Given the integral
[tex]\displaystyle \int \tan(3x) \, dx[/tex]
first substitute y = 3x and dy = 3 dx :
[tex]\displaystyle \int \tan(3x) \, dx = \frac13 \int \tan(y) \, dy[/tex]
then rewrite tan(y) = sin(y)/cos(y) and substitute z = cos(y) and dz = -sin(y) dy :
[tex]\displaystyle \frac13 \int \frac{\sin(y)}{\cos(y)} \, dy = -\frac13 \int \frac{dz}z[/tex]
Recall that 1/z is the derivative of ln|z|. Then back-substitute to get the result in terms of x :
[tex]\displaystyle -\frac13 \int \frac{dz}z = -\frac13 \ln|z| + C[/tex]
[tex]\displaystyle \frac13 \int \tan(y) \, dy = -\frac13 \ln|\cos(y)| + C[/tex]
[tex]\displaystyle \boxed{\int \tan(3x) \, dx = -\frac13 \ln|\cos(3x)| + C}[/tex]
