Answer
$\approx 0.4863853764$ with an error of about $1.9 \times 10^{-10}$
Work Step by Step
We integrate the integral with respect to $ x $ taking a lower limit as $0$ and upper limit as $1$.
$\int_0^{1} \dfrac{1- \cos x}{x^2} dx=\int_0^{1} [0.5-\dfrac{x^2}{4 !}?+\dfrac{x^{4}}{6 !}-...] dx $
or, $=0.5-\dfrac{1}{3 \cdot 4!}+\dfrac{1}{5 \cdot 6!}-....$
But $\dfrac{1}{11 \cdot 12!} \approx 1.9 \times 10^{-10} $ and the proceeding term is greater than $10^{-8}$.
So, the first five terms of the series would give the accuracy.
$\int_0^{1} \dfrac{1- \cos x}{x^2} dx \approx 0.4863853764$ with an error of about $1.9 \times 10^{-10}$.
