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

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx = c x$$$：

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{3}$$$ 與 $$$f{\left(x \right)} = \cos^{2}{\left(x \right)}$$$：

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

套用降冪公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$，令 $$$\alpha=x$$$:

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$：

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx = c x$$$：

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

令 $$$u=2 x$$$。

則 $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步驟見»)，並可得 $$$dx = \frac{du}{2}$$$。

因此，

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

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$：

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

餘弦函數的積分為 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$：

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

回顧一下 $$$u=2 x$$$：

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

因此，

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

加上積分常數：

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

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