# $\sqrt{1+x}$ as summation

It states that: $$\int_0^1(1+(-x^2))^{1/2}dx = \int_0^1\sum_{k=0}^\infty\binom{1/2}{k}(-x^2)^k dx$$ Which I assume gives: $$(1+(-x^2))^{1/2} = \sum_{k=0}^\infty\binom{1/2}{k}(-x^2)^k$$ I have never seen this, and I couldn't find anything about it online, so I was wondering where this comes from. Could it possibly be generalised to: $$(1+a)^{b} = \sum_{k=0}^\infty\binom{b}{k}(a)^k$$
• So this only works when $|x|<1$? Jun 22 '21 at 1:27
• Yeah, only $|x|<1.$ But here, that is also the only real domain of $(1-x^2)^{1/2}$ Jun 22 '21 at 1:43
• What about $(1-0^2)^{1/2} = 1$. Would it still work? Jun 22 '21 at 1:49