Integraal van $$$e^{- 2 x} \sin{\left(e^{- x} \right)}$$$

Bepaal $$$\int e^{- 2 x} \sin{\left(e^{- x} \right)}\, dx$$$.

Zij $$$u=e^{- x}$$$.

Dan $$$du=\left(e^{- x}\right)^{\prime }dx = - e^{- x} dx$$$ (de stappen zijn te zien »), en dan geldt dat $$$e^{- x} dx = - du$$$.

De integraal kan worden herschreven als

$${\color{red}{\int{e^{- 2 x} \sin{\left(e^{- x} \right)} d x}}} = {\color{red}{\int{\left(- u \sin{\left(u \right)}\right)d u}}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=-1$$$ en $$$f{\left(u \right)} = u \sin{\left(u \right)}$$$:

$${\color{red}{\int{\left(- u \sin{\left(u \right)}\right)d u}}} = {\color{red}{\left(- \int{u \sin{\left(u \right)} d u}\right)}}$$

Voor de integraal $$$\int{u \sin{\left(u \right)} d u}$$$, gebruik partiële integratie $$$\int \operatorname{c} \operatorname{dv} = \operatorname{c}\operatorname{v} - \int \operatorname{v} \operatorname{dc}$$$.

Zij $$$\operatorname{c}=u$$$ en $$$\operatorname{dv}=\sin{\left(u \right)} du$$$.

Dan $$$\operatorname{dc}=\left(u\right)^{\prime }du=1 du$$$ (de stappen zijn te zien ») en $$$\operatorname{v}=\int{\sin{\left(u \right)} d u}=- \cos{\left(u \right)}$$$ (de stappen zijn te zien »).

Dus,

$$- {\color{red}{\int{u \sin{\left(u \right)} d u}}}=- {\color{red}{\left(u \cdot \left(- \cos{\left(u \right)}\right)-\int{\left(- \cos{\left(u \right)}\right) \cdot 1 d u}\right)}}=- {\color{red}{\left(- u \cos{\left(u \right)} - \int{\left(- \cos{\left(u \right)}\right)d u}\right)}}$$

Pas de constante-veelvoudregel $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ toe met $$$c=-1$$$ en $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

$$u \cos{\left(u \right)} + {\color{red}{\int{\left(- \cos{\left(u \right)}\right)d u}}} = u \cos{\left(u \right)} + {\color{red}{\left(- \int{\cos{\left(u \right)} d u}\right)}}$$

De integraal van de cosinus is $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$u \cos{\left(u \right)} - {\color{red}{\int{\cos{\left(u \right)} d u}}} = u \cos{\left(u \right)} - {\color{red}{\sin{\left(u \right)}}}$$

We herinneren eraan dat $$$u=e^{- x}$$$:

$$- \sin{\left({\color{red}{u}} \right)} + {\color{red}{u}} \cos{\left({\color{red}{u}} \right)} = - \sin{\left({\color{red}{e^{- x}}} \right)} + {\color{red}{e^{- x}}} \cos{\left({\color{red}{e^{- x}}} \right)}$$

Dus,

$$\int{e^{- 2 x} \sin{\left(e^{- x} \right)} d x} = - \sin{\left(e^{- x} \right)} + e^{- x} \cos{\left(e^{- x} \right)}$$

Voeg de integratieconstante toe:

$$\int{e^{- 2 x} \sin{\left(e^{- x} \right)} d x} = - \sin{\left(e^{- x} \right)} + e^{- x} \cos{\left(e^{- x} \right)}+C$$

$$$\int e^{- 2 x} \sin{\left(e^{- x} \right)}\, dx = \left(- \sin{\left(e^{- x} \right)} + e^{- x} \cos{\left(e^{- x} \right)}\right) + C$$$A