Integral dari $$$\sin{\left(3 t \right)}$$$

Temukan $$$\int \sin{\left(3 t \right)}\, dt$$$.

Misalkan $$$u=3 t$$$.

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

Oleh karena itu,

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

Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=\frac{1}{3}$$$ dan $$$f{\left(u \right)} = \sin{\left(u \right)}$$$:

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

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

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

Ingat bahwa $$$u=3 t$$$:

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

Oleh karena itu,

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

Tambahkan konstanta integrasi:

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

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