int sec^2(3x) dx Find the indefinite integral

Indefinite integral are written in the form of int f(x) dx = F(x) +C
 where: f(x) as the integrand
           F(x) as the anti-derivative function 
           C  as the arbitrary constant known as constant of integration
For the given problem: int sec^2(3x)dx , the integrand function: f(x)=sec^2(3x) is in a form of trigonometric function.
To solve for the indefinite integral, we may apply the basic integration formula for secant function:
int sec^2(u)du=tan(u)+C
We may apply u-substitution when by letting:
u= 3x then du =3 dx or (du)/3 = dx .
Plug-in the values of u =3x and dx= (du)/3 , we get:
int sec^2(3x)dx =int sec^2(u)*(du)/3
                          =int (sec^2(u))/3du
Apply basic integration property: int c*f(x)dx= c int f(x)dx .
int (sec^2(u))/3du =(1/3)int sec^2(u)du
Then following the integral formula for secant, we get:
(1/3)int sec^2(u)du= 1/3tan(u)+C
Plug-in u =3x to solve for the indefinite integral F(x):
int sec^2(3x)dx=1/3tan(3x)+C