$$$\cot^{2}{\left(x \right)}$$$の積分

$$$\int \cot^{2}{\left(x \right)}\, dx$$$ を求めよ。

$$$u=\cot{\left(x \right)}$$$ とする。

すると $$$du=\left(\cot{\left(x \right)}\right)^{\prime }dx = - \csc^{2}{\left(x \right)} dx$$$（手順は»で確認できます）、$$$\csc^{2}{\left(x \right)} dx = - du$$$ となります。

したがって、

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

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=-1$$$ と $$$f{\left(u \right)} = \frac{u^{2}}{u^{2} + 1}$$$ に対して適用する:

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

分数を変形して分解する:

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

項別に積分せよ:

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

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

$$\int{\frac{1}{u^{2} + 1} d u} - {\color{red}{\int{1 d u}}} = \int{\frac{1}{u^{2} + 1} d u} - {\color{red}{u}}$$

$$$\frac{1}{u^{2} + 1}$$$ の不定積分は $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$ です:

$$- u + {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}} = - u + {\color{red}{\operatorname{atan}{\left(u \right)}}}$$

次のことを思い出してください $$$u=\cot{\left(x \right)}$$$:

$$\operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}} = \operatorname{atan}{\left({\color{red}{\cot{\left(x \right)}}} \right)} - {\color{red}{\cot{\left(x \right)}}}$$

したがって、

$$\int{\cot^{2}{\left(x \right)} d x} = - \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}$$

積分定数を加える:

$$\int{\cot^{2}{\left(x \right)} d x} = - \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}+C$$

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