$$$\sqrt{2} \left(\cot^{2}{\left(x \right)} - 1\right)$$$の積分

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

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

項別に積分せよ:

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

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

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

$$$u=\cot{\left(x \right)}$$$ 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。

この積分は次のように書き換えられる

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

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:

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

分数を変形して分解する:

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

項別に積分せよ:

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

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

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

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

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

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

$$\sqrt{2} \left(- x + \operatorname{atan}{\left({\color{red}{u}} \right)} - {\color{red}{u}}\right) = \sqrt{2} \left(- x + \operatorname{atan}{\left({\color{red}{\cot{\left(x \right)}}} \right)} - {\color{red}{\cot{\left(x \right)}}}\right)$$

したがって、

$$\int{\sqrt{2} \left(\cot^{2}{\left(x \right)} - 1\right) d x} = \sqrt{2} \left(- x - \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right)$$

積分定数を加える:

$$\int{\sqrt{2} \left(\cot^{2}{\left(x \right)} - 1\right) d x} = \sqrt{2} \left(- x - \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right)+C$$

$$$\int \sqrt{2} \left(\cot^{2}{\left(x \right)} - 1\right)\, dx = \sqrt{2} \left(- x - \cot{\left(x \right)} + \operatorname{atan}{\left(\cot{\left(x \right)} \right)}\right) + C$$$A