$$$- \frac{x}{2} - \sin{\left(e x \right)} + \cos{\left(x_{1} \right)}$$$ 對 $$$x$$$ 的積分

求$$$\int \left(- \frac{x}{2} - \sin{\left(e x \right)} + \cos{\left(x_{1} \right)}\right)\, dx$$$。

逐項積分：

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

令 $$$u=e x$$$。

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

因此，

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

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

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

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

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

回顧一下 $$$u=e x$$$：

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

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 $$$f{\left(x \right)} = x$$$：

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

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

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

配合 $$$c=\cos{\left(x_{1} \right)}$$$，應用常數法則 $$$\int c\, dx = c x$$$：

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

因此，

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

加上積分常數：

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

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