# Integrating $\int \log(\tan(2x)) dx$

Looking online I see that people are integrating $\int \tan(2x)dx$ by replacing $\tan(2x)$ with $\frac{\sin(2x)}{\cos(2x)}$ and then using $u$-substitution for $2x$. Why can you not simply use $u=2x$ and then integrate $\frac12\int \tan(u)du$?

• If you see that people do X, that does not mean that they can't do Y. – user147263 Jun 20 '15 at 6:47
• the methods are basically identical – robertmartin8 Jun 21 '15 at 0:48

It is correct, your answer should be $\frac 12 \ln|\sec 2x| +C$.