<s>[INST] Integrating $\sqrt{1-x^2}$ without using trigonometry I am a beginning calculus student. Tonight I had a thought. Maybe I could calculate $\pi$ using integration, but no trig.
The problem is that I don't really know where to start. I thought perhaps I could use the Chain Rule for derivatives.
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 [/INST] $\pi$ is a transcendental number. You can't get it in a closed form without using transcendental functions. It's not a value of any algebraic expression (polynomial, rational function, root of polynomial,...) evaluated at any algebraic number. So, you either have to bring $\pi$ into it from the boundaries of the integral, or get it from a transcendental function (trigonometric in this case).
Of course, non-closed form is ok, your original integral is a nice example of that. You could develop your function into an infinite series and try to compute integrals term-by-term. That is actually one of the ways to compute $\pi$ numerically: to construct an infinite sum from such integrals.
We can try the expansion in this case. Consider the series
$$(1+z)^{1/2}=1+\frac{1}{2}z-\frac{1}{2!}\frac{1}{2^2}z^2+\frac{1}{3!}\frac{1\cdot 3}{2^3}z^3-\frac{1}{2!}\frac{1\cdot 3\cdot 5}{2^4}z^4+\cdots$$
$$(1-x^2)^{1/2}=1-\frac{1}{2}x^2-\frac{1}{2!}\frac{1}{2^2}x^4-\frac{1}{3!}\frac{1\cdot 3}{2^3}x^6-\frac{1}{4!}\frac{1\cdot 3\cdot 5}{2^4}x^8+\cdots$$
$$\frac{\pi}{4}=\int_0^1 (1-x^2)^{1/2}dx=1-\frac{1}{2}\frac{1}{3}-\frac{1}{2!}\frac{1}{2^2}\frac{1}{5}-\frac{1}{3!}\frac{1\cdot 3}{2^3}\frac{1}{7}-\frac{1}{4!}\frac{1\cdot 3\cdot 5}{2^4}\frac{1}{9}+\cdots$$
$$\frac{\pi}{4}=1-\sum_{n=1}^\infty \frac{(2n-3)!!}{n! 2^n (2n+1)}$$
where $n!!$ is the double factorial (product of every other term) and by definition $(-1)!!=1$.
The convergence is actually not as bad as in the case of Leibniz formula (https://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80), you can actually get a reasonable value.
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