Answer
$\approx 0.099944461$ with an error of about $2.8 \times 10^{-12}$
Work Step by Step
We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $0.1$.
$\int_0^{0.1}\dfrac{\sin x}{x} dx=\int_0^{0.1} [1-\dfrac{x^2}{3!}+\dfrac{x^{4}}{5!}+...] dx $
or, $=0.1+\dfrac{(0.1)^3}{3 \cdot 3!}+\dfrac{(0.1)^5}{7 \cdot 5!}-...$
But $\dfrac{(0.1)^7}{7 \cdot 7!}\approx 2.8 \times 10^{-12}$ and the proceeding term is greater than $10^{-8}$.
so, the first three terms of the series would give the accuracy.
$\int_0^{0.1}\dfrac{\sin x}{x} dx \approx 0.099944461$ with an error of about $2.8 \times 10^{-12}$
