1. The problem statement, all variables and given/known data Show by appropiate substitutions that ∫ (e2z-1)-0.5 dz from 0 to infinity is equivalent to ∫(1-x2)-0.5 dx from 0 to 1. Thus, show that the answer is π/2. 2. Relevant equations 3. The attempt at a solution Where to begin! I tried the substitution e2z= 2-x2, but this then transforms the integral into ∫(2-x2)/(x(1-x2)0.5) dx, and the limits don't work as it would then require the root of a negative number. I then tried the substitution ez = x, but that gives you ∫1/(x(x2-1)0.5) dx, the limits being infinity and 1. I carried on a bunch of different substitutions involving the ez and a simple x polynomial, which I won't repeat due to them being abysmal failures! In doing the problem I managed to skip the middle step entirely, and got the value of π/2 I believe by re-writing the initial equation as -∫(e2z-1)/(e2z-1)0.5 dz +0.5∫2e2z/(e2z-1)^0.5 dz, both still from 0 to infinity. The first of the pair became -∫(e2z-1)0.5 dz between 0 and infinity, which solves to become -[(e2z-1)0.5)-sec-1(ez)] between infinity and 0. The second part I performed the substitution x=e2z-1, with dx = 2e2z dz. Thus this was transformed into (0.5)∫1/(u0.5) du between 0 and infinity. This integrated to give [u0.5] between 0 and infinity, which was then equal to (e2z-1)0.5. The sum of these two parts gives [sec-1(ez)] between 0 and infinity which is equal to cos-1(1/ez) between 0 and infinity, which gives π/2. Unfortunately the question specifies the middle step has to be achieved :(. Onto my current ideas, I thought using a substitution along the lines of z= (2/π)tan-1(x), which transforms the limits into 1 and 0, and transforms the integral into ∫1/((1-x2)(e4(tan-1(x)/π)0.5) dx. This has the 1-x2 part in it, but it is not square rooted and I feel this is as far as I could go. Any help concerning which substitutions would be appropriate would be very much appreciated :)!!