integrate (e^2x) sin(3x+1) dx
by doing integration by parts- can we use e^2x as first function? if then what will be the answer?

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Solution

$Using ILATE, exponential comes at end so here \sin (3 x + 1) will be first function and e^{2 x} will be second . So \int e^{2 x} \sin (3 x + 1) dx I = \sin (3 x + 1) \int e^{2 x} dx - \int [\frac{d}{dx} \sin (3 x + 1) \int e^{2 x} dx] dx = \frac{\sin (3 x + 1) e^{2 x}}{2} - \frac{3}{2} \int [e^{2 x} \cos (3 x + 1)] dx = \frac{\sin (3 x + 1) e^{2 x}}{2} - \frac{3}{2} [\cos (3 x + 1) \int e^{2 x} dx - \int [\frac{d}{dx} \cos (3 x + 1) \int e^{2 x} dx] dx] = \frac{\sin (3 x + 1) e^{2 x}}{2} - \frac{3}{2} [\frac{1}{2} e^{2 x} \cos (3 x + 1) + \frac{3}{2} \int [e^{2 x} \sin (3 x + 1)] dx] = \frac{\sin (3 x + 1) e^{2 x}}{2} - \frac{3}{4} e^{2 x} \cos (3 x + 1) - \frac{9}{4} I I + \frac{9}{4} I = \frac{\sin (3 x + 1) e^{2 x}}{2} - \frac{3}{4} e^{2 x} \cos (3 x + 1) I = \frac{4}{13} [\frac{\sin (3 x + 1) e^{2 x}}{2} - \frac{3}{4} e^{2 x} \cos (3 x + 1)]$

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