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

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

Untuk integral $$$\int{x \sin{\left(2 x \right)} d x}$$$, gunakan integrasi parsial $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Misalkan $$$\operatorname{u}=x$$$ dan $$$\operatorname{dv}=\sin{\left(2 x \right)} dx$$$.

Maka $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{\sin{\left(2 x \right)} d x}=- \frac{\cos{\left(2 x \right)}}{2}$$$ (langkah-langkah dapat dilihat di »).

Integralnya menjadi

$${\color{red}{\int{x \sin{\left(2 x \right)} d x}}}={\color{red}{\left(x \cdot \left(- \frac{\cos{\left(2 x \right)}}{2}\right)-\int{\left(- \frac{\cos{\left(2 x \right)}}{2}\right) \cdot 1 d x}\right)}}={\color{red}{\left(- \frac{x \cos{\left(2 x \right)}}{2} - \int{\left(- \frac{\cos{\left(2 x \right)}}{2}\right)d x}\right)}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=- \frac{1}{2}$$$ dan $$$f{\left(x \right)} = \cos{\left(2 x \right)}$$$:

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

Misalkan $$$u=2 x$$$.

Kemudian $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{2}$$$.

Integralnya menjadi

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{2}$$$ dan $$$f{\left(u \right)} = \cos{\left(u \right)}$$$:

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

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

$$- \frac{x \cos{\left(2 x \right)}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = - \frac{x \cos{\left(2 x \right)}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

Ingat bahwa $$$u=2 x$$$:

$$- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{4} = - \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

$$$\int x \sin{\left(2 x \right)}\, dx = \left(- \frac{x \cos{\left(2 x \right)}}{2} + \frac{\sin{\left(2 x \right)}}{4}\right) + C$$$A