$$$\sin^{2}{\left(x_{0} \right)}$$$の積分

$$$\int \sin^{2}{\left(x_{0} \right)}\, dx_{0}$$$ を求めよ。

冪低減公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$ を $$$\alpha=x_{0}$$$ に適用する:

$${\color{red}{\int{\sin^{2}{\left(x_{0} \right)} d x_{0}}}} = {\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x_{0} \right)}}{2}\right)d x_{0}}}}$$

定数倍の法則 $$$\int c f{\left(x_{0} \right)}\, dx_{0} = c \int f{\left(x_{0} \right)}\, dx_{0}$$$ を、$$$c=\frac{1}{2}$$$ と $$$f{\left(x_{0} \right)} = 1 - \cos{\left(2 x_{0} \right)}$$$ に対して適用する:

$${\color{red}{\int{\left(\frac{1}{2} - \frac{\cos{\left(2 x_{0} \right)}}{2}\right)d x_{0}}}} = {\color{red}{\left(\frac{\int{\left(1 - \cos{\left(2 x_{0} \right)}\right)d x_{0}}}{2}\right)}}$$

項別に積分せよ:

$$\frac{{\color{red}{\int{\left(1 - \cos{\left(2 x_{0} \right)}\right)d x_{0}}}}}{2} = \frac{{\color{red}{\left(\int{1 d x_{0}} - \int{\cos{\left(2 x_{0} \right)} d x_{0}}\right)}}}{2}$$

$$$c=1$$$ に対して定数則 $$$\int c\, dx_{0} = c x_{0}$$$ を適用する:

$$- \frac{\int{\cos{\left(2 x_{0} \right)} d x_{0}}}{2} + \frac{{\color{red}{\int{1 d x_{0}}}}}{2} = - \frac{\int{\cos{\left(2 x_{0} \right)} d x_{0}}}{2} + \frac{{\color{red}{x_{0}}}}{2}$$

$$$u=2 x_{0}$$$ とする。

すると $$$du=\left(2 x_{0}\right)^{\prime }dx_{0} = 2 dx_{0}$$$（手順は»で確認できます）、$$$dx_{0} = \frac{du}{2}$$$ となります。

したがって、

$$\frac{x_{0}}{2} - \frac{{\color{red}{\int{\cos{\left(2 x_{0} \right)} d x_{0}}}}}{2} = \frac{x_{0}}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2}$$

定数倍の法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ を、$$$c=\frac{1}{2}$$$ と $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ に対して適用する:

$$\frac{x_{0}}{2} - \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{2} = \frac{x_{0}}{2} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{2}$$

余弦の積分は$$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$:

$$\frac{x_{0}}{2} - \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{4} = \frac{x_{0}}{2} - \frac{{\color{red}{\sin{\left(u \right)}}}}{4}$$

次のことを思い出してください $$$u=2 x_{0}$$$:

$$\frac{x_{0}}{2} - \frac{\sin{\left({\color{red}{u}} \right)}}{4} = \frac{x_{0}}{2} - \frac{\sin{\left({\color{red}{\left(2 x_{0}\right)}} \right)}}{4}$$

したがって、

$$\int{\sin^{2}{\left(x_{0} \right)} d x_{0}} = \frac{x_{0}}{2} - \frac{\sin{\left(2 x_{0} \right)}}{4}$$

積分定数を加える:

$$\int{\sin^{2}{\left(x_{0} \right)} d x_{0}} = \frac{x_{0}}{2} - \frac{\sin{\left(2 x_{0} \right)}}{4}+C$$

$$$\int \sin^{2}{\left(x_{0} \right)}\, dx_{0} = \left(\frac{x_{0}}{2} - \frac{\sin{\left(2 x_{0} \right)}}{4}\right) + C$$$A