Evaluate the integral \[ \int tan 3x dx

Re: Evaluate the integral

\(\displaystyle \int \tan{u} \, du = -\ln|\cos{u}| + C\)
 
Re: Evaluate the integral

\(\displaystyle \int \tan{3x} \, dx

This is how the problem is, but I don't know where you are getting the negative sign from?
I got this answer:
ln Icos 3xI +c\)
 
Re: Evaluate the integral

\(\displaystyle \int \tan{x} \, dx = -\int \frac{-\sin{x}}{\cos{x}} \, dx = -\ln|\cos{x}| + C\)
 
Re: Evaluate the integral

\(\displaystyle \int{tan(3x)}dx\)

\(\displaystyle =\int\frac{sin(3x)}{cos(3x)}dx\)

Let \(\displaystyle u=cos(3x), \;\ du = -3sin(x)dx, \;\ \frac{-du}{3}=sin(x)dx\)

\(\displaystyle = \frac{-ln(cos(3x))}{3}+C\)

Since \(\displaystyle ln(cos(x)) = -ln(\frac{1}{cos(x)})\)

we can write it as \(\displaystyle ln(sec(3x))+C\)

Technically, absolute values should be used, but.........
 
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