Taylor series of $\sin(x^2)$

Question

I am stuck on a problem for my calc 2 course. We are being asked to use Taylor series centered around x=0 (Maclaurin series) to approximate $\sin(x^2)$ and we are being asked to calculate the first five (non-zero) terms in the series and then integrate using our approximation. The issue is there are a lot of zero terms therefore by the time I reach my third term (that has a value) I am up to the tenth derivative. We never go this high and I think I must be missing something (?). Sorry if this question is confusing I'm fairly new to this sort of calculus and I couldn't find any examples using this function.

Answer 1

Instead of computing the derivatives of $\sin(x^2)$ to find it's Maclaurin series, it would be easier if we were to substitute $x^2$ into $x$ in the Maclaurin series of $\sin(x)$. We know that $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}\cdots$$ Therefore, we substitute $x^2$ into each $x$ in the equation and we get $$\sin(x^2)=x^2-\frac{({x^2})^3}{6}+\frac{({x^2})^5}{120}\cdots=x^2-\frac{x^6}{6}+\frac{x^{10}}{120}\cdots$$ Can you try to deduce the remaining 2 terms according to this?