TITLE: Geometrical interpretation of $\pi=\int_0^1\frac{4}{1+x^2}dx$.
QUESTION [5 upvotes]: 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?

REPLY [11 votes]: 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 $x=\tan\theta$.)