Solution
= Solution
Use the <Taylor series> $\cos x=1-x^2/2+O(x^4)$. Squaring gives $\cos^2x=1-x^2+O(x^4)$, and hence
$$
\boxed{x^3\cos^2x=x^3-x^5+O(x^7)}.
$$
The first two nonzero terms are therefore $x^3$ and $-x^5$.
= Solution
Use the <Taylor series> $\cos x=1-x^2/2+O(x^4)$. Squaring gives $\cos^2x=1-x^2+O(x^4)$, and hence
$$
\boxed{x^3\cos^2x=x^3-x^5+O(x^7)}.
$$
The first two nonzero terms are therefore $x^3$ and $-x^5$.