With an error of magnitude less than $10^{-5}$ use a series to estimate $$\int_{0}^{0.6} \sin(x^2) dx$$
I have calculated up to the 5th term of the series: $$x^2 - \frac{x^6}{6} + \frac{x^{10}}{120} - \frac{x^{14}}{5040} + \frac{x^{18}}{362880}$$
I don't know how to continue.
Using a Taylor expansion with Lagrangian remainder, one can estimate the error caused by replacing the function with its Taylor expansion.
As we want an error of less than $10^{-5}$, $\frac {f^{(n+1)}(\zeta)} {(n+1)!} (x-x_0)^{n+1}$ must be less than $5*10^{-6}$, while $\zeta$ is some value between $x_0$ and $x$, in this case between $0$ and $0.6$.
Since $0.6$ is very close to $0$, we'd better off assuming $2\zeta\cos(\zeta^2) = 1.2$ here to minimize the error.
So solve for $n$ in the following equation, $$ \frac {1.2*0.6^{n+1}} {(n+1)!}\lt 5*10^{-6} $$ and that is the degree of the highest term in the Taylor expansion you want.