I'm going to assume you mean that it is centered at zero, which would make it a MacLaurin series. In which case the series for sin(x) in Series summation form is represented by:
Sigma from n=0 to infinity of ((-1)^n){[x^(2n+1)]/(2n+1)!}
(that explanation mark is a factorial)
Taking the integral of that summation gives you:
Sigma from n=0 to infinity of ((-1)^n){[x^(2n+2)]/[(2n+2)(2n+1)!]}
Then you simply plug in 10pi and subtract that from the summation with 0 plugged in:
Sigma from n=0 to infinity of ((-1)^n){[(10pi)^(2n+2)]/[(2n+2)(2n+1)!]} MINUS Sigma from n=0 to infinity of ((-1)^n){[0^(2n+2)]/[(2n+2)(2n+1)!]}
The zero cancels out all of the terms in the second summation after the integration so it is irrelavent.
You didn't give me an error bound so to what nth degree would you like me to go out? Although I'm not going to waste my time, I have two finals tomorrow.
Regardless, the problem in question that Warren is asking about is an indefinite integral. It has no endpoints to plug in. Which means it's incredibly simple to simply add +1 and divide by the exponent.
