$$$\theta \tan{\left(2 \right)}$$$の積分

$$$\int \theta \tan{\left(2 \right)}\, d\theta$$$ 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、$$$c=\tan{\left(2 \right)}$$$ と $$$f{\left(\theta \right)} = \theta$$$ に対して適用する:

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

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

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

したがって、

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

積分定数を加える:

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

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