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[✓] Assume real parameters in an integral

Pascale Voirin, HEIA-FR
Posted 5 years ago

Hello, who knows if it's possible to define parameters as real numbers in an integral , like:

Integrate[ r (4a^2-r^2)^(1/2), {r,0,2a Cos[t]}]

Thanks a lot for your input!

And hope you feel good, despite Coronavirus,

Pascale
POSTED BY: Pascale Voirin
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Thanks a lot !
POSTED BY: Pascale Voirin

Henrik's answer is perfect, but if you want the answer without ConditionalExpression, you can do so, for example:

Simplify @@ 
 Integrate[r (4 a^2 - r^2)^(1/2), {r, 0, 2 a Cos[t]}, 
  Assumptions -> {a, t} \[Element] Reals]
POSTED BY: Claudio Chaib

Dear Claudio, I am afraid this way you are just replacing ConditionalExpression with Simplify instead of doing a simplification ...
POSTED BY: Henrik Schachner

Yes, you are right Henrik!! Thanks for showing me that.

Edited:

For now, the easiest way I found to have an answer without ConditionalExpression was:

k = Integrate[r (4 a^2 - r^2)^(1/2), {r, 0, 2 a Cos[t]}, 
   Assumptions -> {a, t} \[Element] Reals];

If[Length@k > 1, k[[1]], k]

First[] was promising but it didn't work out very well..

First@Integrate[r (4 a^2 - r^2)^2, {r, 0, 2 a Cos[t]}, 
  Assumptions -> {a, t} \[Element] Reals]

I wonder if there is a shorter/faster way..?!

POSTED BY: Claudio Chaib

Hi Pascale, you can make Assumptions, try:

Integrate[r (4 a^2 - r^2)^(1/2), {r, 0, 2 a Cos[t]}, Assumptions -> {a, t} \[Element] Reals]
POSTED BY: Henrik Schachner
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