First we must find the series for #sin(x)#
let# f(x) = sin(x) #
#f(0) = sin(0) = 0#
#f'(0) = cos(0) = 1#
#f''(0) = -sin(0) = 0#
#f'''(0) = -cos(0) = -1#
Now we can apply to the macluarin series;
#f(x) = f(0) + f'(0)x + (f''(0)x^2)/(2!) + (f'''(0)x^3)/(3!) + ...#
Hence #sin(x) = x - x^3/(3!) + x^5/(5!) - ...#
Hence for #sin(x^2)# we replace each #x# by #x^2# in the series for #sin(x)#
#sin(x^2) = (x^2) - (x^2)^3/(3!) + (x^2)^5/(5!) - ...#
# = x^2 - x^6/(3!) + x^10/(5!) - ....#
Hence #x^2 - sin(x^2) => #
#x^2 - (x^2-x^6/(3!) +x^10/(5!) - ... ) #
#=> x^6/(3!) - x^10/(5!) + ...#
# = sum_(n=1) ^oo x^(4n+2) / ((2n+1)!) * (-1)^(n+1) #
