<s>[INST] 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?
 [/INST] Let $x = \sin \theta, \int_{-1}^1 \frac{1}{\sqrt{1-x^2}} dx = 2\int_{0}^1 \frac{1}{\sqrt{1-x^2}} dx = 2 \int_0^{\pi/2} \frac{1}{\cos\theta} d\sin\theta =\pi$
$\int_{-\infty}^{\infty}\frac{1}{1+x^2}dx = \int_{-\infty}^{\infty} d \left(\tan^{-1}x \right) =\int_{-\pi/2}^{\pi/2} dy = \pi, where \quad y=\tan^{-1}x$
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