Integrate[Sqrt[(1 - Cos[t])/(Cos[a] - Cos[t])], {t, a, Pi}, Assumptions -> 0 < a < Pi] gives a Complex Expression rather than Pi which is what I expect.

The integral arises in the Tautochrone problem, i.e. this is the integral above the line "(2):" on the page https://proofwiki.org/wiki/Cycloid_has_Tautochrone_Property

I assume the expression is Complex because Mathematica assumes the expression inside the Sqrt is negative at some points but I don't think that is true.

I understand that the integrand has a singularity at t = a making the denominator 0 but I was hoping Mathematica could do the Integration. It doesn't complain about the singularity.

I am using Mathematica 12.1


2 Answers 2


Assumptions are not always applied as constraints. (Took a bit extra massaging to get the result of Integrate into its fully simplified form under the assumptions.)

res = Integrate[Sqrt[(1 - Cos[t])/(Cos[a] - Cos[t])],
  {t, a, Pi}, 
  Assumptions -> 0 < a < Pi]
-I (-1)^Floor[(π + Arg[-1 + Cos[a]])/(
  2 π)] (Log[2] + 2 Log[Cos[a/2]] - Log[-1 - Cos[a]])

Assuming[0 < a < Pi,
  Method -> {"ConditionSimplifier" -> 
    (Reduce[# && $Assumptions, ##2] &)}]

(*  Pi  *)
  • 2
    $\begingroup$ +1 Add the redundant assumption that Cos[a] < 1, i.e., Assuming[0 < a < Pi && Cos[a] < 1, Integrate[Sqrt[(1 - Cos[t])/(Cos[a] - Cos[t])], {t, a, Pi}] // FullSimplify] $\endgroup$
    – Bob Hanlon
    Sep 20, 2022 at 13:18
  • $\begingroup$ @BobHanlon Thanks. I thought of that (Cos[a] < 1), which appears in the output of FullSimplify[]; but I wanted to see if I could get Mathematica to deduce it. I don't know why it took such an effort to get it to do that. Simplify[] or FullSimplify[] (with the assumption 0 < a < Pi) ought to be enough, imo. $\endgroup$
    – Michael E2
    Sep 20, 2022 at 13:21
  • $\begingroup$ I agree, that is why I refer to it as a redundant assumption. $\endgroup$
    – Bob Hanlon
    Sep 20, 2022 at 13:23
  • $\begingroup$ @BobHanlon I noticed and understood "redundant". I was explaining why I didn't use Cos[a] < 1. I also realize it simplifies the code. Still, for whatever reason, I was stubbornly focused on taming Mma. $\endgroup$
    – Michael E2
    Sep 20, 2022 at 13:24
  • $\begingroup$ Wow, that was fast! Thanks Michael and Bob! This is a great site! $\endgroup$ Sep 20, 2022 at 15:01

$a$ needs to be between 0 and Pi/2 so that Cos[a] remain positive in order to simplify the sqrt which comes into play using the substitution made in the page.

Starting from

enter image description here

Then using Mathematica

Integrate[Sin[1/2*t]/Sqrt[Cos[1/2*a]^2 - Cos[1/2*t]^2], {t, a, Pi}, 
 Assumptions -> 0 < a < Pi/2]

Mathematica graphics

  • $\begingroup$ Thanks Naseer but I wanted a solution for my original integral as I want Mathematica to do as much work as possible. $\endgroup$ Sep 20, 2022 at 15:05

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