TITLE: Integral of $\cos(\sqrt{x^3})$
QUESTION [5 upvotes]: As the title explains, I can't seem to get anywhere on this one, so could someone show me how I would proceed with solving this?
$$\int\cos\left(\sqrt{x^3}\right)\,dx$$
I can't shrug off the feeling that I'll end up somewhere in the complex realm.

REPLY [2 votes]: Another form of solution is to express the integrand in a power series and integrate term by term. 
$$ \cos(\sqrt{x^3}) = 1 - \frac{x^3}{2!} + \frac{x^6}{4!} - \frac{x^9}{6!} +\ldots, \quad x \ge 0.$$
Because this power series is convergent for all $x \ge 0$, the power series of the resulting integral is also convergent for all $x\ge0$. Hence,
$$\begin{align}
\int \cos(\sqrt{x^3})\,dx &= x - \frac{x^4}{2!4} + \frac{x^7}{4!7} - \frac{x^{10}}{6!10} +\ldots \\
&= \sum_{n=0}^{\infty} \frac{(-1)^nx^{3n+1}}{(2n)!(3n+1)}, \quad x >= 0.
\end{align}
$$