$$$\frac{\cos{\left(x \right)} \cot{\left(x \right)}}{\sin{\left(x \right)}}$$$の積分

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

被積分関数を簡単化する:

$${\color{red}{\int{\frac{\cos{\left(x \right)} \cot{\left(x \right)}}{\sin{\left(x \right)}} d x}}} = {\color{red}{\int{\left(-1 + \frac{1}{\sin^{2}{\left(x \right)}}\right)d x}}}$$

項別に積分せよ:

$${\color{red}{\int{\left(-1 + \frac{1}{\sin^{2}{\left(x \right)}}\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{\frac{1}{\sin^{2}{\left(x \right)}} d x}\right)}}$$

$$$c=1$$$ に対して定数則 $$$\int c\, dx = c x$$$ を適用する:

$$\int{\frac{1}{\sin^{2}{\left(x \right)}} d x} - {\color{red}{\int{1 d x}}} = \int{\frac{1}{\sin^{2}{\left(x \right)}} d x} - {\color{red}{x}}$$

被積分関数を余割関数を用いて書き換えなさい:

$$- x + {\color{red}{\int{\frac{1}{\sin^{2}{\left(x \right)}} d x}}} = - x + {\color{red}{\int{\csc^{2}{\left(x \right)} d x}}}$$

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

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

したがって、

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

積分定数を加える:

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

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