$$$2 \tan^{2}{\left(\theta \right)}$$$の積分

$$$\int 2 \tan^{2}{\left(\theta \right)}\, d\theta$$$ を求めよ。

定数倍の法則 $$$\int c f{\left(\theta \right)}\, d\theta = c \int f{\left(\theta \right)}\, d\theta$$$ を、$$$c=2$$$ と $$$f{\left(\theta \right)} = \tan^{2}{\left(\theta \right)}$$$ に対して適用する:

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

$$$u=\tan{\left(\theta \right)}$$$ とする。

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

積分は次のようになります

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

分数を変形して分解する:

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

項別に積分せよ:

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

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

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

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

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

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

$$- 2 \operatorname{atan}{\left({\color{red}{u}} \right)} + 2 {\color{red}{u}} = - 2 \operatorname{atan}{\left({\color{red}{\tan{\left(\theta \right)}}} \right)} + 2 {\color{red}{\tan{\left(\theta \right)}}}$$

したがって、

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

簡単化せよ:

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

積分定数を加える:

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

$$$\int 2 \tan^{2}{\left(\theta \right)}\, d\theta = 2 \left(- \theta + \tan{\left(\theta \right)}\right) + C$$$A