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

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

逐項積分：

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

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

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

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 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$：

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

逐項積分：

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

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

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

令 $$$u=2 x$$$。

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

因此，

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

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

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

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

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

回顧一下 $$$u=2 x$$$：

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

令 $$$u=\cos{\left(x \right)}$$$。

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

所以，

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

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

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

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

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

回顧一下 $$$u=\cos{\left(x \right)}$$$：

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

因此，

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

加上積分常數：

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

$$$\int \left(- \frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + \cos^{2}{\left(x \right)}\right)\, dx = \left(\frac{x}{2} + \frac{\sin{\left(2 x \right)}}{4} - \frac{1}{\cos{\left(x \right)}}\right) + C$$$A