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

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

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

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

Então $$$\operatorname{du}=\left(\cos{\left(2 x \right)}\right)^{\prime }dx=- 2 \sin{\left(2 x \right)} dx$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{x} d x}=e^{x}$$$ (os passos podem ser vistos »).

Assim,

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=-2$$$ e $$$f{\left(x \right)} = e^{x} \sin{\left(2 x \right)}$$$:

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

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

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

Então $$$\operatorname{du}=\left(\sin{\left(2 x \right)}\right)^{\prime }dx=2 \cos{\left(2 x \right)} dx$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{x} d x}=e^{x}$$$ (os passos podem ser vistos »).

A integral torna-se

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=2$$$ e $$$f{\left(x \right)} = e^{x} \cos{\left(2 x \right)}$$$:

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

Chegamos a uma integral que já vimos.

Assim, obtivemos a seguinte equação simples em relação à integral:

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

Resolvendo, obtemos que

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

Portanto,

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

Adicione a constante de integração:

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

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