Integral dari $$$- x \sin{\left(x \right)} \tan{\left(1 \right)}$$$

Temukan $$$\int \left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)\, dx$$$.

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

$${\color{red}{\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x}}} = {\color{red}{\left(- \tan{\left(1 \right)} \int{x \sin{\left(x \right)} d x}\right)}}$$

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

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

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

Oleh karena itu,

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

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

$$- \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} - {\color{red}{\int{\left(- \cos{\left(x \right)}\right)d x}}}\right) = - \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} - {\color{red}{\left(- \int{\cos{\left(x \right)} d x}\right)}}\right)$$

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

$$- \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} + {\color{red}{\int{\cos{\left(x \right)} d x}}}\right) = - \tan{\left(1 \right)} \left(- x \cos{\left(x \right)} + {\color{red}{\sin{\left(x \right)}}}\right)$$

Oleh karena itu,

$$\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x} = - \left(- x \cos{\left(x \right)} + \sin{\left(x \right)}\right) \tan{\left(1 \right)}$$

Sederhanakan:

$$\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x} = \left(x \cos{\left(x \right)} - \sin{\left(x \right)}\right) \tan{\left(1 \right)}$$

Tambahkan konstanta integrasi:

$$\int{\left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)d x} = \left(x \cos{\left(x \right)} - \sin{\left(x \right)}\right) \tan{\left(1 \right)}+C$$

$$$\int \left(- x \sin{\left(x \right)} \tan{\left(1 \right)}\right)\, dx = \left(x \cos{\left(x \right)} - \sin{\left(x \right)}\right) \tan{\left(1 \right)} + C$$$A