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

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

设$$$u=2 x$$$。

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

因此，

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

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

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

使用公式 $$$\cos{\left( u \right)}=\frac{1 - \tan^{2}{\left(\frac{ u }{2} \right)}}{\tan^{2}{\left(\frac{ u }{2} \right)} + 1}$$$ 改写被积函数:

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

设$$$v=\tan{\left(\frac{u}{2} \right)}$$$。

则 $$$u=2 \operatorname{atan}{\left(v \right)}$$$ 且 $$$du=\left(2 \operatorname{atan}{\left(v \right)}\right)^{\prime }dv = \frac{2}{v^{2} + 1} dv$$$（步骤见»）。

因此，

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

化简:

$$- \frac{{\color{red}{\int{\frac{2}{\left(v^{2} + 1\right) \left(\frac{1 - v^{2}}{v^{2} + 1} - 2\right)} d v}}}}{2} = - \frac{{\color{red}{\int{\left(- \frac{2}{3 v^{2} + 1}\right)d v}}}}{2}$$

对 $$$c=-2$$$ 和 $$$f{\left(v \right)} = \frac{1}{3 v^{2} + 1}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$$- \frac{{\color{red}{\int{\left(- \frac{2}{3 v^{2} + 1}\right)d v}}}}{2} = - \frac{{\color{red}{\left(- 2 \int{\frac{1}{3 v^{2} + 1} d v}\right)}}}{2}$$

设$$$w=\sqrt{3} v$$$。

则$$$dw=\left(\sqrt{3} v\right)^{\prime }dv = \sqrt{3} dv$$$ (步骤见»)，并有$$$dv = \frac{\sqrt{3} dw}{3}$$$。

所以，

$${\color{red}{\int{\frac{1}{3 v^{2} + 1} d v}}} = {\color{red}{\int{\frac{\sqrt{3}}{3 \left(w^{2} + 1\right)} d w}}}$$

对 $$$c=\frac{\sqrt{3}}{3}$$$ 和 $$$f{\left(w \right)} = \frac{1}{w^{2} + 1}$$$ 应用常数倍法则 $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$：

$${\color{red}{\int{\frac{\sqrt{3}}{3 \left(w^{2} + 1\right)} d w}}} = {\color{red}{\left(\frac{\sqrt{3} \int{\frac{1}{w^{2} + 1} d w}}{3}\right)}}$$

$$$\frac{1}{w^{2} + 1}$$$ 的积分为 $$$\int{\frac{1}{w^{2} + 1} d w} = \operatorname{atan}{\left(w \right)}$$$:

$$\frac{\sqrt{3} {\color{red}{\int{\frac{1}{w^{2} + 1} d w}}}}{3} = \frac{\sqrt{3} {\color{red}{\operatorname{atan}{\left(w \right)}}}}{3}$$

回忆一下 $$$w=\sqrt{3} v$$$:

$$\frac{\sqrt{3} \operatorname{atan}{\left({\color{red}{w}} \right)}}{3} = \frac{\sqrt{3} \operatorname{atan}{\left({\color{red}{\sqrt{3} v}} \right)}}{3}$$

回忆一下 $$$v=\tan{\left(\frac{u}{2} \right)}$$$:

$$\frac{\sqrt{3} \operatorname{atan}{\left(\sqrt{3} {\color{red}{v}} \right)}}{3} = \frac{\sqrt{3} \operatorname{atan}{\left(\sqrt{3} {\color{red}{\tan{\left(\frac{u}{2} \right)}}} \right)}}{3}$$

回忆一下 $$$u=2 x$$$:

$$\frac{\sqrt{3} \operatorname{atan}{\left(\sqrt{3} \tan{\left(\frac{{\color{red}{u}}}{2} \right)} \right)}}{3} = \frac{\sqrt{3} \operatorname{atan}{\left(\sqrt{3} \tan{\left(\frac{{\color{red}{\left(2 x\right)}}}{2} \right)} \right)}}{3}$$

因此，

$$\int{\frac{1}{2 - \cos{\left(2 x \right)}} d x} = \frac{\sqrt{3} \operatorname{atan}{\left(\sqrt{3} \tan{\left(x \right)} \right)}}{3}$$

加上积分常数：

$$\int{\frac{1}{2 - \cos{\left(2 x \right)}} d x} = \frac{\sqrt{3} \operatorname{atan}{\left(\sqrt{3} \tan{\left(x \right)} \right)}}{3}+C$$

$$$\int \frac{1}{2 - \cos{\left(2 x \right)}}\, dx = \frac{\sqrt{3} \operatorname{atan}{\left(\sqrt{3} \tan{\left(x \right)} \right)}}{3} + C$$$A