y
     ∫(ex - 1)-0.5 dx = pi/6 
      ln(4/3)

where ln x denotes the natural logarithm of x.

You are correct. And I am correct too.

Let y = ln(8-4abs(sqrt(3))). Considering that y < ln(4/3), 
-(2 arccos(e^-0.5y) - 2 arccos(±sqrt(3)/2)) = pi/6
2 arccos(e^-0.5y) - 2 arccos(±sqrt(3)/2) = -pi/6
2 arccos(e^-0.5y) - 2 ({±pi/6, 5pi/6, 7pi/6} + 2n pi) = -pi/6
where {...} denotes a list, and n represents any integer

2 arccos(e^-0.5y) - ({±pi/3, 5pi/3, 7pi/3} + 4n pi) = -pi/6
2 arccos(e^-0.5y) = -pi/6 + {±pi/3, 5pi/3, 7pi/3} + 4n pi
2 arccos(e^-0.5y) = {-pi/2, pi/6} + 4n pi
arccos(e^-0.5y) = {-pi/4, pi/12} + 2n pi
e^-0.5y = cos({-pi/4, pi/12} + 2n pi)
e^-0.5y = cos({-pi/4, pi/12})
e^-0.5y = {abs(sqrt(2)/2), abs(sqrt(2+abs(sqrt(3))))/2}
-0.5y = {ln abs(sqrt(2)/2), ln abs(sqrt(2+abs(sqrt(3))))/2}
-y = {ln 1/2, ln (2+abs(aqrt(3)))/4}
y = {ln 2, ln(8-4abs(sqrt(3)))} 

so that y = ln(8-4abs(sqrt(3))) is a valid solution. Indeed, I believe that this can also be validated by means of the numerical method.

y = ln(8-4abs(sqrt(3))) ~= 0.0693364641950739

Posted by Kurious on 2007-11-28 07:28:10