Why in the world would one use a trig substitution to find a Taylor's series??
Of course, you could do this "the old-fashioned way": use the definition of Taylor's series. (By the way, you say "Taylor Series" but don't say about what point. I assume you mean the McLaurin series: about x0= 0.)
f(x)= 1/(1-x2) so f(0)= 1
f'(x)= ((1- x2)-1)'
= (2x)(1-x2)-2 so f'(0)= 0
f"(x)= 2(1-x2)-2-2(2x)(1-x2)-3 so f"(0)= 2, etc.
So far we would have f(0)+ f'(0)(x)+ f"(0)/2(x2= 1+ x2 but the derivatives quickly become very difficult to find!
I would be inclined to remember the formula for sum of a geometric series: If |r|< 1, then Σrn= 1/(1-r).
The right hand side looks a lot like 1(1- x2) if we were to take r= x2!
Σ x2n= 1/(1-x2)
Since an infinitely differentiable function only has one power series, that IS the Taylor's series.
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