an interesring integral

[@Metaverse-d9f]

really "interesring" :)

[@maxvangulik1988]

sin(2pi-t)=-sin(t)

[@holyshit922]

Is this really the result
My program for trapezoid method gives me numerical value of
ln(1+sqrt(3)/2)*Pi
This result is confirmed by CAS programs like Maple

[@seegeeaye]

@@holyshit922 as for this particular integral, the result should be right

[@holyshit922]

@@seegeeaye Are you sure ?
Numerical value of this integral is 1.9597591638
but your result is 1.273806205

[@seegeeaye]

@@holyshit922 are they the same integral? my answer should be ok unless you find a mistake in the process

[@holyshit922]

@@seegeeaye I tried different approach
A little bit more difficult than yours
\int_{0}^{\pi}\ln{(2+cos(x))}dx = \int_{0}^{\frac{\pi}{2}}\ln{(2+cos(x))}dx + \int_{\frac{\pi}{2}}^{\pi}\ln{(2+cos(x))}dx
\int_{0}^{\frac{\pi}{2}}\ln{(2+\sin(x))}d(\frac{\pi}{2}-x) + \int_{0}^{\frac{\pi}{2}}\ln{(2-\cos(x))}dx
\int_{0}^{\frac{\pi}{2}}\ln{(3+\cos^2(x))}dx
\int_{0}^{\frac{\pi}{2}}\ln{(\frac{3}{\cos^2{x}}+1)}dx +2\int_{0}^{\frac{\pi}{2}}{\ln{(cos(x))}dx}
t = tan(x)
\int_{0}^{\infty}\frac{\ln(4+3x^2)}{1+x^2}dx +2\int_{0}^{\frac{\pi}{2}}{\ln{(cos(x))}dx}
We know that \int{\frac{6x^2y}{4+3x^2y^2},y=0..1} = ln(4+3x^2)-ln(4)
so we can convert integral \int_{0}^{\infty}\frac{\ln(4+3x^2)}{1+x^2}dx
to double integral then change order of integration and convert to iterated integral
We probabliy could use also complex integration to calculate \int_{0}^{\infty}\frac{\ln(4+3x^2)}{1+x^2}dx
I like your approach because it is as simple as possible but there must be somewhere a mistake

[@seegeeaye]

@@holyshit922 I try to follow yours, it's too complicated. I just double checked mine again just now, I didn't see any mistake, I believe my answer is correct, not even approximation