Indefinite integrals 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 sin(2x)cos(4x) dx` or `intcos(4x)sin(2x) dx`   has a integrand in a form of trigonometric function. To evaluate this, we apply the identity:

`cos(A)sin(B) =[sin(A+B) -sin(A-B)]/2`

The integral becomes:

`int cos(4x)sin(2x) dx = int[sin(4x+2x) -sin(4x-2x)]/2dx`

 Apply the basic properties of integration: `int c*f(x) dx= c int f(x) dx` .

`int [sin(4x+2x) -sin(4x-2x)]/2dx = 1/2int[sin(4x+2x) -sin(4x-2x)]dx`

 Apply the basic integration property: `int (u+v) dx = int (u) dx + int (v) dx` .

`1/2 *[int sin(4x+2x)dx+int sin(4x-2x)dx]`

Then apply u-substitution to be able to apply integration formula for cosine function: `int sin(u) du= -cos(u) +C` .

For the integral:`int sin(4x+2x)dx` , we let `u = 4x+2x =6x` then `du= 6 dx` or `(du)/6 =dx` .

`int sin(4x+2x)dx=intsin(6x) dx`

                                  `=intsin(u) *(du)/6`

                                  `= 1/6 int sin(u)du`

                                  `=-1/6cos(u) +C`

Plug-in `u =6x ` on `-1/6 cos(u) +C` , we get:

`int sin(4x+2x)dx= -1/6 cos(6x) +C`

For the integral: `intsin(4x-2x)dx` , we let` u = 4x-2x =2x` then `du= 2 dx` or `(du)/2 =dx` .

`intsin(4x-2x)dx=intsin(2x) dx`

                                 `=intsin(u) *(du)/2`

                                 `= 1/2 int sin(u)du`

                                 `= -1/2cos(u) +C`

Plug-in `u =2x` on `-1/2 cos(u) +C` , we get:

`intsin(4x-2x)dx= -1/2 cos(2x) +C`

Combing the results, we get the indefinite integral as:

`intcos(4x)sin(2x) dx= 1/2*[ -1/6 cos(6x) -(-1/2 cos(2x))] +C`

or   `-1/12 cos(6x) +1/4 cos(2x) +C`