A Taylor series question The Taylor series for $cos(x)$ about $x=0$ is $1-x^2/(2!)+x^4/(4!)-x^6/(6!)+...$
If $h$ is a function such that $h'(x) = cos(x^3)$, then the coefficient of $x^7$ in the Taylor series for $h(x)$ about $x = 0$ is?

The Taylor expansion of $\cos t$ is 
$$1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+\cdots.$$
Let us cross our fingers and treat this as a "long" polynomial. Then substituting $x^3$ for $t$, we get
$$\cos(x^3)=1-\frac{x^6}{2!}+\frac{x^{12}}{4!}-\frac{x^{18}}{6!}+\cdots.$$
This is $h'(x)$. Integrate term by term, again treating the expression as if it were a long polynomial. We get 
$$h(x)=C-\frac{x^7}{7\cdot 2!}+\frac{x^{13}}{13\cdot 4!}-\frac{x^{19}}{19\cdot 6!}+\cdots.$$
