What's the the integral of $\tan(4x)$?

Question

How is $\displaystyle \int \tan(4x) dx = \cos^2 (4x)$? Shouldn't it be $\ln(\sec(4x))$? I don't understand... Please, help.

Thank you.

Answer 1

$$\int\tan x\,dx=-\log\cos x +C=\log\sec x+C\implies$$

$$\int\tan 4x\,dx=\frac14\int (4dx)\tan 4x=\frac14\log\sec 4x+C$$

Answer 2

This is a correct method for the function as stated above,

\begin{eqnarray} \int \tan(4x)dx &=& \frac{1}{4}\ln (\sec(4x)) + C. \end{eqnarray}

Answer 3

$$ \int{\tan(4x)dx}=\int\frac{\sin(4x)dx}{\cos(4x)}=|u=\cos(4x)\Rightarrow -4\sin(4x)dx=dt\Rightarrow \sin(4x)dx=-\frac{1}{4}dt| $$ $$ =-\frac{1}{4}\int\frac{dt}{t}=-\frac{1}{4}\ln|t|=-\frac{1}{4}\ln|\cos(4x)|=\frac{1}{4}(-1)\ln|\cos(4x)|=\frac{1}{4}\ln|\cos(4x)|^{-1} $$ $$ =\frac{1}{4}\ln|\frac{1}{\cos(4x)}|=\frac{1}{4}\ln|\sec(4x)|+C $$