$$$\frac{1}{1 - \sin{\left(x \right)}}$$$ 的積分

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

將 $$$1$$$ 重寫為 $$$\sin^2\left(\frac{x}{2}\right)+\cos^2\left(\frac{x}{2}\right)$$$，並應用正弦的二倍角公式 $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

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

配方法（步驟見»）:

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

將分子與分母同時乘以 $$$\sec^2\left(\frac{x}{2}\right)$$$:

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

令 $$$u=\tan{\left(\frac{x}{2} \right)} - 1$$$。

則 $$$du=\left(\tan{\left(\frac{x}{2} \right)} - 1\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} \right)}}{2} dx$$$ (步驟見»)，並可得 $$$\sec^{2}{\left(\frac{x}{2} \right)} dx = 2 du$$$。

所以，

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

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

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

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

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

回顧一下 $$$u=\tan{\left(\frac{x}{2} \right)} - 1$$$：

$$- 2 {\color{red}{u}}^{-1} = - 2 {\color{red}{\left(\tan{\left(\frac{x}{2} \right)} - 1\right)}}^{-1}$$

因此，

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

加上積分常數：

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

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