<s>[INST] 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?
 [/INST] So, what happened was when I took the indefinite integral, to cover all potential functions which satisfy the indefinite integral, + C needs to be included.
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