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Re: help with integration

  • To: mathgroup at smc.vnet.net
  • Subject: [mg121652] Re: help with integration
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Sat, 24 Sep 2011 22:32:13 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • Reply-to: hanlonr at cox.net

Integrate[
  r^2*ArcCos[x/r] - x*Sqrt[r^2 - x^2], {x, Sqrt[3] y, 
   Sqrt[r^2 - y^2]}, Assumptions -> {y > 0, r > 2 y}] // Simplify

y^3/3 + (2/3)*r^2*Sqrt[r^2 - 3*y^2] + y^2*Sqrt[r^2 - 3*y^2] - 
   r^2*y*(1 + Sqrt[3]*ArcCos[(Sqrt[3]*y)/r]) + r^2*Sqrt[r^2 - y^2]*
     ArcSec[r/Sqrt[r^2 - y^2]]

Assuming[{y > 0, r > 2 y}, 
 Integrate[
   r^2*ArcCos[x/r] - x*Sqrt[r^2 - x^2], {x, Sqrt[3] y, 
    Sqrt[r^2 - y^2]}] // Simplify]

y^3/3 + (2/3)*r^2*Sqrt[r^2 - 3*y^2] + y^2*Sqrt[r^2 - 3*y^2] - 
   r^2*y*(1 + Sqrt[3]*ArcCos[(Sqrt[3]*y)/r]) + r^2*Sqrt[r^2 - y^2]*
     ArcSec[r/Sqrt[r^2 - y^2]]

% === %%

True


Bob Hanlon

---- Salman Durrani <dsalman96 at yahoo.com> wrote: 

=============
I am trying to do the following integration:

Integrate[r^2*ArcCos[x/r] - x*Sqrt[r^2 - x^2], {x, Sqrt[3] y, Sqrt[r^2 - y^2]} ]

When I try, Mathematica comes back to be with a very long conditional expression.

How do I tell mathematica that r is positive, real in the above equation not complex.

Thanks

Kahless






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