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Scholar
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posted by  jackma12 on 8/24/2009 8:09:40 PM  |  status: Closed  |  Earned Karma: 352

integral

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Tags: Calculus
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Sage
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posted by Solvemath on 8/24/2009 8:24:42 PM  |  status: Live
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A
 
 
Let
Then:
  
 
 
 
Oracle
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posted by Kingle on 8/24/2009 8:36:07 PM  |  status: Live
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   we know that   so we have
 


let



  

 

 

 

Oracle
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posted by Altair on 8/24/2009 8:41:48 PM  |  status: Live
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consider:
(secx)' = secx*tanx,
sec2x-1 = (1/cos2x)-1=(1-cos2x)/cos2x = sin2x/sin2x = tan2x
and
sec3x*tan5x = (secx*tanx)(sec2x*tan4x) = (secx*tanx)(sec2x*(1-sec2x)2=
=secx*tanx*(sec2x*(1-2sex2x+sec4x))=secx*tanx*(sec2x-2sec4x+sec6x)
we have:

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