Integrating $\Int \Log(\Tan(2X)) Dx$
$\Begingroup$ 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...
$\begingroup$
$\endgroup$
2
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 $\frac12\int \tan(u)du$?
2 Answers
$\begingroup$
$\endgroup$
That's perfectly fine as long as you can find an antiderivative.
$\begingroup$
$\endgroup$
0
It is correct, your answer should be $\frac 12 \ln|\sec 2x| +C$.
Robert Thorne covers electric vehicle innovations, autonomous driving systems, global mobility trends, and automotive engineering developments.
