How would I find the Taylor series around $x = 0$ for this integral?
$$\int \cos(x^2)$$
My first point of confusion is if it is around $x = 0$, doesn't that make it a Maclaurin series?
Would I go about finding the higher order derivatives of $\int \cos(x^2)$ while substituting $x=0$ for each derivative until I find a pattern to then substitute into the Taylor series formula?
$\endgroup$21 Answer
$\begingroup$\begin{align*} \int\cos(x^2)dx&=\int\left(\sum_{n=0}^\infty\frac{(-1)^nx^{4n}}{(2n)!}\right)dx\\ &=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!}\int x^{4n}dx\\ &=\sum_{n=0}^\infty\frac{(-1)^n}{(2n)!(4n+1)}x^{4n+1}+C. \end{align*}
$\endgroup$1
