<s>[INST] Prove $e^{x \cos(x)}=1+x+\frac{x^2}{2} - \frac{x^3}{3}-\frac{11x^4}{24}- \frac{x^5}{5} + \cdots$ How do we do this question using Maclaurin's Series. I tried expanding it by putting $x \cos x$ in place of $x$ in Maclaurin's expansion of $e^x$, and then using multinomial theorem to open the squares, cubes etc of $\cos x$. Is my approach correct?
 [/INST] There's a sign error in your $x^3$ coefficient. With $\equiv$ denoting equality up to $x^5$ terms, $$\exp(x\cos x)\equiv\exp\left(x-\frac{x^3}{2}+\frac{x^5}{24}\right)\\\equiv\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\frac{x^5}{120}\right)\left(1-\frac{x^3}{2}\right)\left(1+\frac{x^5}{24}\right)\\\equiv\left(1+x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{11x^4}{24}-\frac{29x^5}{120}\right)\left(1+\frac{x^5}{24}\right)\\\equiv 1+x+\frac{x^2}{2}-\frac{x^3}{3}-\frac{11x^4}{24}-\frac{x^5}{5}.$$
</s>