<s>[INST] what is the taylor expansion of $\ln(\sin x/x)$ what is the Taylor expansion of the function
$$f(x) = \ln \frac{\sin x}x$$
around  the point $x=0$? Ignore powers of $x$ which are greater than $6$.
Here is my method:
$$\ln(1+x)=x-\frac{x^2}2 + \frac{x^3}3 -\frac{x^4}4,$$
so we should get the function
$$g(x)=\frac{\sin x}x$$
in the form of $1 + x$ to get the answer … or is there another way to consider the Taylor expansion of $\ln (u)$ by considering $f$ to be the answer of $$\frac{\sin x}x\cdots$$
Which way is the best?? What is the final answer???  
 [/INST] Your instincts are correct.  Write
$$\frac{\sin{x}}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots$$
We wish to maintain the series to $O(x^6)$.  Then
$$\begin{align}\ln{\left ( \frac{\sin{x}}{x}\right)} &=  \left (\frac{\sin{x}}{x} - 1\right ) - \frac12 \left (\frac{\sin{x}}{x} - 1\right )^2 + \frac13 \left (\frac{\sin{x}}{x} - 1\right )^3-\cdots \\ &= \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right ) - \frac12 \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right )^2+\\ &\frac13 \left ( - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots \right )^3-\cdots \\ &= - \frac{x^2}{6} + \frac{x^4}{120} - \frac{x^6}{5040} + \cdots - \frac{x^4}{72} \left (1 - \frac{x^2}{10}+\cdots \right ) - \frac{x^6}{648}+\cdots \\ &= - \frac{x^2}{6} - \frac{x^4}{180}-\frac{x^6}{2835}+\cdots \end{align}$$
Note that
$$-\frac{1}{180} = \frac{1}{120} - \frac{1}{72}$$
$$-\frac{1}{2835} = -\frac{1}{5040}+\frac{1}{720}-\frac{1}{648}$$
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