Determine the value of the constant y, whenever:

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

where ln x denotes the natural logarithm of x.

We can easily solve the indefinite integral by substituting e^x = u, and then u = sec^2 theta, to yield: 2 arccos(e^-0.5x) + C

We use this to solve the definite integral and obtain:
2 arccos(e^-0.5y) - 2 arccos(ħsqrt(3)/2) = pi/6
2 arccos(e^-0.5y) - 2 ({ħpi/6, 5pi/6, 7pi/6} + 2n pi) = pi/6
where {...} denotes a list, and n represents any integer

2 arccos(e^-0.5y) - ({ħpi/3, 5pi/3, 7pi/3} + 4n pi) = pi/6
2 arccos(e^-0.5y) = pi/6 + {ħpi/3, 5pi/3, 7pi/3} + 4n pi
2 arccos(e^-0.5y) = {pi/2, -pi/6, 11pi/6, 15pi/6} + 4n pi
arccos(e^-0.5y) = {pi/4, -pi/12, 11pi/12, 15pi/12} + 2n pi
e^-0.5y = cos({pi/4, -pi/12, 11pi/12, 15pi/12} + 2n pi)
e^-0.5y = cos({pi/4, -pi/12, 11pi/12, 15pi/12})
e^-0.5y = {ħsqrt(2)/2, ħsqrt(2+abs(sqrt(3)))/2}
-0.5y = {ln ħsqrt(2)/2, ln(ħsqrt(2+abs(sqrt(3)))/2)}
-0.5y = {0.5 ln 0.5, 0.5 ln((2+abs(sqrt(3)))/4)}
y = {ln 2, ln(8-4abs(sqrt(3)))}

Both are correct. Please note that ln(8-4abs(sqrt(3))) < ln(4/3) < ln 2.

Posted by Kurious on 2007-11-27 18:54:23