Answer
$$| Error| \lt 0.00011$$
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} \cos t^2 dt=\int_0^{1} [1-\dfrac{t^2}{2}+\dfrac{t^{8}}{4 !}-...] dx $
or, $=[t-\dfrac{t^5}{10}+\dfrac{t^{9}}{9 \cdot 4 !}-...]_0^{10} ....$
So, the error can be calculated as:
$| Error| \lt \dfrac{1}{13 \cdot 6 !} \approx 0.00011$
