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

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=-1$$$ e $$$f{\left(x \right)} = e^{2 x} \cos{\left(e^{x} \right)}$$$:

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

Seja $$$u=2 x$$$.

Então $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (veja os passos »), e obtemos $$$dx = \frac{du}{2}$$$.

A integral pode ser reescrita como

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

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{2}$$$ e $$$f{\left(u \right)} = e^{u} \cos{\left(e^{\frac{u}{2}} \right)}$$$:

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

Seja $$$v=e^{\frac{u}{2}}$$$.

Então $$$dv=\left(e^{\frac{u}{2}}\right)^{\prime }du = \frac{e^{\frac{u}{2}}}{2} du$$$ (veja os passos »), e obtemos $$$e^{\frac{u}{2}} du = 2 dv$$$.

Assim,

$$- \frac{{\color{red}{\int{e^{u} \cos{\left(e^{\frac{u}{2}} \right)} d u}}}}{2} = - \frac{{\color{red}{\int{2 v \cos{\left(v \right)} d v}}}}{2}$$

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

$$- \frac{{\color{red}{\int{2 v \cos{\left(v \right)} d v}}}}{2} = - \frac{{\color{red}{\left(2 \int{v \cos{\left(v \right)} d v}\right)}}}{2}$$

Para a integral $$$\int{v \cos{\left(v \right)} d v}$$$, use integração por partes $$$\int \operatorname{m} \operatorname{dy} = \operatorname{m}\operatorname{y} - \int \operatorname{y} \operatorname{dm}$$$.

Sejam $$$\operatorname{m}=v$$$ e $$$\operatorname{dy}=\cos{\left(v \right)} dv$$$.

Então $$$\operatorname{dm}=\left(v\right)^{\prime }dv=1 dv$$$ (os passos podem ser vistos ») e $$$\operatorname{y}=\int{\cos{\left(v \right)} d v}=\sin{\left(v \right)}$$$ (os passos podem ser vistos »).

A integral pode ser reescrita como

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

A integral do seno é $$$\int{\sin{\left(v \right)} d v} = - \cos{\left(v \right)}$$$:

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

Recorde que $$$v=e^{\frac{u}{2}}$$$:

$$- \cos{\left({\color{red}{v}} \right)} - {\color{red}{v}} \sin{\left({\color{red}{v}} \right)} = - \cos{\left({\color{red}{e^{\frac{u}{2}}}} \right)} - {\color{red}{e^{\frac{u}{2}}}} \sin{\left({\color{red}{e^{\frac{u}{2}}}} \right)}$$

Recorde que $$$u=2 x$$$:

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

Portanto,

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

Adicione a constante de integração:

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

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