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I am trying to solve the below mentioned integral. Is it possible to solve this with mathematica. If not, can I get an approximation for this by introducting the variable not to take certain values ?

enter image description here

Integrate[
 4 x/π^2 ArcCos[(y^2 + 3)/(4 y)] ArcCos[(x^2 + y^2 - 
      1)/(2 x y)], {x, 0, 1}, {y, 1, 2}]
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  • $\begingroup$ Some spaces would do a lot to clarify your expression. Even better would be Mathematica code - have you tried anything yet? $\endgroup$ – Yves Klett Sep 22 '14 at 7:59
  • $\begingroup$ I have edited the post with a newer image. I do not have the software installed yet but I am looking to find out whether this solvable or not with mathmetica. I want to have value/solution to put in an expression $\endgroup$ – Waqas Sep 22 '14 at 8:03
  • $\begingroup$ Better than an image would be TeX code. That would at least eliminate some work for us. $\endgroup$ – Yves Klett Sep 22 '14 at 8:05
  • $\begingroup$ I am sorry for not providing the code since I am in the process of installing and getting to use mathemtica only solve this integral problem. If I can have an information about the solvability, it would be great. I tried the online integral calculator but it says it does not have a solution. $\endgroup$ – Waqas Sep 22 '14 at 8:09
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    $\begingroup$ You can try the free Wolfram cloud, it has Mathematica there for free wolfram.com/programming-cloud/pricing click on the free option. Also can try Wolfram alpha, it is supposed to be able to do integration as well. $\endgroup$ – Nasser Sep 22 '14 at 8:13
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Judging from the shape,

Plot3D[4 (x y)/\[Pi]^2 ArcCos[(y^2 + 3)/(4 y)] ArcCos[(x^2 + y^2 - 
      1)/(2 x y)], {x, 0, 1}, {y, 1, 2}]

Mathematica graphics

and from

Reduce[
     {
      4 (x y)/\[Pi]^2 ArcCos[(y^2 + 3)/(4 y)] ArcCos[(x^2 + y^2 - 1)/(2 x y)] == 0
      , 0 <= x <= 1
      , 1 <= y <= 2
      }, {x, y}
     ]

0 < x <= 1 && (y == 1 || y == 1 + x)

the expression should be integrable in the range {x, 0, 1}, {y, 1, 1 + x}

NIntegrate[
 4 (x y)/\[Pi]^2 ArcCos[(y^2 + 3)/(4 y)] ArcCos[(x^2 + y^2 - 
      1)/(2 x y)], {x, 0, 1}, {y, 1, 1 + x}, WorkingPrecision -> 20]

0.053240519638085381131

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    $\begingroup$ I think you're missing a y in the inner integral. $\endgroup$ – Netsie Sep 22 '14 at 9:31
  • $\begingroup$ @Netsie, you are correct, I have corrected that now. $\endgroup$ – rhermans Sep 22 '14 at 9:32
  • $\begingroup$ Thank you so much for the solution. I have actually observed that the actual integral equation has always the 1+x upper range for y instead of the value which I had earlier shared. I want to ask if the upper limit of the y integral is x+1, then this integral is integrable over the entire range ? $\endgroup$ – Waqas Sep 23 '14 at 0:27
  • $\begingroup$ @Waqas, The expression is real and well behaved in {x, 0, 1}, {y, 1, 1 + x} , and complex for y>1+x. Does this answer your question? $\endgroup$ – rhermans Sep 23 '14 at 6:12
  • $\begingroup$ Yes it surely does and I am pretty relieved.As I am not that efficient in mathematical language, can you elaborate a bit more what does well behaved mean ? $\endgroup$ – Waqas Sep 23 '14 at 10:56

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