$$$\frac{\cos^{2}{\left(x \right)}}{3} - 1$$$の積分

$$$\int \left(\frac{\cos^{2}{\left(x \right)}}{3} - 1\right)\, dx$$$ を求めよ。

項別に積分せよ:

$${\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{3} - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{\frac{\cos^{2}{\left(x \right)}}{3} d x}\right)}}$$

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

$$\int{\frac{\cos^{2}{\left(x \right)}}{3} d x} - {\color{red}{\int{1 d x}}} = \int{\frac{\cos^{2}{\left(x \right)}}{3} d x} - {\color{red}{x}}$$

定数倍の法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ を、$$$c=\frac{1}{3}$$$ と $$$f{\left(x \right)} = \cos^{2}{\left(x \right)}$$$ に対して適用する:

$$- x + {\color{red}{\int{\frac{\cos^{2}{\left(x \right)}}{3} d x}}} = - x + {\color{red}{\left(\frac{\int{\cos^{2}{\left(x \right)} d x}}{3}\right)}}$$

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:

$$- x + \frac{{\color{red}{\int{\cos^{2}{\left(x \right)} d x}}}}{3} = - x + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{3}$$

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:

$$- x + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{3} = - x + \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}}{3}$$

項別に積分せよ:

$$- x + \frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}}}{6} = - x + \frac{{\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}}{6}$$

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

$$- x + \frac{\int{\cos{\left(2 x \right)} d x}}{6} + \frac{{\color{red}{\int{1 d x}}}}{6} = - x + \frac{\int{\cos{\left(2 x \right)} d x}}{6} + \frac{{\color{red}{x}}}{6}$$

$$$u=2 x$$$ 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、

$$- \frac{5 x}{6} + \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{6} = - \frac{5 x}{6} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{6}$$

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:

$$- \frac{5 x}{6} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{6} = - \frac{5 x}{6} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{6}$$

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

$$- \frac{5 x}{6} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{12} = - \frac{5 x}{6} + \frac{{\color{red}{\sin{\left(u \right)}}}}{12}$$

次のことを思い出してください $$$u=2 x$$$:

$$- \frac{5 x}{6} + \frac{\sin{\left({\color{red}{u}} \right)}}{12} = - \frac{5 x}{6} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{12}$$

したがって、

$$\int{\left(\frac{\cos^{2}{\left(x \right)}}{3} - 1\right)d x} = - \frac{5 x}{6} + \frac{\sin{\left(2 x \right)}}{12}$$

積分定数を加える:

$$\int{\left(\frac{\cos^{2}{\left(x \right)}}{3} - 1\right)d x} = - \frac{5 x}{6} + \frac{\sin{\left(2 x \right)}}{12}+C$$

$$$\int \left(\frac{\cos^{2}{\left(x \right)}}{3} - 1\right)\, dx = \left(- \frac{5 x}{6} + \frac{\sin{\left(2 x \right)}}{12}\right) + C$$$A