"Let $~F(x)~$ be the unique function that satisfies $~F(0) = 0~$ and $~F′(x) = \dfrac{\sin(x^3)}{x}~$ for all $~x$. Find the Taylor series for $~F(x)~$ about $~x = 0$."
Wouldn't the value of every derivative at $~0~$ just be $~0~$? So how does a Taylor series even exist?
If $~F(0)=0$, then can the Taylor series of $~F(x)~$ be the same as that of $~F'(x)~$?
You have $$F′(x) = \dfrac{\sin(x^3)}{x}$$
Now use the Taylor series $$\sin(t)=\sum_{n=0}^\infty\frac{(-1)^n }{(2 n+1)!}t^{2 n+1}$$ Make $t=x^3$ to get $$F'(x)=\sum_{n=0}^\infty\frac{(-1)^n }{(2 n+1)!}x^{6 n+2}$$ and integrate to get $$F(x)=\sum_{n=0}^\infty\frac{(-1)^n }{(6n+3)(2 n+1)!}x^{6 n+3}+C$$