$$$- \tan{\left(x \right)} + \sec{\left(x \right)}$$$ 的积分

该计算器将求出$$$- \tan{\left(x \right)} + \sec{\left(x \right)}$$$的积分/原函数,并显示步骤。

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您的输入

$$$\int \left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)\, dx$$$

解答

逐项积分:

$${\color{red}{\int{\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right)d x}}} = {\color{red}{\left(- \int{\tan{\left(x \right)} d x} + \int{\sec{\left(x \right)} d x}\right)}}$$

将正切表示为 $$$\tan\left(x\right)=\frac{\sin\left(x\right)}{\cos\left(x\right)}$$$:

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

$$$u=\cos{\left(x \right)}$$$

$$$du=\left(\cos{\left(x \right)}\right)^{\prime }dx = - \sin{\left(x \right)} dx$$$ (步骤见»),并有$$$\sin{\left(x \right)} dx = - du$$$

该积分可以改写为

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

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

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

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

$$\int{\sec{\left(x \right)} d x} + {\color{red}{\int{\frac{1}{u} d u}}} = \int{\sec{\left(x \right)} d x} + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$

回忆一下 $$$u=\cos{\left(x \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{\sec{\left(x \right)} d x} = \ln{\left(\left|{{\color{red}{\cos{\left(x \right)}}}}\right| \right)} + \int{\sec{\left(x \right)} d x}$$

将正割改写为 $$$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$$$:

$$\ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\sec{\left(x \right)} d x}}} = \ln{\left(\left|{\cos{\left(x \right)}}\right| \right)} + {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}}$$

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。:

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

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

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

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

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

积分变为

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

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

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

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

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

因此,

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

加上积分常数:

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

答案

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