Integral of $$$\frac{\tan{\left(x \right)}}{\cos{\left(x \right)}}$$$

Find $$$\int \frac{\tan{\left(x \right)}}{\cos{\left(x \right)}}\, dx$$$.

Rewrite the integrand:

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

Let $$$u=\sec{\left(x \right)}$$$.

Then $$$du=\left(\sec{\left(x \right)}\right)^{\prime }dx = \tan{\left(x \right)} \sec{\left(x \right)} dx$$$ (steps can be seen »), and we have that $$$\tan{\left(x \right)} \sec{\left(x \right)} dx = du$$$.

So,

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

$${\color{red}{\int{1 d u}}} = {\color{red}{u}}$$

Recall that $$$u=\sec{\left(x \right)}$$$:

$${\color{red}{u}} = {\color{red}{\sec{\left(x \right)}}}$$

Therefore,

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

Add the constant of integration:

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

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