Answer
$\int_0^{0.1}\dfrac{\sin x}{x} dx \approx 0.099944461$ (Error of about $2.8 \times 10^{-12})$
Work Step by Step
Integrate the integral with respect to $ x $ as follows:
$\int_0^{0.1}\dfrac{\sin x}{x} dx=\int_0^{0.1} [1-\dfrac{x^2}{3!}+\dfrac{x^{4}}{5!}+...] \space dx \\=0.1+\dfrac{(0.1)^3}{3 \cdot 3!}+\dfrac{(0.1)^5}{7 \cdot 5!}-...$
But $\dfrac{(0.1)^7}{7 \cdot 7!}\approx 2.8 \times 10^{-12}$ and this means that the proceeding term is greater than $10^{-5}$ and the first five terms of the series yield the accuracy.
Thus, $\int_0^{0.1}\dfrac{\sin x}{x} dx \approx 0.099944461$ (Error of about $2.8 \times 10^{-12})$
