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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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