$$$\frac{1}{\tan^{2}{\left(x \right)}}$$$の積分

$$$\int \frac{1}{\tan^{2}{\left(x \right)}}\, dx$$$ を求めよ。

$$$u=\tan{\left(x \right)}$$$ とする。

すると $$$x=\operatorname{atan}{\left(u \right)}$$$ および $$$dx=\left(\operatorname{atan}{\left(u \right)}\right)^{\prime }du = \frac{du}{u^{2} + 1}$$$（手順は»で確認できます）。

したがって、

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

部分分数分解を行う (手順は»で確認できます):

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

項別に積分せよ:

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

$$$n=-2$$$ を用いて、べき乗の法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ を適用します:

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

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

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

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

$$- \operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}}^{-1} = - \operatorname{atan}{\left({\color{red}{\tan{\left(x \right)}}} \right)} - {\color{red}{\tan{\left(x \right)}}}^{-1}$$

したがって、

$$\int{\frac{1}{\tan^{2}{\left(x \right)}} d x} = - \operatorname{atan}{\left(\tan{\left(x \right)} \right)} - \frac{1}{\tan{\left(x \right)}}$$

簡単化せよ:

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

積分定数を加える:

$$\int{\frac{1}{\tan^{2}{\left(x \right)}} d x} = - x - \frac{1}{\tan{\left(x \right)}}+C$$

$$$\int \frac{1}{\tan^{2}{\left(x \right)}}\, dx = \left(- x - \frac{1}{\tan{\left(x \right)}}\right) + C$$$A