Efficient way to compute the ninth order Maclaurin polynomial of $\sin(\arctan(x^2))$

Question

Substitution is a favourite of mine but this really gives me an headache. We had a discussion in class about the Maclaurin evaluation up to degree 9 on this function:

$$f(x) = \sin(\arctan(x^2))$$

There are endless possibilities. What would be your best shot?

Answer 1

You need simply to compose the expansions

$$\arctan x = x -\frac{x^3}{3}+o(x^{\frac92})$$

$$\arctan x^2 =x^2 -\frac{x^6}{3}+o(x^9)$$

$$\sin t = t -\frac{t^3}{6}+o(t^3)$$

Thus for $t= \arctan x^2 =x^2 -\frac{x^6}{3}+o(x^9)$

$$\sin (\arctan x^2) = x^2 -\frac{x^6}{3}-\frac16\left(x^2 -\frac{x^6}{3}\right)^3+o(x^9)=\\=x^2 -\frac{x^6}{3}-\frac{x^6}{6}+o(x^9)=x^2 -\frac{x^6}{2}+o(x^9)$$

Answer 2

Hint: $$\sin(\arctan(x^2))=\frac{x^2}{\sqrt{1+x^4}},$$ using elementary trigonometric identities. Then use the binomial expansion $$(1+x^4)^{-1/2}=1+\binom{-1/2}{1}x^4+\binom{-1/2}{2}x^8+\ldots$$