Integral of $$$n \tan{\left(x \right)} \sec{\left(x \right)}$$$ with respect to $$$x$$$

Find $$$\int n \tan{\left(x \right)} \sec{\left(x \right)}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=n$$$ and $$$f{\left(x \right)} = \tan{\left(x \right)} \sec{\left(x \right)}$$$:

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

The integral of $$$\tan{\left(x \right)} \sec{\left(x \right)}$$$ is $$$\int{\tan{\left(x \right)} \sec{\left(x \right)} d x} = \sec{\left(x \right)}$$$:

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

Therefore,

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

Add the constant of integration:

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

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