By binomial series,
#1/{sqrt{1-x^2}}=sum_{n=0}^{infty}{1cdot3cdot5cdot cdots cdot(2n-1)}/{2^n n!}x^{2n}#
Let us first find the binomial series for
#1/{sqrt{1+x}}=(1+x)^{-1/2}#
Its binomial coefficient is
#C(-1/2,n)={(-1/2)(-3/2)(-5/2)cdot cdots cdot(-{2n-1}/2)}/{n!}#
by factoring out all #-#'s and #1/2#'s,
#={(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}#
So, we have the binomial series
#1/{sqrt{1+x}}=sum_{n=0}^{infty}{(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}x^n#
Now, we can find the binomial series for the posted function by replacing #x# by #-x^2#.
#1/{sqrt{1-x^2}}#
#=sum_{n=0}^{infty}{(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}(-x^2)^n#
which simplifies to
#=sum_{n=0}^{infty}{(-1)^n[1cdot3cdot5cdot cdots cdot(2n-1)]}/{2^n n!}(-1)^nx^{2n}#
since #(-1)^ncdot(-1^n)=(-1)^{2n}=1#,
#=sum_{n=0}^{infty}{1cdot3cdot5cdot cdots cdot(2n-1)}/{2^n n!}x^{2n}#
