Integral dari $$$e^{- 2 x} \sin{\left(e^{- x} \right)}$$$

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

Misalkan $$$u=e^{- x}$$$.

Kemudian $$$du=\left(e^{- x}\right)^{\prime }dx = - e^{- x} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$e^{- x} dx = - du$$$.

Oleh karena itu,

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=-1$$$ dan $$$f{\left(u \right)} = u \sin{\left(u \right)}$$$:

$${\color{red}{\int{\left(- u \sin{\left(u \right)}\right)d u}}} = {\color{red}{\left(- \int{u \sin{\left(u \right)} d u}\right)}}$$

Untuk integral $$$\int{u \sin{\left(u \right)} d u}$$$, gunakan integrasi parsial $$$\int \operatorname{c} \operatorname{dv} = \operatorname{c}\operatorname{v} - \int \operatorname{v} \operatorname{dc}$$$.

Misalkan $$$\operatorname{c}=u$$$ dan $$$\operatorname{dv}=\sin{\left(u \right)} du$$$.

Maka $$$\operatorname{dc}=\left(u\right)^{\prime }du=1 du$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{\sin{\left(u \right)} d u}=- \cos{\left(u \right)}$$$ (langkah-langkah dapat dilihat di »).

Dengan demikian,

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=-1$$$ dan $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

Integral dari kosinus adalah $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

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

Ingat bahwa $$$u=e^{- x}$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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