In order to find the Maclaurin series of #(sin(x^2))/(x)#, we first need to take note of some prominent, already-derived Maclaurin series.
We know that the Maclaurin series of #sin(x)# is the following:
#sin(x) = x-(x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + ... = sum_(n=0)^(∞) (-1)^(n) (x^(2n+1))/((2n+1)!)#
Given this known Maclaurin series, we can now see that
#sin(x^2) = sum_(n=0)^(∞) (-1)^(n) ((x^2)^(2n+1))/((2n+1)!) = sum_(n=0)^(∞) (-1)^(n) (x^(4n+2))/((2n+1)!) #
Note that we still have another factor we have to consider, namely the division by #x#, which we can easily incorporate into our series.
Since we now have the series
#sin(x^2)/x = 1/x * sum_(n=0)^(∞) (-1)^(n) (x^(4n+2))/((2n+1)!)#
All we need to do is to divide by an #x#, giving us
#sum_(n=0)^(∞) (-1)^(n) (x^(4n+1))/((2n+1)!) = x/(1!) - x^(5)/(3!)+x^(9)/(5!)-x^(13)/(7!) + ...#
We can check the first #4# terms of this Maclaurin series, as shown in the graph below.
Graph of #sin(x^2)/x#
graph{sin(x^2)/x [-10, 10, -5, 5]}
Graph of #x/(1!) -x^(5)/(3!)+x^(9)/(5!)-x^(13)/(7!)#
graph{x-x^(5)/(3!)+x^(9)/(5!)-x^(13)/(7!) [-10, 10, -5, 5]}
Overlapping both graphs produces the following:
