$$$\frac{2 \sin{\left(x \right)}}{5}$$$の積分

$$$\int \frac{2 \sin{\left(x \right)}}{5}\, dx$$$ を求めよ。

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{2}{5}$$$ と $$$f{\left(x \right)} = \sin{\left(x \right)}$$$ に対して適用する:

$${\color{red}{\int{\frac{2 \sin{\left(x \right)}}{5} d x}}} = {\color{red}{\left(\frac{2 \int{\sin{\left(x \right)} d x}}{5}\right)}}$$

正弦関数の不定積分は$$$\int{\sin{\left(x \right)} d x} = - \cos{\left(x \right)}$$$です:

$$\frac{2 {\color{red}{\int{\sin{\left(x \right)} d x}}}}{5} = \frac{2 {\color{red}{\left(- \cos{\left(x \right)}\right)}}}{5}$$

したがって、

$$\int{\frac{2 \sin{\left(x \right)}}{5} d x} = - \frac{2 \cos{\left(x \right)}}{5}$$

積分定数を加える:

$$\int{\frac{2 \sin{\left(x \right)}}{5} d x} = - \frac{2 \cos{\left(x \right)}}{5}+C$$

$$$\int \frac{2 \sin{\left(x \right)}}{5}\, dx = - \frac{2 \cos{\left(x \right)}}{5} + C$$$A