Integral dari $$$- 6 x \cos{\left(4 x \right)}$$$

Temukan $$$\int \left(- 6 x \cos{\left(4 x \right)}\right)\, dx$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=-6$$$ dan $$$f{\left(x \right)} = x \cos{\left(4 x \right)}$$$:

$${\color{red}{\int{\left(- 6 x \cos{\left(4 x \right)}\right)d x}}} = {\color{red}{\left(- 6 \int{x \cos{\left(4 x \right)} d x}\right)}}$$

Untuk integral $$$\int{x \cos{\left(4 x \right)} d x}$$$, gunakan integrasi parsial $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Misalkan $$$\operatorname{u}=x$$$ dan $$$\operatorname{dv}=\cos{\left(4 x \right)} dx$$$.

Maka $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{\cos{\left(4 x \right)} d x}=\frac{\sin{\left(4 x \right)}}{4}$$$ (langkah-langkah dapat dilihat di »).

Jadi,

$$- 6 {\color{red}{\int{x \cos{\left(4 x \right)} d x}}}=- 6 {\color{red}{\left(x \cdot \frac{\sin{\left(4 x \right)}}{4}-\int{\frac{\sin{\left(4 x \right)}}{4} \cdot 1 d x}\right)}}=- 6 {\color{red}{\left(\frac{x \sin{\left(4 x \right)}}{4} - \int{\frac{\sin{\left(4 x \right)}}{4} d x}\right)}}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=\frac{1}{4}$$$ dan $$$f{\left(x \right)} = \sin{\left(4 x \right)}$$$:

$$- \frac{3 x \sin{\left(4 x \right)}}{2} + 6 {\color{red}{\int{\frac{\sin{\left(4 x \right)}}{4} d x}}} = - \frac{3 x \sin{\left(4 x \right)}}{2} + 6 {\color{red}{\left(\frac{\int{\sin{\left(4 x \right)} d x}}{4}\right)}}$$

Misalkan $$$u=4 x$$$.

Kemudian $$$du=\left(4 x\right)^{\prime }dx = 4 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{4}$$$.

Integralnya menjadi

$$- \frac{3 x \sin{\left(4 x \right)}}{2} + \frac{3 {\color{red}{\int{\sin{\left(4 x \right)} d x}}}}{2} = - \frac{3 x \sin{\left(4 x \right)}}{2} + \frac{3 {\color{red}{\int{\frac{\sin{\left(u \right)}}{4} d u}}}}{2}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{4}$$$ dan $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$- \frac{3 x \sin{\left(4 x \right)}}{2} + \frac{3 {\color{red}{\int{\frac{\sin{\left(u \right)}}{4} d u}}}}{2} = - \frac{3 x \sin{\left(4 x \right)}}{2} + \frac{3 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{4}\right)}}}{2}$$

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

$$- \frac{3 x \sin{\left(4 x \right)}}{2} + \frac{3 {\color{red}{\int{\sin{\left(u \right)} d u}}}}{8} = - \frac{3 x \sin{\left(4 x \right)}}{2} + \frac{3 {\color{red}{\left(- \cos{\left(u \right)}\right)}}}{8}$$

Ingat bahwa $$$u=4 x$$$:

$$- \frac{3 x \sin{\left(4 x \right)}}{2} - \frac{3 \cos{\left({\color{red}{u}} \right)}}{8} = - \frac{3 x \sin{\left(4 x \right)}}{2} - \frac{3 \cos{\left({\color{red}{\left(4 x\right)}} \right)}}{8}$$

Oleh karena itu,

$$\int{\left(- 6 x \cos{\left(4 x \right)}\right)d x} = - \frac{3 x \sin{\left(4 x \right)}}{2} - \frac{3 \cos{\left(4 x \right)}}{8}$$

Sederhanakan:

$$\int{\left(- 6 x \cos{\left(4 x \right)}\right)d x} = - \frac{3 \left(4 x \sin{\left(4 x \right)} + \cos{\left(4 x \right)}\right)}{8}$$

Tambahkan konstanta integrasi:

$$\int{\left(- 6 x \cos{\left(4 x \right)}\right)d x} = - \frac{3 \left(4 x \sin{\left(4 x \right)} + \cos{\left(4 x \right)}\right)}{8}+C$$

$$$\int \left(- 6 x \cos{\left(4 x \right)}\right)\, dx = - \frac{3 \left(4 x \sin{\left(4 x \right)} + \cos{\left(4 x \right)}\right)}{8} + C$$$A