Calculus: Integration by Parts (Loop Method)

First parts: let $u = e^{2x} \implies u' = 2e^{2x}$; $v' = \cos(3x) \implies v = \frac{1}{3}\sin(3x)$

$I = \frac{1}{3}e^{2x}\sin(3x) - \int \frac{2}{3}e^{2x}\sin(3x)\,dx$

Second parts on $\int e^{2x}\sin(3x)dx$: let $u = e^{2x} \implies u' = 2e^{2x}$; $v' = \sin(3x) \implies v = -\frac{1}{3}\cos(3x)$

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

Substitute back: $I = \frac{1}{3}e^{2x}\sin(3x) - \frac{2}{3}\left(-\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{3}I\right) = \frac{1}{3}e^{2x}\sin(3x) + \frac{2}{9}e^{2x}\cos(3x) - \frac{4}{9}I$

Group $I$ terms: $I + \frac{4}{9}I = \frac{13}{9}I = e^{2x}\left(\frac{1}{3}\sin(3x) + \frac{2}{9}\cos(3x)\right) = \frac{e^{2x}}{9}(3\sin(3x) + 2\cos(3x))$

Multiply by $\frac{9}{13}$: $I = \frac{e^{2x}}{13}(2\cos(3x) + 3\sin(3x)) + C$ (AG)