looks like you've just about got it. It's easier here to use 0 as the sum starting index.

$$\sum_{i=0}^{n-1}f\left(-5+\frac{4i}{n}\right) \frac{4}{n}$$

now make use of the fact that

$$f(x)=x \sin(x)$$

and note that

$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$

you'll end up with an approximation formula for the integral in terms of n. Then take the limit as n goes to infinity.