I tried to evaluate this using the substitution method: making u=sint and du=costdt. Then that created a big jumbled incorrect mess. I don't know how to attack this problem.
One of the problems with this particular integral, is that you cannot solve it. So don't bother with solving this integral, it won't work. But there is a way to make this integral disappear: differentiate it. Now, I wonder, is there a way that you can solve a limit using derivatives...
L'hospital rule: so take the derivative of the top and the bottom of (x/x-3), thus giving you 1/1 times the integral. How will this make the integral disappear?
Ok, so after applying L'Hopitals rule, I ended up with the limit as x approaches 3 of [sin(x)/x + sinx]. That can't be right. I'm sorry that I am mathematically incompetent.
According to the fundamental theorem of calculus, F'(x) equals f(x) or sin(x)/x. Now, upon calculating the derivative of xF(x), I use the product rule which will be x'F(x) + xF'(x). That gives me the integral from 3 to x of sin(t)/t dt + sinx.