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   sci.math.symbolic      Symbolic algebra discussion      10,432 messages   

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   Message 9,215 of 10,432   
   clicliclic@freenet.de to Nasser M. Abbasi   
   Re: integration of sec(t)   
   22 Nov 16 23:27:12   
   
   "Nasser M. Abbasi" schrieb:   
   >   
   > It is interesting how different CAS outputs differ sometimes.   
   > I was trying to verify book result for int(sec(t),t) and this   
   > is what some CAS systems gave   
   >   
   > Mathematica:     ln(cos(t/2)+sin(t/2))-ln(cos(t/2)-sin(t/2))   
   > Rubi:            arctanh(sin(t))   
   > Fricas:          1/2 ( ln(1+sin(t)) - ln(1-sin(t)) )   
   > Maple & Maxima:  ln(sec(t)+tan(t))   
   >   
   > Text Book happens to give result shown by Maple&Maxima.   
   >   
   > Ofcourse all other results are also correct, I am sure,   
   > and these can all be converted to each others.   
   >   
   > Was Wondering if there is any preference, math-wise, to any one   
   > of the above results. From the Mathematica result, it seems   
   > to hint that it used the Tangent half-angle substitution,   
   > also called Weierstrass substitution.   
   >   
   > What does you CAS give to int(sec(t),t) if different from   
   > the above?   
   >   
      
   Derive 6.10 evaluates INT(SEC(t), t) to LN(TAN((2*t + pi)/4)). Stepwise   
   evaluation gives:   
      
   INT(SEC(t),t)   
      
   " SEC(z) -> 1/COS(z) "   
      
   INT(1/COS(t),t)   
      
   " INT(1/COS(a*x+b),x) -> LN(COS(a*x+b))/a-LN(1-SIN(a*x+b))/a "   
      
   LN(COS(t))-LN(-SIN(t)+1)   
      
   " one final step "   
      
   LN(TAN((2*t+pi)/4))   
      
   The mechanics of the final step is a mystery to me.   
      
   Martin.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)   

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