jkwii said:
had another quick query,
pi/6
∫ln(tanx + Sqrt(3)) dx
0
i think u use f(x) = f(a-x) to do it but i neek working pls.
Using the following fact
pi/6 pi/6
∫f(x)dx = ∫f(pi/6-x)dx
0 0
pi/6
let I = ∫ln(tanx + Sqrt(3)) dx
0
pi/6
∫ln(tanx + Sqrt(3)) dx =
0
pi/6
∫ln(tan(pi/6-x) + Sqrt(3)) dx =
0
pi/6
∫ln( (1- sqrt(3)tanx)/(sqrt(3)+tanx) + Sqrt(3)) dx =
0
(common denom inside ln)
pi/6
∫ ln(4/(sqrt(3)+tanx)) dx =
0
pi/6
∫ ln4 - ln(sqrt(3)+tanx)dx =
0
pi/6
∫ ln4 - I
0
Therefore 2I = (pi*ln4)/6
hence I=(pi*ln4)/12
Q.E.D.
This is an interesting question to do. Which book did you find that?