Integral de $$$e^{x} \cos{\left(2 x \right)}$$$

Halla $$$\int e^{x} \cos{\left(2 x \right)}\, dx$$$.

Para la integral $$$\int{e^{x} \cos{\left(2 x \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\cos{\left(2 x \right)}$$$ y $$$\operatorname{dv}=e^{x} dx$$$.

Entonces $$$\operatorname{du}=\left(\cos{\left(2 x \right)}\right)^{\prime }dx=- 2 \sin{\left(2 x \right)} dx$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{e^{x} d x}=e^{x}$$$ (los pasos pueden verse »).

La integral puede reescribirse como

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

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=-2$$$ y $$$f{\left(x \right)} = e^{x} \sin{\left(2 x \right)}$$$:

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

Para la integral $$$\int{e^{x} \sin{\left(2 x \right)} d x}$$$, utiliza la integración por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sean $$$\operatorname{u}=\sin{\left(2 x \right)}$$$ y $$$\operatorname{dv}=e^{x} dx$$$.

Entonces $$$\operatorname{du}=\left(\sin{\left(2 x \right)}\right)^{\prime }dx=2 \cos{\left(2 x \right)} dx$$$ (los pasos pueden verse ») y $$$\operatorname{v}=\int{e^{x} d x}=e^{x}$$$ (los pasos pueden verse »).

Por lo tanto,

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

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=2$$$ y $$$f{\left(x \right)} = e^{x} \cos{\left(2 x \right)}$$$:

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

Hemos llegado a una integral que ya hemos visto.

Así, hemos obtenido la siguiente ecuación simple con respecto a la integral:

$$\int{e^{x} \cos{\left(2 x \right)} d x} = 2 e^{x} \sin{\left(2 x \right)} + e^{x} \cos{\left(2 x \right)} - 4 \int{e^{x} \cos{\left(2 x \right)} d x}$$

Al resolverlo, obtenemos que

$$\int{e^{x} \cos{\left(2 x \right)} d x} = \frac{\left(2 \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) e^{x}}{5}$$

Por lo tanto,

$$\int{e^{x} \cos{\left(2 x \right)} d x} = \frac{\left(2 \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) e^{x}}{5}$$

Añade la constante de integración:

$$\int{e^{x} \cos{\left(2 x \right)} d x} = \frac{\left(2 \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) e^{x}}{5}+C$$

$$$\int e^{x} \cos{\left(2 x \right)}\, dx = \frac{\left(2 \sin{\left(2 x \right)} + \cos{\left(2 x \right)}\right) e^{x}}{5} + C$$$A