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# Double Integration runs a long time

Posted 9 years ago
 Hi, I am trying to solve following expression int(int((1-exp(-5.5/cos(x)))*sin(x), x = 0 .. arctan(300*cos(y)+sqrt(12.25-90000*sin(y)^2))), y = 0 .. Pi)  But I am not getting results. Please help me!!! Thanks
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Posted 9 years ago
 An additional reply that was here should have been a new discussion. It was moved to
Posted 9 years ago
 Thanks Bruce. I'll make sure to do this in future posts.
Posted 9 years ago
 I think this is more a mathematical question, but anyway...In this case one is lucky, because the inner integral can be done analytically! What remains is then just a single integral which can be calculated numerically at least. Here is my result: In:= ClearAll["Global*"]; In:= fInt[x_] = Integrate[(1 - Exp[-5.5/Cos[x]]) Sin[x], x] Out= -1. Cos[x] + E^(-5.5 Sec[x]) Cos[x] + 5.5 ExpIntegralEi[-5.5 Sec[x]] In:= NIntegrate[ fInt[ArcTan[300 Cos[y] + Sqrt[12.25 - 90000 Sin[y]^2]]] - fInt, {y, 0, Pi}] Out= 3.13316 + 0. I In:= Chop[%] Out= 3.13316 `Cheers Henrik
Posted 9 years ago
 Thank you for including code and not just a picture of code.
Posted 9 years ago
 I tried what you tried... and did not get a result. The kernel just keeps running without evaluating. I then tried a numeric integration and got a result. I could be wrong here. Anyone else have any ideas? Attachments:
Posted 9 years ago
 Same thing is happening with me. But I have solved analytically the inner integral.Now the problem is with second integral, i.e. integration with respect to y, from 0 to pi. I got following result by first integral Posted 9 years ago
 I apologize. I missed a parenthesis in my code. I fixed it. I am getting the same result as Henrik. Here is it...I believe Henrik is correct. The inner integral is solvable analytically. But the outer integral might not be (or it's too large for Mathematica to calculate), hence using numerical integration on the outer integral. I'm leave it running for an hour or so just in case and get back to you if Mathematica is able to do the outer integral analytically. Attachments: