Integral de $$$e^{- 2 t} \sin{\left(4 t \right)}$$$

Encontre $$$\int e^{- 2 t} \sin{\left(4 t \right)}\, dt$$$.

Para a integral $$$\int{e^{- 2 t} \sin{\left(4 t \right)} d t}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=\sin{\left(4 t \right)}$$$ e $$$\operatorname{dv}=e^{- 2 t} dt$$$.

Então $$$\operatorname{du}=\left(\sin{\left(4 t \right)}\right)^{\prime }dt=4 \cos{\left(4 t \right)} dt$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{- 2 t} d t}=- \frac{e^{- 2 t}}{2}$$$ (os passos podem ser vistos »).

Assim,

$${\color{red}{\int{e^{- 2 t} \sin{\left(4 t \right)} d t}}}={\color{red}{\left(\sin{\left(4 t \right)} \cdot \left(- \frac{e^{- 2 t}}{2}\right)-\int{\left(- \frac{e^{- 2 t}}{2}\right) \cdot 4 \cos{\left(4 t \right)} d t}\right)}}={\color{red}{\left(- \int{\left(- 2 e^{- 2 t} \cos{\left(4 t \right)}\right)d t} - \frac{e^{- 2 t} \sin{\left(4 t \right)}}{2}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=-2$$$ e $$$f{\left(t \right)} = e^{- 2 t} \cos{\left(4 t \right)}$$$:

$$- {\color{red}{\int{\left(- 2 e^{- 2 t} \cos{\left(4 t \right)}\right)d t}}} - \frac{e^{- 2 t} \sin{\left(4 t \right)}}{2} = - {\color{red}{\left(- 2 \int{e^{- 2 t} \cos{\left(4 t \right)} d t}\right)}} - \frac{e^{- 2 t} \sin{\left(4 t \right)}}{2}$$

Para a integral $$$\int{e^{- 2 t} \cos{\left(4 t \right)} d t}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=\cos{\left(4 t \right)}$$$ e $$$\operatorname{dv}=e^{- 2 t} dt$$$.

Então $$$\operatorname{du}=\left(\cos{\left(4 t \right)}\right)^{\prime }dt=- 4 \sin{\left(4 t \right)} dt$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{- 2 t} d t}=- \frac{e^{- 2 t}}{2}$$$ (os passos podem ser vistos »).

Assim,

$$2 {\color{red}{\int{e^{- 2 t} \cos{\left(4 t \right)} d t}}} - \frac{e^{- 2 t} \sin{\left(4 t \right)}}{2}=2 {\color{red}{\left(\cos{\left(4 t \right)} \cdot \left(- \frac{e^{- 2 t}}{2}\right)-\int{\left(- \frac{e^{- 2 t}}{2}\right) \cdot \left(- 4 \sin{\left(4 t \right)}\right) d t}\right)}} - \frac{e^{- 2 t} \sin{\left(4 t \right)}}{2}=2 {\color{red}{\left(- \int{2 e^{- 2 t} \sin{\left(4 t \right)} d t} - \frac{e^{- 2 t} \cos{\left(4 t \right)}}{2}\right)}} - \frac{e^{- 2 t} \sin{\left(4 t \right)}}{2}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=2$$$ e $$$f{\left(t \right)} = e^{- 2 t} \sin{\left(4 t \right)}$$$:

$$- 2 {\color{red}{\int{2 e^{- 2 t} \sin{\left(4 t \right)} d t}}} - \frac{e^{- 2 t} \sin{\left(4 t \right)}}{2} - e^{- 2 t} \cos{\left(4 t \right)} = - 2 {\color{red}{\left(2 \int{e^{- 2 t} \sin{\left(4 t \right)} d t}\right)}} - \frac{e^{- 2 t} \sin{\left(4 t \right)}}{2} - e^{- 2 t} \cos{\left(4 t \right)}$$

Chegamos a uma integral que já vimos.

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

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

Resolvendo, obtemos que

$$\int{e^{- 2 t} \sin{\left(4 t \right)} d t} = \frac{\left(- \sin{\left(4 t \right)} - 2 \cos{\left(4 t \right)}\right) e^{- 2 t}}{10}$$

Portanto,

$$\int{e^{- 2 t} \sin{\left(4 t \right)} d t} = \frac{\left(- \sin{\left(4 t \right)} - 2 \cos{\left(4 t \right)}\right) e^{- 2 t}}{10}$$

Adicione a constante de integração:

$$\int{e^{- 2 t} \sin{\left(4 t \right)} d t} = \frac{\left(- \sin{\left(4 t \right)} - 2 \cos{\left(4 t \right)}\right) e^{- 2 t}}{10}+C$$

$$$\int e^{- 2 t} \sin{\left(4 t \right)}\, dt = \frac{\left(- \sin{\left(4 t \right)} - 2 \cos{\left(4 t \right)}\right) e^{- 2 t}}{10} + C$$$A