# `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`

Approved by eNotes Editorial Team