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Question Number 23108 by Kathie last updated on 26/Oct/17

∫e^(−x) sin2x dx

$$\int\mathrm{e}^{−\mathrm{x}} \mathrm{sin2x}\:\mathrm{dx} \\ $$

Answered by sma3l2996 last updated on 26/Oct/17

I=∫e^(−x) sin2xdx u=sin2x⇒u′=2cos2x v′=e^(−x) ⇒v=−e^(−x) I=−e^(−x) sin2x+2∫e^(−x) cos2xdx+c u=cos2x⇒u′=−2sin2x v′=e^(−x) ⇒v=−e^(−x) I=−e^(−x) sin2x−2e^(−x) cos2x−4∫e^(−x) sin2xdx+c_1 I=−e^(−x) (sin2x+2cos2x)−4I+c_1 5I=−e^(−x) (sin2x+2cos2x)+c_1 I=−(1/5)e^(−x) (sin2x+2cos2x)+C

$${I}=\int{e}^{−{x}} {sin}\mathrm{2}{xdx} \\ $$$${u}={sin}\mathrm{2}{x}\Rightarrow{u}'=\mathrm{2}{cos}\mathrm{2}{x} \\ $$$${v}'={e}^{−{x}} \Rightarrow{v}=−{e}^{−{x}} \\ $$$${I}=−{e}^{−{x}} {sin}\mathrm{2}{x}+\mathrm{2}\int{e}^{−{x}} {cos}\mathrm{2}{xdx}+{c} \\ $$$${u}={cos}\mathrm{2}{x}\Rightarrow{u}'=−\mathrm{2}{sin}\mathrm{2}{x} \\ $$$${v}'={e}^{−{x}} \Rightarrow{v}=−{e}^{−{x}} \\ $$$${I}=−{e}^{−{x}} {sin}\mathrm{2}{x}−\mathrm{2}{e}^{−{x}} {cos}\mathrm{2}{x}−\mathrm{4}\int{e}^{−{x}} {sin}\mathrm{2}{xdx}+{c}_{\mathrm{1}} \\ $$$${I}=−{e}^{−{x}} \left({sin}\mathrm{2}{x}+\mathrm{2}{cos}\mathrm{2}{x}\right)−\mathrm{4}{I}+{c}_{\mathrm{1}} \\ $$$$\mathrm{5}{I}=−{e}^{−{x}} \left({sin}\mathrm{2}{x}+\mathrm{2}{cos}\mathrm{2}{x}\right)+{c}_{\mathrm{1}} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{5}}{e}^{−{x}} \left({sin}\mathrm{2}{x}+\mathrm{2}{cos}\mathrm{2}{x}\right)+{C} \\ $$

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