$$$- \sec^{2}{\left(x \right)}$$$ 的积分

求$$$\int \left(- \sec^{2}{\left(x \right)}\right)\, dx$$$。

对 $$$c=-1$$$ 和 $$$f{\left(x \right)} = \sec^{2}{\left(x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

$$$\sec^{2}{\left(x \right)}$$$ 的积分为 $$$\int{\sec^{2}{\left(x \right)} d x} = \tan{\left(x \right)}$$$:

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

因此，

$$\int{\left(- \sec^{2}{\left(x \right)}\right)d x} = - \tan{\left(x \right)}$$

加上积分常数：

$$\int{\left(- \sec^{2}{\left(x \right)}\right)d x} = - \tan{\left(x \right)}+C$$

$$$\int \left(- \sec^{2}{\left(x \right)}\right)\, dx = - \tan{\left(x \right)} + C$$$A