How to demonstrate the equality of these integral representations of $\pi$?

Each of the three following definite integrals are well known to have the same value of $\pi$: $$\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x=2\int_{-1}^1\sqrt{1-x^2}\,\mathrm{d}x=\int_{-\infty}^{\infty}\frac{1}{1+x^2}\,\mathrm{d}x=\pi.$$

I like taking the first definite integral as the definition of $\pi$, since it represents half the circumference of the unit circle. The second integral obviously represents the area of the unit circle, but as an exercise I wanted to prove its equality with integral #1 using just elementary integration rules. I was successful once I tried integrating by parts:

$$\begin{align} \int\sqrt{1-x^2}\,\mathrm{d}x &=x\sqrt{1-x^2}-\int\frac{-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\ &=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\frac{1-x^2}{\sqrt{1-x^2}}\,\mathrm{d}x\\ &=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-\int\sqrt{1-x^2}\,\mathrm{d}x\\ \implies2\int\sqrt{1-x^2}\,\mathrm{d}x&=x\sqrt{1-x^2}+\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x\\ \implies2\int_{-1}^1\sqrt{1-x^2}\,\mathrm{d}x&=\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x. \end{align}$$

Having accomplished this much, I decided I'd like to demonstrate the equality of these two integrals to $\int_{-\infty}^{\infty}\frac{1}{1+x^2}\mathrm{d}x$ as well, in a similarly elementary manner, but I'm stumped as to what to try. Can anyone suggest a substitution or transformation that demonstrates their equality?

These are both immediate using two subsitutions. Everything is even, so split all of them at $0$.

$$\begin{aligned} t=\sqrt{1-x^2}:\quad\int_0^1 \frac{dx}{\sqrt{1-x^2}}= \int_0^1 2\sqrt{1-t^2}\,dt\end{aligned}$$

And:

$$\begin{aligned} t=\frac{x}{\sqrt{1-x^2}}:\quad\int_0^{1} \frac{dx}{\sqrt{1-x^2}}= \int_0^{\infty}\frac{dt}{1+t^2}\end{aligned}$$

If in the third integral you let $x=\tan(t)$, you get

$$ I_3 = \int_{-\pi/2}^{\pi/2} 1 dt. $$

If, in the second integral, you do the substitution $x=\sin(u)$, you get $$ I_2 = 2\int_{-\pi/2}^{\pi/2} \cos^2 t ~ dt. $$

Alternatively, substitute $x = \cos(u)$ to get $$ I_2 = 2\int_{-\pi/2}^{\pi/2} \sin^2 t ~ dt. $$

So combining these last two,

$$ I_2 = \frac{1}{2} \left (2\int_{-\pi/2}^{\pi/2} \sin^2 t ~ dt + 2\int_{-\pi/2}^{\pi/2} \sin^2 t ~ dt \right) \\ = \frac{1}{2} \left (2\int_{-\pi/2}^{\pi/2} \sin^2 t + \cos^2 t~ dt \right) \\ = \int_{-\pi/2}^{\pi/2} 1~ dt $$

hence the second and third integrals are equal, as you wanted.

Not completely satisfactory, but a shift from integrating from $-1$ to $1$ to integrating from $-\infty$ to $\infty$ is likely to involve something like a tangent substitution in general, so it's not too surprising to see this route work out.

Let $x = \sin \theta, \int_{-1}^1 \frac{1}{\sqrt{1-x^2}} dx = 2\int_{0}^1 \frac{1}{\sqrt{1-x^2}} dx = 2 \int_0^{\pi/2} \frac{1}{\cos\theta} d\sin\theta =\pi$

$\int_{-\infty}^{\infty}\frac{1}{1+x^2}dx = \int_{-\infty}^{\infty} d \left(\tan^{-1}x \right) =\int_{-\pi/2}^{\pi/2} dy = \pi, where \quad y=\tan^{-1}x$