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

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

改写被积函数:

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

Expand the expression:

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

逐项积分：

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

使用公式$$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$将余弦用正弦表示，然后使用二倍角公式$$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$将正弦改写。:

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

将分子和分母同时乘以 $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:

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

设$$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$。

则$$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (步骤见»)，并有$$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$。

所以，

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

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

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

回忆一下 $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$:

$$- \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{2 \cos{\left(x \right)} d x} = - \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)} + \int{2 \cos{\left(x \right)} d x}$$

对 $$$c=2$$$ 和 $$$f{\left(x \right)} = \cos{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$- \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + {\color{red}{\int{2 \cos{\left(x \right)} d x}}} = - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + {\color{red}{\left(2 \int{\cos{\left(x \right)} d x}\right)}}$$

余弦函数的积分为 $$$\int{\cos{\left(x \right)} d x} = \sin{\left(x \right)}$$$：

$$- \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + 2 {\color{red}{\int{\cos{\left(x \right)} d x}}} = - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + 2 {\color{red}{\sin{\left(x \right)}}}$$

因此，

$$\int{\frac{\cos{\left(2 x \right)}}{\cos{\left(x \right)}} d x} = - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + 2 \sin{\left(x \right)}$$

加上积分常数：

$$\int{\frac{\cos{\left(2 x \right)}}{\cos{\left(x \right)}} d x} = - \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + 2 \sin{\left(x \right)}+C$$

$$$\int \frac{\cos{\left(2 x \right)}}{\cos{\left(x \right)}}\, dx = \left(- \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right) + 2 \sin{\left(x \right)}\right) + C$$$A