$$$\frac{1}{- x \sin{\left(\frac{1}{2} \right)} + 1}$$$ 的积分

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

设$$$u=- x \sin{\left(\frac{1}{2} \right)} + 1$$$。

则$$$du=\left(- x \sin{\left(\frac{1}{2} \right)} + 1\right)^{\prime }dx = - \sin{\left(\frac{1}{2} \right)} dx$$$ (步骤见»)，并有$$$dx = - \frac{du}{\sin{\left(\frac{1}{2} \right)}}$$$。

因此，

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

对 $$$c=- \frac{1}{\sin{\left(\frac{1}{2} \right)}}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$- \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{\sin{\left(\frac{1}{2} \right)}} = - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{\sin{\left(\frac{1}{2} \right)}}$$

回忆一下 $$$u=- x \sin{\left(\frac{1}{2} \right)} + 1$$$:

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{\sin{\left(\frac{1}{2} \right)}} = - \frac{\ln{\left(\left|{{\color{red}{\left(- x \sin{\left(\frac{1}{2} \right)} + 1\right)}}}\right| \right)}}{\sin{\left(\frac{1}{2} \right)}}$$

因此，

$$\int{\frac{1}{- x \sin{\left(\frac{1}{2} \right)} + 1} d x} = - \frac{\ln{\left(\left|{x \sin{\left(\frac{1}{2} \right)} - 1}\right| \right)}}{\sin{\left(\frac{1}{2} \right)}}$$

加上积分常数：

$$\int{\frac{1}{- x \sin{\left(\frac{1}{2} \right)} + 1} d x} = - \frac{\ln{\left(\left|{x \sin{\left(\frac{1}{2} \right)} - 1}\right| \right)}}{\sin{\left(\frac{1}{2} \right)}}+C$$

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