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find-e-x-sinxdx-




Question Number 172012 by Mikenice last updated on 23/Jun/22
find  ∫e^x sinxdx
$${find} \\ $$$$\int{e}^{{x}} {sinxdx} \\ $$
Answered by puissant last updated on 23/Jun/22
J=∫e^x sinxdx   { ((u′=e^x )),((v=sinx)) :} ⇒  { ((u=e^x )),((v′=cosx)) :}  J = e^x sinx −∫e^x cosxdx  L=∫e^x cosxdx   { ((u′=e^x )),((v=cosx)) :} ⇒  { ((u=e^x )),((v′=−sinx)) :}  L = e^x cosx + ∫e^x sinxdx  ⇒ J= e^x sinx − e^x cosx−J   ⇒ J  =  (e^x /2)sinx  − (e^x /2)cosx + C
$${J}=\int{e}^{{x}} {sinxdx} \\ $$$$\begin{cases}{{u}'={e}^{{x}} }\\{{v}={sinx}}\end{cases}\:\Rightarrow\:\begin{cases}{{u}={e}^{{x}} }\\{{v}'={cosx}}\end{cases} \\ $$$${J}\:=\:{e}^{{x}} {sinx}\:−\int{e}^{{x}} {cosxdx} \\ $$$${L}=\int{e}^{{x}} {cosxdx} \\ $$$$\begin{cases}{{u}'={e}^{{x}} }\\{{v}={cosx}}\end{cases}\:\Rightarrow\:\begin{cases}{{u}={e}^{{x}} }\\{{v}'=−{sinx}}\end{cases} \\ $$$${L}\:=\:{e}^{{x}} {cosx}\:+\:\int{e}^{{x}} {sinxdx} \\ $$$$\Rightarrow\:{J}=\:{e}^{{x}} {sinx}\:−\:{e}^{{x}} {cosx}−{J}\: \\ $$$$\Rightarrow\:{J}\:\:=\:\:\frac{{e}^{{x}} }{\mathrm{2}}{sinx}\:\:−\:\frac{{e}^{{x}} }{\mathrm{2}}{cosx}\:+\:{C} \\ $$

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