Let be f(x) = 2x sin(4x).
We want to find int 2x sin(4x) dx. We know the integral of the function sin or cos so it would be good for us to get rid of 2x by differentiating it.
Let's say g(x) = 2x and h'(x) = sin(4x).
So g'(x) = 2 and h(x) = (-1)/4 cos(4x).
int g(x) h'(x) dx = g(x)h(x) - int g'(x) h(x) dx using integration by parts.
int 2x sin(4x) dx = 2x*(-1)/4 cos(4x) - int 2 *(-1)/4 cos(4x) dx
=(-1)/2x cos(4x) + 1/2 int cos(4x) dx
=(-1)/2x cos(4x) + 1/2 * 1/4 sin(4x)
=(-1)/2x cos(4x) + 1/8 sin(4x)
Therefore, int 2x sin(4x) dx = (-1)/2x cos(4x) + 1/8 sin(4x) + C, where C is the constant of integration.
