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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 $\frac12\int \tan(u)du$?

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2 Answers

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That's perfectly fine as long as you can find an antiderivative.

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

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