From: Albert_Rich@msn.com   
      
   On Wednesday, December 21, 2016 at 5:13:50 PM UTC-10, Nasser M. Abbasi wrote:   
   > On 11/22/2016 4:27 PM, clicliclic@freenet.de wrote:   
   >   
   > > 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.   
   > >   
   >   
   > I also tried to see how LN(TAN((2*t+pi)/4)) same about, but   
   > gave up after few tries. But I just saw this is also given   
   > in Schaum's outlines, "mathematical handbook of formulas   
   > and tables", second edition, page 65. Here is screen shot of the page   
   > fyi.  #16.15 on the page   
   >   
   > http://12000.org/tmp/122116/sec.jpg   
   >   
   > If someone knows how this was derived, it will interesting   
   > to know.   
   >   
   > --Nasser   
      
   It's easy.  By default Derive assumes variables real.  Therefore 1-sin(t) is   
   nonnegative, log(x)-log(y) equals log(x/y) if y is nonnegative, and   
   cos(z)/(1+sin(z) equals tan(z/2+pi/4).   
      
   Albert   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)