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?

  • $\begingroup$ Maclaurin series are Taylor series. $\endgroup$
    – Kenta S
    Apr 21, 2021 at 20:38
  • $\begingroup$ Oh okay, thanks. $\endgroup$ Apr 21, 2021 at 20:42

1 Answer 1


\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*}

  • $\begingroup$ Oh, I think I see. You can just insert the power series for cosx into the integral, and then compute? And the series excluding x^4n is taken out of the integral because its a coefficient? $\endgroup$ Apr 21, 2021 at 20:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .