Geometrical interpretation of $\pi=\int_0^1\frac{4}{1+x^2}dx$.

Question

How to show that $$\pi=\int_0^1\frac{4}{1+x^2}dx?$$ I know how to do it symbolically by using that $\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$. But is there a geometrical interpretation of this result?

Answer 1

Consider the unit semicircle in the upper half-plane and the line $y=1$.

With a little geometry, one can show that $$\mathrm dl=\frac{\mathrm dx}{1+x^2}.$$ Therefore, integrating $\mathrm dx/(1+x^2)$ over $x\in[0,1]$ gives you the circumference of the $45^\circ$ arc bounded by $x=0$ and $x=y$.

(This is just the geometrical interpretation of the substitution $x=\tan\theta$.)

Answer 2

The substitution

$$ x = \tan\left[\frac{ t\sqrt{1-t^2} + \arcsin t }{2}\right] $$ turns the integral into $$ \int_0^1 4\sqrt{1-t^2} dt. $$ But this is exactly four times the area of a quarter unit circle, hence equals $\pi$.