Approximating pi using the $\int_0^1\sqrt{1-x^2}dx$ The first step would be to find the series expansion for $\sqrt{1-x^2}$ using the binomial theorem
$$\sqrt{1-x^2} = \sum_{n=0}^\infty {1/2\choose n}(-x^2)^n $$
Expanding and simplifying the first terms
$$\sqrt{1-x^2} = 1 -\left(\frac{1\cdot x^2}{2 \cdot1!} + \frac{1 \cdot 3 \cdot x^4}{2^2 \cdot 2! } + \frac{1 \cdot 3 \cdot 5 \cdot x^6}{2^3 \cdot 3!} +... \right)$$
Here I run into a problem. How come all the terms are negative while $1$ isn't? This would mean I can't add it to the infinite sum. So continuing:
$$\sqrt{1-x^2} = 1- \sum_{n=0}^\infty \frac{(2n+1)!!}{2^n n!}x^{2n} = 1-\sum_{n=0}^\infty \frac{(2n+1)!!}{(2n)!!}x^{2n} $$
When I plot this on desmos my approximation is a) shifted by $1$ unit down b) inaccurate. On the image below I have used 50 terms so I doubt the error is due to a lack of terms.

Where have I made a mistake. Thank you

We can use
$$\frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty {2n \choose n} x^n$$
Since both
$$\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}$$
and
$$\int_{-1}^1 2\sqrt{1-x^2}dx$$
converge to $\pi$, you can either use
$$\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty {2n \choose n} \left(\frac{x}{2}\right)^{2n}$$
or
$$2\sqrt{1-x^2} = \frac{2(1-x^2)}{\sqrt{1-x^2}} = 2+2\sum_{n=1}^\infty \left[{2n \choose n} - {2n-2 \choose n-1}\right]\left(\frac{x}{2}\right)^{2n}$$
$$= 2+2\sum_{n=1}^\infty {2n-2 \choose n-1}\left[3-\frac{2}{n}\right]\left(\frac{x}{2}\right)^{2n}$$
