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

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

項別に積分せよ:

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

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

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

$$$u=\tan{\left(x \right)}$$$ 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、

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

分数を変形して分解する:

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

項別に積分せよ:

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

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

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

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

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

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

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

したがって、

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

簡単化せよ:

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

積分定数を加える:

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

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