<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] Faster converging series
$\sqrt{1-x^2} = 1 - \frac{1}{1!} \frac{1}{2} x^2 - \frac{1}{2!} \frac{1}{2} \frac{1}{2} x^4 - \frac{1}{3!} \frac{1}{2} \frac{1}{2} \frac{3}{2} x^6 + \cdots$  [by Binomial theorem]
$\ = 1 - \sum_{k=1}^\infty \frac{(2k-3)!!}{2^k k!} x^{2k}$  [where $n!!$ is the double-factorial of $n$]
$\int \sqrt{1-x^2}\ dx = x - \sum_{k=1}^\infty \frac{(2k-3)!!}{2^k k! (2k+1)} x^{2k+1}$  [which converges faster when $x = \frac{1}{2}$ than $x = 1$]
$π = 12 \cdot \left( \int_{0}^{\frac{1}{2}} \sqrt{1-x^2}\ dx - \frac{\sqrt{3}}{8} \right)$  [the integral is $\frac{1}{12}$ of the pie plus a triangle]
$\ = 12 \cdot \left( \frac{1}{2} - \sum_{k=1}^\infty \frac{(2k-3)!!}{2^{3k+1} k! (2k+1)} \right ) - \frac{3\sqrt{3}}{2} = 6 - \frac{3\sqrt{3}}{2} - 12 \sum_{k=1}^\infty \frac{(2k-3)!!}{2^{3k+1} k! (2k+1)}$
Notes
Binomial theorem for non-integer powers can be proven by using Taylor series of $x \mapsto (1+x)^r$ on $(0,1)$. But that requires knowledge of the derivatives, for which the simplest proof is by product rule and chain rule for integer powers, and by implicit differentiation for extending to rational powers, followed by limits for extending to real powers.
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