$$$\left(- \tan{\left(x \right)} + \sec{\left(x \right)}\right) \sec{\left(x \right)}$$$の積分

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

Expand the expression:

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

項別に積分せよ:

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

$$$\sec^{2}{\left(x \right)}$$$ の不定積分は $$$\int{\sec^{2}{\left(x \right)} d x} = \tan{\left(x \right)}$$$ です:

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

$$$\tan{\left(x \right)} \sec{\left(x \right)}$$$ の不定積分は $$$\int{\tan{\left(x \right)} \sec{\left(x \right)} d x} = \sec{\left(x \right)}$$$ です:

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

したがって、

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

積分定数を加える:

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

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