Integral of $$$\tan{\left(u \right)} \sec{\left(u \right)}$$$
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Your Input
Find $$$\int \tan{\left(u \right)} \sec{\left(u \right)}\, du$$$.
Solution
The integral of $$$\tan{\left(u \right)} \sec{\left(u \right)}$$$ is $$$\int{\tan{\left(u \right)} \sec{\left(u \right)} d u} = \sec{\left(u \right)}$$$:
$${\color{red}{\int{\tan{\left(u \right)} \sec{\left(u \right)} d u}}} = {\color{red}{\sec{\left(u \right)}}}$$
Therefore,
$$\int{\tan{\left(u \right)} \sec{\left(u \right)} d u} = \sec{\left(u \right)}$$
Add the constant of integration:
$$\int{\tan{\left(u \right)} \sec{\left(u \right)} d u} = \sec{\left(u \right)}+C$$
Answer
$$$\int \tan{\left(u \right)} \sec{\left(u \right)}\, du = \sec{\left(u \right)} + C$$$A