Menu Close

Question-216799




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

Leave a Reply

Your email address will not be published. Required fields are marked *