Answer
$\int_0^{1} \dfrac{1- \cos x}{x^2} dx \approx 0.4863853764$ (Error of about $1.9 \times 10^{-10}$.)
Work Step by Step
Integrate the integral with respect to $ x $ as follows:
$\int_0^{1} \dfrac{1- \cos x}{x^2} dx=\int_0^{1} [0.5-\dfrac{x^2}{4 !}?+\dfrac{x^{4}}{6 !}-...] \space dx \\=0.5-\dfrac{1}{3 \cdot 4!}+\dfrac{1}{5 \cdot 6!}-....$
But $\dfrac{1}{11 \cdot 12!} \approx 1.9 \times 10^{-10} $ and this means that the proceeding term is greater than $10^{-8}$ and the first five terms of the series yield the accuracy.
Thus, $\int_0^{1} \dfrac{1- \cos x}{x^2} dx \approx 0.4863853764$ (Error of about $1.9 \times 10^{-10}$.)
