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Question Number 170878 by 0731619 last updated on 02/Jun/22

	Answered by thfchristopher last updated on 02/Jun/22

	$\int e^{2 x} \sin 3 x d x$ $= \frac{1}{2} \int \sin 3 x d (e^{2 x})$ $= \frac{1}{2} e^{2 x} \sin 3 x - \frac{1}{2} \int e^{2 x} d (\sin 3 x)$ $= \frac{1}{2} e^{2 x} \sin 3 x - \frac{3}{2} \int e^{2 x} \cos 3 x d x$ $= \frac{1}{2} e^{2 x} \sin 3 x - \frac{3}{4} \int \cos 3 x d (e^{2 x})$ $= \frac{1}{2} e^{2 x} \sin 3 x - \frac{3}{4} e^{2 x} \cos 3 x + \frac{3}{4} \int e^{2 x} d (\cos 3 x)$ $= \frac{1}{2} e^{2 x} \sin 3 x - \frac{3}{4} e^{2 x} \cos 3 x - \frac{9}{4} \int e^{2 x} \sin 3 x d x$ $\Rightarrow \frac{13}{4} \int e^{2 x} \sin 3 x d x = \frac{e^{2 x}}{4} (2 \sin 3 x - 3 \cos 3 x) + C$ $\Rightarrow \int e^{2 x} \sin 3 x d x = \frac{e^{2 x}}{13} (2 \sin 3 x - 3 \cos 3 x) + C$
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