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Question Number 216799 by OPAVdx last updated on 20/Feb/25

Commented by MathematicalUser2357 last updated on 22/Feb/25

sen?

$${sen}? \\ $$

Answered by MATHEMATICSAM last updated on 22/Feb/25

∫ e^x sinx dx u = sinx ⇒ du = cosx dx dv = e^x dx ⇒ v = e^x ∫udv = uv − ∫vdu = e^x sinx − ∫e^x cosx dx ....(i) Now lets integrate ∫e^x cosx dx Now u = cosx ⇒ du = − sinx dx dv = e^x dx ⇒ v = e^x ∫udv = uv − ∫vdu = e^x cosx + ∫e^x sinx dx ... (ii) From (i) and (ii) ∫e^x sinx dx = e^x sinx − e^x cosx − ∫e^x sinx dx ⇒ 2∫e^x sinx dx = e^x (sinx − cosx) ⇒ ∫e^x sinx dx = ((e^x (sinx − cosx))/2) + C

$$\int\:{e}^{{x}} \mathrm{sin}{x}\:{dx} \\ $$$${u}\:=\:\mathrm{sin}{x} \\ $$$$\Rightarrow\:{du}\:=\:\mathrm{cos}{x}\:{dx} \\ $$$${dv}\:=\:{e}^{{x}} \:{dx} \\ $$$$\Rightarrow\:{v}\:=\:{e}^{{x}} \\ $$$$ \\ $$$$\int{udv}\:=\:{uv}\:−\:\int{vdu} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:{e}^{{x}} \mathrm{sin}{x}\:−\:\int{e}^{{x}} \mathrm{cos}{x}\:{dx}\:....\left(\mathrm{i}\right) \\ $$$$ \\ $$$$\mathrm{Now}\:\mathrm{lets}\:\mathrm{integrate}\:\int{e}^{{x}} \mathrm{cos}{x}\:{dx} \\ $$$$\:\mathrm{Now}\:{u}\:=\:\mathrm{cos}{x} \\ $$$$\Rightarrow\:{du}\:=\:−\:\mathrm{sin}{x}\:{dx} \\ $$$${dv}\:=\:{e}^{{x}} \:{dx} \\ $$$$\Rightarrow\:{v}\:=\:{e}^{{x}} \\ $$$$\int{udv}\:=\:{uv}\:−\:\int{vdu} \\ $$$$\:\:\:\:\:\:\:\:\:=\:{e}^{{x}} \mathrm{cos}{x}\:+\:\int{e}^{{x}} \mathrm{sin}{x}\:{dx}\:...\:\left(\mathrm{ii}\right) \\ $$$$ \\ $$$$\mathrm{From}\:\left(\mathrm{i}\right)\:\mathrm{and}\:\left(\mathrm{ii}\right) \\ $$$$\int{e}^{{x}} \mathrm{sin}{x}\:{dx}\:=\:{e}^{{x}} \mathrm{sin}{x}\:−\:{e}^{{x}} \mathrm{cos}{x}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\int{e}^{{x}} \mathrm{sin}{x}\:{dx} \\ $$$$\Rightarrow\:\mathrm{2}\int{e}^{{x}} \mathrm{sin}{x}\:{dx}\:=\:{e}^{{x}} \left(\mathrm{sin}{x}\:−\:\mathrm{cos}{x}\right) \\ $$$$\Rightarrow\:\int{e}^{{x}} \mathrm{sin}{x}\:{dx}\:=\:\frac{{e}^{{x}} \left(\mathrm{sin}{x}\:−\:\mathrm{cos}{x}\right)}{\mathrm{2}}\:+\:\mathrm{C} \\ $$

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