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

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

逐項積分：

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

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

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

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

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

逐項積分：

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

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

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

令 $$$u=2 x$$$。

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

該積分可改寫為

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

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 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$：

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

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

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

回顧一下 $$$u=2 x$$$：

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

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

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

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

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

逐項積分：

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

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

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

積分 $$$\int{\cos{\left(2 x \right)} d x}$$$ 已經計算過：

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

因此，

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

因此，

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

加上積分常數：

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

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