Integral dari $$$\frac{\sin{\left(5 x - 3 \right)}}{t}$$$ terhadap $$$x$$$

Temukan $$$\int \frac{\sin{\left(5 x - 3 \right)}}{t}\, dx$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=\frac{1}{t}$$$ dan $$$f{\left(x \right)} = \sin{\left(5 x - 3 \right)}$$$:

$${\color{red}{\int{\frac{\sin{\left(5 x - 3 \right)}}{t} d x}}} = {\color{red}{\frac{\int{\sin{\left(5 x - 3 \right)} d x}}{t}}}$$

Misalkan $$$u=5 x - 3$$$.

Kemudian $$$du=\left(5 x - 3\right)^{\prime }dx = 5 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = \frac{du}{5}$$$.

Jadi,

$$\frac{{\color{red}{\int{\sin{\left(5 x - 3 \right)} d x}}}}{t} = \frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{t}$$

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{5}$$$ dan $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

$$\frac{{\color{red}{\int{\frac{\sin{\left(u \right)}}{5} d u}}}}{t} = \frac{{\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{5}\right)}}}{t}$$

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

$$\frac{{\color{red}{\int{\sin{\left(u \right)} d u}}}}{5 t} = \frac{{\color{red}{\left(- \cos{\left(u \right)}\right)}}}{5 t}$$

Ingat bahwa $$$u=5 x - 3$$$:

$$- \frac{\cos{\left({\color{red}{u}} \right)}}{5 t} = - \frac{\cos{\left({\color{red}{\left(5 x - 3\right)}} \right)}}{5 t}$$

Oleh karena itu,

$$\int{\frac{\sin{\left(5 x - 3 \right)}}{t} d x} = - \frac{\cos{\left(5 x - 3 \right)}}{5 t}$$

Tambahkan konstanta integrasi:

$$\int{\frac{\sin{\left(5 x - 3 \right)}}{t} d x} = - \frac{\cos{\left(5 x - 3 \right)}}{5 t}+C$$

$$$\int \frac{\sin{\left(5 x - 3 \right)}}{t}\, dx = - \frac{\cos{\left(5 x - 3 \right)}}{5 t} + C$$$A