Integral de $$$e^{- 2 x} \sin{\left(e^{- x} \right)}$$$
Calculadora relacionada: Calculadora de integrales definidas e impropias
Tu entrada
Halla $$$\int e^{- 2 x} \sin{\left(e^{- x} \right)}\, dx$$$.
Solución
Sea $$$u=e^{- x}$$$.
Entonces $$$du=\left(e^{- x}\right)^{\prime }dx = - e^{- x} dx$$$ (los pasos pueden verse »), y obtenemos que $$$e^{- x} dx = - du$$$.
Entonces,
$${\color{red}{\int{e^{- 2 x} \sin{\left(e^{- x} \right)} d x}}} = {\color{red}{\int{\left(- u \sin{\left(u \right)}\right)d u}}}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = u \sin{\left(u \right)}$$$:
$${\color{red}{\int{\left(- u \sin{\left(u \right)}\right)d u}}} = {\color{red}{\left(- \int{u \sin{\left(u \right)} d u}\right)}}$$
Para la integral $$$\int{u \sin{\left(u \right)} d u}$$$, utiliza la integración por partes $$$\int \operatorname{c} \operatorname{dv} = \operatorname{c}\operatorname{v} - \int \operatorname{v} \operatorname{dc}$$$.
Sean $$$\operatorname{c}=u$$$ y $$$\operatorname{dv}=\sin{\left(u \right)} du$$$.
Entonces $$$\operatorname{dc}=\left(u\right)^{\prime }du=1 du$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{\sin{\left(u \right)} d u}=- \cos{\left(u \right)}$$$ (los pasos pueden verse »).
La integral puede reescribirse como
$$- {\color{red}{\int{u \sin{\left(u \right)} d u}}}=- {\color{red}{\left(u \cdot \left(- \cos{\left(u \right)}\right)-\int{\left(- \cos{\left(u \right)}\right) \cdot 1 d u}\right)}}=- {\color{red}{\left(- u \cos{\left(u \right)} - \int{\left(- \cos{\left(u \right)}\right)d u}\right)}}$$
Aplica la regla del factor constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ con $$$c=-1$$$ y $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:
$$u \cos{\left(u \right)} + {\color{red}{\int{\left(- \cos{\left(u \right)}\right)d u}}} = u \cos{\left(u \right)} + {\color{red}{\left(- \int{\cos{\left(u \right)} d u}\right)}}$$
La integral del coseno es $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:
$$u \cos{\left(u \right)} - {\color{red}{\int{\cos{\left(u \right)} d u}}} = u \cos{\left(u \right)} - {\color{red}{\sin{\left(u \right)}}}$$
Recordemos que $$$u=e^{- x}$$$:
$$- \sin{\left({\color{red}{u}} \right)} + {\color{red}{u}} \cos{\left({\color{red}{u}} \right)} = - \sin{\left({\color{red}{e^{- x}}} \right)} + {\color{red}{e^{- x}}} \cos{\left({\color{red}{e^{- x}}} \right)}$$
Por lo tanto,
$$\int{e^{- 2 x} \sin{\left(e^{- x} \right)} d x} = - \sin{\left(e^{- x} \right)} + e^{- x} \cos{\left(e^{- x} \right)}$$
Añade la constante de integración:
$$\int{e^{- 2 x} \sin{\left(e^{- x} \right)} d x} = - \sin{\left(e^{- x} \right)} + e^{- x} \cos{\left(e^{- x} \right)}+C$$
Respuesta
$$$\int e^{- 2 x} \sin{\left(e^{- x} \right)}\, dx = \left(- \sin{\left(e^{- x} \right)} + e^{- x} \cos{\left(e^{- x} \right)}\right) + C$$$A