Maclaurin Series $ \int \frac{\sin(x)}{5x} $ I am supposed to evaluate the indefinite integral as an infinite series centered at $ x=0 $ and give the first five non-zero terms of the series.
$ \int \frac{\sin(x)}{5x} $
Here is what I have done so far:
$ g(x) = \sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} $
$ \int \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)!} \frac{x^{2n+1}}{5x} $
$ \frac{1}{5}\int \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)!} x^{2n} $
$ \frac{1}{5} \sum_{n=0}^{\infty} \frac{(-1)^n }{(2n+1)!} \frac{x^{2n+1}}{2n+1} $
For $ n=0 $, $ \frac{1}{5}(x)$
$ n=1 $, $ \frac{1}{5}(\frac{-1}{3!} \frac{x^3}{3})$
$ n=2 $, $ \frac{1}{5}(\frac{1}{5!} \frac{x^5}{5})$
$ n=3 $, $ \frac{1}{5}(\frac{-1}{7!} \frac{x^7}{7})$
$ n=4 $, $ \frac{1}{5}(\frac{1}{9!} \frac{x^9}{9})$
And sum them.
I've done this a few different ways and am getting the same result. Can anyone shed some light on where I am going wrong?

Your result is perfectly good (before the expansion of the factorials, what I cannot check). 
Forgetting the signs, the denominators are (as you wrote)
$$5 \qquad 5\times 3!\times 3=90\qquad 5\times 5!\times 5=3000 \qquad 5\times 7!\times 7=176400$$ $$\qquad 5\times 9!\times 9=16329600$$ Making then $$\frac{x}{5}-\frac{x^3}{90}+\frac{x^5}{3000}-\frac{x^7}{176400}+\frac{x^9}{16329600}
   +O\left(x^{11}\right)$$
