You should use the property of linearity of integral, such that:

`int (3 - 5x)cos 4x dx = int 3*cos 4x dx - int 5x*cos 4x dx`

You need to evaluate the integral `int x*cos 4x dx` using parts, such that:

`int f(x)*g'(x) dx = f(x)g(x) - int f'(x)g(x)dx`

Selecting` f(x) = x` and `g'(x) = cos 4x` yields:

`f(x) = x => f'(x) = 1 `

`g'(x) = cos 4x => g(x) = (sin 4x)/4`

`int x*cos 4x dx = x*(sin 4x)/4 - int (sin 4x)/4 dx`

`int x*cos 4x dx = x*(sin 4x)/4 - (1/4)*(-cos 4x)/4 `

`int (3 - 5x)cos 4x dx = (3/4)*sin 4x - (5/4)x*(sin 4x) - (5/16)*(cos 4x) + c`

**Hence, evaluating the indefinite integral using the property of linearity and parts yields **`int (3 - 5x)cos 4x dx = (3/4)*sin 4x - (5/4)x*(sin 4x) - (5/16)*(cos 4x) + c.`

One of the techniques for evaluating integrals is integration by parts.

We'll express the formula of integration by parts using differentials:

Int u dv = u*v - Int v du

We'll put u = 3 - 5x

We'll differentiate both sides:

du = -5dx

We'll put dv = cos(4x) dx.

We'll integrate both sides:

Int dv = Int cos(4x) dx

v = (sin 4x)/4

We'll substitute in the formula of integral:

Int (3 - 5x) cos(4x) dx = (3 - 5x)*(sin 4x)/4 + 5 Int (sin 4x)dx/4

Int (3 - 5x) cos(4x) dx = (3 - 5x)*(sin 4x)/4 + (5/4)Int (sin 4x)dx

Int (3 - 5x) cos(4x) dx = (3 - 5x)*(sin 4x)/4 + (5/4)[(1/4)(- cos 4x)] + C

**Int (3 - 5x) cos(4x) dx = (3 - 5x)*(sin 4x)/4 - (5/16)[(cos 4x)] + C**