$$$\frac{1}{\cos^{2}{\left(x \right)}}$$$の積分

$$$\int \frac{1}{\cos^{2}{\left(x \right)}}\, dx$$$ を求めよ。

被積分関数を正割関数で表しなさい:

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

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

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

したがって、

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

積分定数を加える:

$$\int{\frac{1}{\cos^{2}{\left(x \right)}} d x} = \tan{\left(x \right)}+C$$

$$$\int \frac{1}{\cos^{2}{\left(x \right)}}\, dx = \tan{\left(x \right)} + C$$$A