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?
 A: 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:
$$\begin{aligned} t=\frac{x}{\sqrt{1-x^2}}:\quad\int_0^{1} \frac{dx}{\sqrt{1-x^2}}= \int_0^{\infty}\frac{dt}{1+t^2}\end{aligned}$$
A: 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. 
A: Let $x = \sin \theta, \int_{-1}^1 \frac{1}{\sqrt{1-x^2}} dx = 2\int_{0}^1 \frac{1}{\sqrt{1-x^2}} dx = 2 \int_0^{\pi/2} \frac{1}{\cos\theta} d\sin\theta =\pi$
$\int_{-\infty}^{\infty}\frac{1}{1+x^2}dx = \int_{-\infty}^{\infty} d \left(\tan^{-1}x \right) =\int_{-\pi/2}^{\pi/2} dy = \pi, where \quad y=\tan^{-1}x$
