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I'm trying to integrate $e^{2x} \cdot \sin 3x dx$ using integration by parts technique. According to some websites my answer is incorrect.

My attempt was:

$\int e^{2x}\sin (3x) dx = [u=e^{2x}, du = 2e^{2x} , dv= \sin 3x, v= -1/3\cos 3x] = uv -\int vdu = -\frac{1}{3}e^{2x}\cos 3x + \frac {2}{3}\int e^{2x} \cos (3x) dx = [u=e^{2x}, du = 2e^{2x}, dv = \cos (3x), v = \frac{1}{3}\sin (3x)] = -\frac{1}{3}e^{2x} \cos3x + \frac{1}{3} e^{2x}\sin 3x -\frac{2}{3}\int e^2x \sin (3x) dx $

And so I get: $\frac {5}{3}\int e^{2x} \sin (3x)dx = -\frac{1}{3}e^{2x}\cos 3x + \frac{1}{3}e^{2x}\sin 3x$

And finally:

$\int e^{2x} \sin (3x)dx = \frac{1}{5}e^{2x} (\sin(3x)-\cos(3x))+C$.

According to the websites I checked the answer should be $\frac{1}{13}e^{2x}(2\sin(3x)-3\cos(3x) + C$.

Where is my mistake?

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In your second integration by parts you forgot the factor of $\frac{2}{3}$ from the first one, so \begin{align*} ... &= -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3}\left(\frac{1}{3}e^{2x}\sin 3x - \frac{2}{3}\int e^{2x}\sin 3x\,dx\right) \\ &= -\frac{4}{9}\int e^{2x}\sin 3x\,dx - \frac{1}{3}e^{2x}\cos 3x + \frac{2}{9}e^{2x}\sin 3x. \end{align*} Now simplify.

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