int_0^1 sinx/x dx Use a power series to approximate the value of the integral with an error of less than 0.0001.

From the Power Series table for trigonometric function, we have:
sin(x) =sum_(n=0)^oo (-1)^n x^(2n+1)/((2n+1)!)
             = x -x^3/(3!) +x^5/(5!) - x^7/(7!) +...
Applying it on the integral int_0^1 sin(x)/x dx where the integrand is f(x)=sin(x)/x, we get:
int_0^1 sin(x)/x dx = int_0^1 sin(x)*1/x dx
                 =int_0^1 sum_(n=0)^oo (-1)^n x^(2n+1)/((2n+1)!) *1/xdx
                   =int_0^1sum_(n=0)^oo (-1)^n x^(2n+1)/((2n+1)!) *x^(-1)dx
                    =int_0^1sum_(n=0)^oo (-1)^n x^(2n+1-1)/((2n+1)!) dx
                    =int_0^1sum_(n=0)^oo (-1)^n x^(2n)/((2n+1)!) dx
                   =int_0^1 [1 -x^2/(3!) +x^4/(5!) - x^6/(7!) +...] dx
 Or 
int_0^1 sin(x)/x dx =int_0^1 sin(x)/x dx
                   =int_0^1 1/x* [x -x^3/(3!) +x^5/(5!) - x^7/(7!) +...] dx
                    = int_0^1 [x/x -x^3/(3!x) +x^5/(5!x) - x^7/(7!x) +...] dx
                    =int_0^1 [1 -x^2/(3!) +x^4/(5!) - x^6/(7!) +...] dx
To determine the indefinite integral, we integrate each term using Power rule for integration: int x^n dx = x^(n+1)/(n+1) .
int_0^1 [1 -x^2/3! +x^4/5! - x^6/7! +...] dx= [x -x^3/(3!*3) +x^5/(5!*5) - x^7/(7!*7) +...]_0^1
           = [x -x^3/(1*2*3*3) +x^5/(1*2*3*4*5*5) - x^7/(1*2*3*4*5*6*7*7) +...]_0^1
           = [x -x^3/(6*3) +x^5/(120*5) - x^7/(5040*7) +...]_0^1
           = [x -x^3/18 +x^5/600- x^7/35280+...]_0^1
Apply definite integral formula: F(x)|a^b =F(b)-F(a) .
F(1)= 1-1^3/18 +1^5/600- 1^7/35280+...
         =1-1/18 +1/600- 1/35280+...
F(0)= 0-0^3/18 +0^5/600- 0^7/35280+...
         =0-0+0-0+...
All the terms are 0 then F(0)=0 .
We can stop on the 4th term  (1/35280 ~~2.8345x10^(-5)) since we only need an error less than 0.0001.
F(1)-F(0)= [1-1/18 +1/600- 1/35280]- [0]
                     =1-1/18 +1/600- 1/35280
                      = 0.9460827664
 Then, the approximated integral value will be:
int_0^1 sin(x)/x dx~~0.9461