Help with Maclaurin series of $\cos(\ln(x+1))$? Hi I've almost completed a maths question I am stuck on I just can't seem to get to the final result. The question is:
Find the Maclaurin series of $g(x) = \cos(\ln(x+1))$ up to order 3.
I have used the formulas which I won't type out as I'm not great with Mathjax yet sorry. But I have obtained:
$$ \ln(x+1) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} $$
$$ \cos x =   1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} $$
I'm sure what I have done above is correct however I can't seem to get to the final solution shown in the answers and would like some help please. Thanks 

Since$$\cos x=1-\frac{x^2}2+\cdots,$$you have\begin{align}\cos\bigl(\log(x+1)\bigr)&=1-\frac12\left(x-\frac{x^2}2+\frac{x^3}3-\cdots\right)^2+\cdots\\&=1-\frac{x^2}2+\frac{x^3}2+\cdots\end{align}Note that this argument works because the series for $\log(x+1)$ has no constant term. A consequence of this is that no matter how more terms you use from both series, you will get no new term whose degree is smaller than $4$.
