# How to solve $\int e^{2x}\sin(3x)\ dx$?

Integration by parts - unless I'm not thinking straight - doesn't seem to help here.. If I pick either of the functions $e^{2x}$ or $\sin(3x)$ to be $u$ or $dv$, they don't change to anything easier... $e^{2x}$ stays in the form $e^x$ and $\sin(3x)$ flip-flops between $\sin(3x)$ and $\cos(3x)$, neglecting the constant that gets introduced.

So how can I approach solving this? u-substitution doesn't seem to be something I can use either.

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This is almost identical to a couple of questions asked before. See: math.stackexchange.com/questions/307995/… and also: math.stackexchange.com/questions/136595/… –  Tom Oldfield Feb 27 at 22:24
I guess you could also use $\int e^{2x}\sin(3x)\ dx = \int e^{2x}\Im(\cos(3x)+i\sin(3x))\ dx =\int e^{2x}\Im(e^{3ix})\ dx=\int \Im(e^{2x}e^{3ix})\ dx =\Im(\int e^{(2+3i)x})\ dx$. Quite sure it's not the most efficient way though. –  xavierm02 Feb 27 at 22:38

The key here is to set $v' = e^{2x}$ and integrate by parts twice. This will make $\sin$ become $\cos$ and then $\sin$ again. The same integral will show up on the right side but with a different factor (so it won't cancel out).

Call the integral $I$ and integrate by parts twice to get: \begin{align} I &= \int e^{2x} \sin(3x) \,dx = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2} \int e^{2x} \cos(3x) \,dx \\ &= \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2} \left(\frac{1}{2} e^{2x}\cos(3x) + \frac{3}{2} \int e^{2x} \sin(3x) \,dx\right) \end{align}

Thus: $$I = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2} \left(\frac{1}{2} e^{2x}\cos(3x) + \frac{3}{2} I \right)$$

Solve for $I$ to get:

$$I = \frac{1}{13}e^{2x}\left(2\sin(3x) -3\cos(3x)\right)$$

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