I'm using the standard Maclaurin expansion for #cos z# and going from there. You could do this by differentiating the function, which would obviously involve much more work.
The Maclaurin series for #cos z# is:
#M(cos z) =sum _{n=0}^{\infty } {(-1)^{n}}/{(2n)!} z^{2n} #
For #z = x^2/2#, this becomes:
#M (cos (x^2/2)) =sum _{n=0}^{\infty } {(-1)^{n}}/{(2n)!} (x^2/2)^{2n} #
# =sum _{n=0}^{\infty } {(-1)^{n}}/{(2n)!} x^(4n)/(2^{2n}) #
And so for: #f(x)= x cos (1/2)x^2#, this is:
#M (x cos (1/2)x^2) =sum _{n=0}^{\infty } {(-1)^{n}}/{(2n)!} x^((4ncolor(red)(+1)))/(2^{2n}) #
Pluggin in #n = 0, 1, 2#:
#= x- x^{5}/{8}+x^{9}/{384}-\cdots #
