\int e^{2x}\sin4x\,dx$. Parts gives $I=\frac12e^{2x}\cos4x+2J$. A second parts step gives $J=\frac12e^{2x}\sin4x-2I$. Therefore $I=\frac12e^{2x}\cos4x+e^{2x}\sin4x-4I$, so $5I=\frac12e^{2x}(\cos4x+2\sin4x)+C\prime$ and $I=\frac1{10}e^{2x}(\cos4x+2\sin4x)+C$.

Parts gives $I=\frac14e^{2x}\sin4x-\frac12J$. Taking the corresponding second parts route gives $J=-\frac14e^{2x}\cos4x+\frac12I$. Thus $I=\frac14e^{2x}\sin4x+\frac18e^{2x}\cos4x-\frac14I$. Collecting the returning integral yields $\frac54I=\frac18e^{2x}(\cos4x+2\sin4x)+C\prime$, and the same final antiderivative follows.