∫(a+bx)sin(x) dx

First of all, we can use the partial integral to solve this integral. The formula of integral by parts is:

∫u dv = uv - ∫v du

Here we can make u = (a+bx) and dv = sin(x)dx. Then, find out Du Hefu:

du = bdx

v = -cos(x)

Now, apply the partial integral formula to the original integral:

∫(a+bx)sin(x)dx =(a+bx)(-cos(x))-∫(-cos(x))bdx

= -(a + bx)cos(x) + b∫cos(x)dx

Next, we calculate ∫cos(x)dx, which is a common trigonometric function integral:

∫cos(x)dx = sin(x)

Now, substitute this result into the previous formula:

-(a + bx)cos(x) + bsin(x)

Now, we can deal with the first term-(a+bx) cos (x) by partial integration again, let u = -(a+bx) and dv = cos(x)dx, and then find out du and v:

du = -bdx

v = sin(x)

Application of partial integral;

-(a+bx)cos(x)+b∫cos(x)dx =-(a+bx)cos(x)+bsin(x)

Now, we have the final integral result:

-(a + bx)cos(x) + bsin(x) + C

This is the answer we need, where c is the integral constant. So ∫ (a+bx) sin (x) dx =-(a+bx) cos (x)+bsin (x)+c, which is consistent with your deduction.