The Maclaurin series for sin x is

$\displaystyle \frac {\sin x}x = \sum\limits_{k = 0}^\infty {\frac{{( - 1)^k x^{2k} }}

{{(2k + 1)!}}} = 1 - \frac {x^2}{3!} + \frac {x^4}{5!} - \frac {x^6}{7!} + ..$

use the corresponding Maclaurin polynomial of degree 5 to approximate

$\displaystyle \int\frac{\sin x}{x} dx $

integration from -1 to 1

i dont have any idea to solve it..