`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`