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

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

項別に積分せよ:

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

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

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

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

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

項別に積分せよ:

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

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

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

$$$u=2 x$$$ とする。

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

積分は次のようになります

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

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:

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

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

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

次のことを思い出してください $$$u=2 x$$$:

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

$$$u=2 x$$$ とする。

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

したがって、

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

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 $$$f{\left(u \right)} = \cos^{2}{\left(u \right)}$$$ に対して適用する:

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

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

$$\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} - \frac{{\color{red}{\int{\cos^{2}{\left(u \right)} d u}}}}{2} = \frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{2}$$

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 $$$f{\left(u \right)} = \cos{\left(2 u \right)} + 1$$$ に対して適用する:

$$\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 u \right)}}{2} + \frac{1}{2}\right)d u}}}}{2} = \frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} - \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}{2}\right)}}}{2}$$

積分 $$$\int{\left(\cos{\left(2 u \right)} + 1\right)d u}$$$ はすでに計算されています:

$$\int{\left(\cos{\left(2 u \right)} + 1\right)d u} = u + \frac{\sin{\left(2 u \right)}}{2}$$

したがって、

$$\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} - \frac{{\color{red}{\int{\left(\cos{\left(2 u \right)} + 1\right)d u}}}}{4} = \frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} - \frac{{\color{red}{\left(u + \frac{\sin{\left(2 u \right)}}{2}\right)}}}{4}$$

次のことを思い出してください $$$u=2 x$$$:

$$\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(2 {\color{red}{u}} \right)}}{8} - \frac{{\color{red}{u}}}{4} = \frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(2 {\color{red}{\left(2 x\right)}} \right)}}{8} - \frac{{\color{red}{\left(2 x\right)}}}{4}$$

したがって、

$$\int{\left(\cos^{2}{\left(x \right)} - \cos^{2}{\left(2 x \right)}\right)d x} = \frac{\sin{\left(2 x \right)}}{4} - \frac{\sin{\left(4 x \right)}}{8}$$

簡単化せよ:

$$\int{\left(\cos^{2}{\left(x \right)} - \cos^{2}{\left(2 x \right)}\right)d x} = \sin^{3}{\left(x \right)} \cos{\left(x \right)}$$

積分定数を加える:

$$\int{\left(\cos^{2}{\left(x \right)} - \cos^{2}{\left(2 x \right)}\right)d x} = \sin^{3}{\left(x \right)} \cos{\left(x \right)}+C$$

$$$\int \left(\cos^{2}{\left(x \right)} - \cos^{2}{\left(2 x \right)}\right)\, dx = \sin^{3}{\left(x \right)} \cos{\left(x \right)} + C$$$A