Integral dari $$$\cos{\left(4 x - 2 \right)} - 1$$$

Temukan $$$\int \left(\cos{\left(4 x - 2 \right)} - 1\right)\, dx$$$.

Integralkan suku demi suku:

$${\color{red}{\int{\left(\cos{\left(4 x - 2 \right)} - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{\cos{\left(4 x - 2 \right)} d x}\right)}}$$

Terapkan aturan konstanta $$$\int c\, dx = c x$$$ dengan $$$c=1$$$:

$$\int{\cos{\left(4 x - 2 \right)} d x} - {\color{red}{\int{1 d x}}} = \int{\cos{\left(4 x - 2 \right)} d x} - {\color{red}{x}}$$

Misalkan $$$u=4 x - 2$$$.

Kemudian $$$du=\left(4 x - 2\right)^{\prime }dx = 4 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{4}$$$.

Oleh karena itu,

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

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

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

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

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

Ingat bahwa $$$u=4 x - 2$$$:

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

Oleh karena itu,

$$\int{\left(\cos{\left(4 x - 2 \right)} - 1\right)d x} = - x + \frac{\sin{\left(4 x - 2 \right)}}{4}$$

Tambahkan konstanta integrasi:

$$\int{\left(\cos{\left(4 x - 2 \right)} - 1\right)d x} = - x + \frac{\sin{\left(4 x - 2 \right)}}{4}+C$$

$$$\int \left(\cos{\left(4 x - 2 \right)} - 1\right)\, dx = \left(- x + \frac{\sin{\left(4 x - 2 \right)}}{4}\right) + C$$$A