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. $~~~~~~~~~$
On
You could try a Taylor expansion and then integrate.
For example
$$\sqrt{1 - x^2}= 1-\dfrac{1}{2}{{x}^{2}}-\dfrac{1}{8} {{x}^{4}} -\dfrac{1}{16} {{x}^{6}} - \dfrac{5}{128} {{x}^{8}} - \dfrac{7}{256} {{x}^{10}} - \cdots$$
which has an indefinite integral of
$$x-\dfrac{1}{6}{{x}^{3}}-\dfrac{1}{40} {{x}^{5}} -\dfrac{1}{112} {{x}^{7}} - \dfrac{5}{1152} {{x}^{9}} - \dfrac{7}{2816} {{x}^{11}} - \cdots$$
so $$\pi = 4-\dfrac{2}{3}-\dfrac{1}{10} -\dfrac{1}{28} - \dfrac{5}{288} - \dfrac{7}{704} - \cdots$$
but convergence is fairly slow with the sum of those particular terms giving about $3.17$
On
Integrating by parts,
$$I_n=\int(1-x^2)^ndx=(1-x^2)^n\int dx-\int\left[\dfrac{d\{(1-x^2)^n\}}{dx}\int dx\right]dx$$
$$=(1-x^2)^nx+2n\int(1-x^2)^{n-1}x^2\ dx$$
$$=(1-x^2)^nx+2n\int(1-x^2)^{n-1}\{1-(1-x^2)\}\ dx$$
$$I_n=(1-x^2)^nx+2n[I_{n-1}-I_n]$$
$$\implies(2n+1)I_n=(1-x^2)^nx+2nI_{n-1}$$
Set $n=\dfrac12$
But I'm not sure about solving $\int\dfrac{dx}{\sqrt{1-x^2}}$ w/o using Trigonometry as $\dfrac{d(\arcsin x)}{dx}=\dfrac1{\sqrt{1-x^2}}$
On
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.
$\pi$ 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.