Not using Taylor series or Maclaurin series, prove $\frac{1}{\sqrt{1-x^2}}=1+\frac{1\cdot3}{2\cdot4}x^2+\frac{1\cdot3\cdot5}{2\cdot4\cdot6}x^4+\cdots$

Question

I’m trying to find a proof of
$$\frac{1}{\sqrt{1-x^2}} = 1+\frac{1\cdot3}{2\cdot4}x^2+\frac{1\cdot3\cdot5}{2\cdot4\cdot6}x^4+\cdots,$$ which doesn’t need Taylor or Maclaurin series, like the proof of Mercator series and Leibniz series.
I tried to prove it by using calculus, but I couldn’t hit upon a good proof.

Answer 1

Since

$$\frac{(2n-1)!!}{(2n)!!}=\frac{1}{4^n}\binom{2n}{n}=\frac{1}{\pi}\int_{0}^{\pi}\cos^{2n}(\theta)\,d\theta$$ due to $\cos\theta=\frac{e^{i\theta}-e^{-i\theta}}{2}$, the binomial theorem (standard form) and $\int_{-\pi}^{\pi}e^{im\theta}\,d\theta = 2\pi\,\delta(m)$, for any $x$ such that $|x|<1$ we have

$$ \sum_{n\geq 0}\frac{(2n-1)!!}{(2n)!!}x^{2n} = \frac{1}{\pi}\int_{0}^{\pi}\sum_{n\geq 0} x^{2n}\cos^{2n}(\theta)\,d\theta = \frac{2}{\pi}\int_{0}^{\pi/2}\frac{d\theta}{1-x^2\cos^2\theta} $$ and by enforcing the substitution $\theta=\arctan t$ in the last integral we get $$ \sum_{n\geq 0}\frac{(2n-1)!!}{(2n)!!}x^{2n} =\frac{2}{\pi}\int_{0}^{+\infty}\frac{dt}{(1-x^2)+t^2}=\frac{1}{\sqrt{1-x^2}}$$ as wanted.