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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.

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    $\begingroup$ If the Maclaurin series of $\sin(z)$ looks like $z-z^3/6+z^5/120-\cdots,$ what does the Maclaurin series of $\sin(x^2)$ look like? $\endgroup$
    – Integrand
    Jun 23 '20 at 2:15
  • $\begingroup$ It would be that but squared? $\endgroup$
    – Jackson
    Jun 23 '20 at 2:16
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    $\begingroup$ Careful: what do you mean by 'that'? $\endgroup$
    – Integrand
    Jun 23 '20 at 2:17
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    $\begingroup$ Sorry not exactly "that" but the first five terms of the Maclaurin series of sin(x) and then I would square them? $\endgroup$
    – Jackson
    Jun 23 '20 at 2:18
  • $\begingroup$ Are you computing the Maclaurin series by differentiating $\cos(x^2)$ lots of times? $\endgroup$ Jun 23 '20 at 2:18
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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?

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  • $\begingroup$ Yes, thank you very much this helps a lot $\endgroup$
    – Jackson
    Jun 23 '20 at 2:59

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