Integral dari $$$- 6 \sin{\left(2 t \right)}$$$

Temukan $$$\int \left(- 6 \sin{\left(2 t \right)}\right)\, dt$$$.

Terapkan aturan pengali konstanta $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ dengan $$$c=-6$$$ dan $$$f{\left(t \right)} = \sin{\left(2 t \right)}$$$:

$${\color{red}{\int{\left(- 6 \sin{\left(2 t \right)}\right)d t}}} = {\color{red}{\left(- 6 \int{\sin{\left(2 t \right)} d t}\right)}}$$

Misalkan $$$u=2 t$$$.

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

Integralnya menjadi

$$- 6 {\color{red}{\int{\sin{\left(2 t \right)} d t}}} = - 6 {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}}$$

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

$$- 6 {\color{red}{\int{\frac{\sin{\left(u \right)}}{2} d u}}} = - 6 {\color{red}{\left(\frac{\int{\sin{\left(u \right)} d u}}{2}\right)}}$$

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

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

Ingat bahwa $$$u=2 t$$$:

$$3 \cos{\left({\color{red}{u}} \right)} = 3 \cos{\left({\color{red}{\left(2 t\right)}} \right)}$$

Oleh karena itu,

$$\int{\left(- 6 \sin{\left(2 t \right)}\right)d t} = 3 \cos{\left(2 t \right)}$$

Tambahkan konstanta integrasi:

$$\int{\left(- 6 \sin{\left(2 t \right)}\right)d t} = 3 \cos{\left(2 t \right)}+C$$

$$$\int \left(- 6 \sin{\left(2 t \right)}\right)\, dt = 3 \cos{\left(2 t \right)} + C$$$A