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

Answered by thfchristopher last updated on 02/Jun/22

∫e^(2x) sin 3xdx =(1/2)∫sin 3xd(e^(2x) ) =(1/2)e^(2x) sin 3x−(1/2)∫e^(2x) d(sin 3x) =(1/2)e^(2x) sin 3x−(3/2)∫e^(2x) cos 3xdx =(1/2)e^(2x) sin 3x−(3/4)∫cos 3xd(e^(2x) ) =(1/2)e^(2x) sin 3x−(3/4)e^(2x) cos 3x+(3/4)∫e^(2x) d(cos 3x) =(1/2)e^(2x) sin 3x−(3/4)e^(2x) cos 3x−(9/4)∫e^(2x) sin 3xdx ⇒((13)/4)∫e^(2x) sin 3xdx=(e^(2x) /4)(2sin 3x−3cos 3x)+C ⇒∫e^(2x) sin 3xdx=(e^(2x) /(13))(2sin 3x−3cos 3x)+C

$$\int{e}^{\mathrm{2}{x}} \mathrm{sin}\:\mathrm{3}{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{sin}\:\mathrm{3}{xd}\left({e}^{\mathrm{2}{x}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} \mathrm{sin}\:\mathrm{3}{x}−\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{\mathrm{2}{x}} {d}\left(\mathrm{sin}\:\mathrm{3}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} \mathrm{sin}\:\mathrm{3}{x}−\frac{\mathrm{3}}{\mathrm{2}}\int{e}^{\mathrm{2}{x}} \mathrm{cos}\:\mathrm{3}{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} \mathrm{sin}\:\mathrm{3}{x}−\frac{\mathrm{3}}{\mathrm{4}}\int\mathrm{cos}\:\mathrm{3}{xd}\left({e}^{\mathrm{2}{x}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} \mathrm{sin}\:\mathrm{3}{x}−\frac{\mathrm{3}}{\mathrm{4}}{e}^{\mathrm{2}{x}} \mathrm{cos}\:\mathrm{3}{x}+\frac{\mathrm{3}}{\mathrm{4}}\int{e}^{\mathrm{2}{x}} {d}\left(\mathrm{cos}\:\mathrm{3}{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{x}} \mathrm{sin}\:\mathrm{3}{x}−\frac{\mathrm{3}}{\mathrm{4}}{e}^{\mathrm{2}{x}} \mathrm{cos}\:\mathrm{3}{x}−\frac{\mathrm{9}}{\mathrm{4}}\int{e}^{\mathrm{2}{x}} \mathrm{sin}\:\mathrm{3}{xdx} \\ $$$$\Rightarrow\frac{\mathrm{13}}{\mathrm{4}}\int{e}^{\mathrm{2}{x}} \mathrm{sin}\:\mathrm{3}{xdx}=\frac{{e}^{\mathrm{2}{x}} }{\mathrm{4}}\left(\mathrm{2sin}\:\mathrm{3}{x}−\mathrm{3cos}\:\mathrm{3}{x}\right)+{C} \\ $$$$\Rightarrow\int{e}^{\mathrm{2}{x}} \mathrm{sin}\:\mathrm{3}{xdx}=\frac{{e}^{\mathrm{2}{x}} }{\mathrm{13}}\left(\mathrm{2sin}\:\mathrm{3}{x}−\mathrm{3cos}\:\mathrm{3}{x}\right)+{C} \\ $$

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