[VA] SRC #009 - Pi Day 2021 Special

[Albert Chan]

(03-21-2021 06:48 PM)robve Wrote:  Proof (I've simplified this somewhat):

$$ 4 \int_0^1 \sqrt{1-x^2} \,dx = 4 \int_0^{\frac{\pi}{2}} \sqrt{1-\sin^2 \theta} \cos \theta \,d\theta = 4 \int_0^{\frac{\pi}{2}} \sqrt{\cos^2 \theta} \cos \theta \,d\theta = 4 \int_0^{\frac{\pi}{2}} \cos^2 \theta \,d\theta = \pi $$

A comment about the last (missing) step.
Instead of using half-angle formula, cos(x/2)^2 = (1+cos(x))/2, then integrate, fold the integral.

\(\displaystyle \int _a^b f(x)\;dx = \int _a^b {f(x) + f(a+b-x) \over 2}\;dx \)

\(\displaystyle 4 \int_0^{\pi \over 2} \cos^2 θ\;dθ
= 2 \int_0^{\pi \over 2} (\cos^2 θ \;+\; \sin^2 θ)\;dθ
= 2 \int_0^{\pi \over 2} 1\;dθ
= \pi \)

This is the same trick used in [VA] Short & Sweet Math Challenge #25.

(02-28-2021 02:18 AM)Valentin Albillo Wrote:  My original solution for "Concoction the Third: Weird Integral"



What's so weird about this integral?