I was trying to (re)calculate a problem of an older Wolfram blog post (Problem 11457, by M. L. Glasser) with Mathematica 9.0.0.0 (on OS X 10.8.2).
Assuming[0 < a < b, Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], {x, a, b}]]
Instead of the expected solution, it just returns the integral unevaluated. Is this a regression?
More details: As pointed out in the commentes, the indefinite integral
Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], x]
still gives the same result in Mathematica 8 and 9.
The next two each returned ConditionalExpression in Mathematica 8 but return unevaluated in Mathematica 9:
Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], {x, a, b}]
Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], {x, a, b}, Assumptions -> 0 <= a <= b]
The actual problem
Integrate[ArcCos[x/Sqrt[(a + b) x - a b]], {x, a, b}, Assumptions -> 0 < a < b]
computes correctly to ((a - b)^2 \[Pi])/(4 (a + b)) in Mathematica 8 but still returns unevaulated in Mathematica 9.
((a - b)^2 \[Pi])/(4 (a + b))– Ajasja Jan 24 at 12:16ver.9while they are inver.8mathematica.stackexchange.com/questions/18327/… – Artes Jan 24 at 17:18