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: $1-x+x^2-\frac{5x^3}{6}+\frac{2x^4}{3}$

Your answer is correct, but there is a somewhat simpler method. You are looking for a polynomial $P=a+bx+cx^2+dx^3+ex^4$ such that $P\times(1+x-\frac{x^3}6)\equiv 1\pmod{x^5}$. This gives (by comparing coefficients of $x^0,x^1,\ldots,x^4$) the equations
$$
  \begin{align}a&=1\\a+b&=0\\b+c&=0\\-\tfrac16a+c+d&=0\\-\tfrac16b+d+e&=0,
  \end{align}
$$
which you can solve straight away as $a=1$, $b=-1$, $c=1$, $d=-\tfrac56$, $e=\tfrac23$.
