$$$\frac{\csc^{2}{\left(x \right)}}{9}$$$の積分

$$$\int \frac{\csc^{2}{\left(x \right)}}{9}\, dx$$$ 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、$$$c=\frac{1}{9}$$$ と $$$f{\left(x \right)} = \csc^{2}{\left(x \right)}$$$ に対して適用する:

$${\color{red}{\int{\frac{\csc^{2}{\left(x \right)}}{9} d x}}} = {\color{red}{\left(\frac{\int{\csc^{2}{\left(x \right)} d x}}{9}\right)}}$$

$$$\csc^{2}{\left(x \right)}$$$ の不定積分は $$$\int{\csc^{2}{\left(x \right)} d x} = - \cot{\left(x \right)}$$$ です:

$$\frac{{\color{red}{\int{\csc^{2}{\left(x \right)} d x}}}}{9} = \frac{{\color{red}{\left(- \cot{\left(x \right)}\right)}}}{9}$$

したがって、

$$\int{\frac{\csc^{2}{\left(x \right)}}{9} d x} = - \frac{\cot{\left(x \right)}}{9}$$

積分定数を加える:

$$\int{\frac{\csc^{2}{\left(x \right)}}{9} d x} = - \frac{\cot{\left(x \right)}}{9}+C$$

$$$\int \frac{\csc^{2}{\left(x \right)}}{9}\, dx = - \frac{\cot{\left(x \right)}}{9} + C$$$A