<s>[INST] Maclaurin series of $\frac{1}{1+\sin x}$ Find the terms through degree four of the Maclaurin series of $f(x)$.
$$f(x) = \frac{1}{1+\sin x}$$
My work:
The Maclaurin series for $\sin x$ up to degree $4$ is $x - \frac{x^3}{6} + \frac{x^5}{120}$
The Maclaurin series for $\frac{1}{1+x}$ up to degree $4$ is $1 - x + x^2 - x^3 + x^4$
I substituted $x - \frac{x^3}{6} + \frac{x^5}{120}$ for $x$ in $1 - x + x^2 - x^3 + x^4$
Did I do this right? 
Plugging this into WolframAlpha, I get this: http://goo.gl/SKddyh
Which doesn't seem like the answer in the text: $1-x+x^2-\frac{5x^3}{6}+\frac{2x^4}{3}$
 [/INST] You did it right and nothing has to be added to the answers you received.
I do not know if you were obliged to use these steps since there is a direct way of doing this expansion applying the basic rules, that is to say that the Taylor series of $f(x)$ built at $x=0$ just write (up to the fourth degree)
$$f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+\frac{1}{6} f^{(3)}(0) x^3+\frac{1}{24}
   f^{(4)}(0) x^4+O\left(x^5\right)$$ The value of the function and its first four derivatives are $1,-1 ,2,-5,16$ (they are easy to evaluate bcause $sin(0)=0$ and $cos(0)=1$). 
For sure, you arrive to your result.  
You also have another way : replace in the denominator $sin(x)$ by its Taylor expansion and perform the long division of numerator by denominator.
You have done a good job ! Congratulations.
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