Use 2-2*cos(x) = 4*(sin(x/2))^2, then B-A = 2*pi*log(2) + 2*\int_0^{\pi} log(sin(x/2)) dx. For the last integral, see https://socratic.org/questions/integrate-log-sinx-from-0-to-pi-2 On Tue, Aug 21, 2018 at 8:28 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Thanks! (That HAD to be true since no two real numbers can be as close as these were.)
—Dan
According to maple they are symbolically equal. I did not dig into it to try to see why.
On Mon, Aug 20, 2018 at 5:19 PM Dan Asimov <dasimov@earthlink.net> wrote:
Let f(x) = ln(2-2*cos(x))
Then are these two numbers equal?
A = -Integral from 0 to π/3 of f(t) dt
B = Integral from π/3 to π of f(t) dt
I can't tell. But they're extremely close.
—Dan
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