From: Albert_Rich@msn.com   
      
   On Sunday, March 2, 2014 12:29:28 AM UTC-10, clicl...@freenet.de wrote:   
      
   > I have done some extra work and arrived at the following alternative   
   > evaluations for the Examples 62, 64, 66 (p. 268) and 118 (p. 309) from   
   > Chapter 5:   
   >   
   > INT(SQRT(TAN(x)),x)=-1/SQRT(2)*ATANH((1+TAN(x))/(SQRT(2)*SQRT(TA~   
   > N(x))))+1/SQRT(2)*ATAN(1+SQRT(2)*SQRT(TAN(x)))-1/SQRT(2)*ATAN(1-~   
   > SQRT(2)*SQRT(TAN(x)))=-1/SQRT(2)*LN((1+TAN(x)+SQRT(2)*SQRT(TAN(x~   
   > )))/SEC(x))+1/SQRT(2)*ATAN(1+SQRT(2)*SQRT(TAN(x)))-1/SQRT(2)*ATA~   
   > N(1-SQRT(2)*SQRT(TAN(x)))   
   >   
   > [...]   
      
   For example 62, you combined the two logs into an inverse hyperbolic tangent.   
   The two inverse tangents can also be combined to yield the elegant   
   antiderivative for sqrt(tan(x)):   
      
   -ArcTan[(1-Tan[x])/(Sqrt[2]*Sqrt[Tan[x]])]/Sqrt[2] -   
   ArcTanh[(1+Tan[x])/(Sqrt[2]*Sqrt[Tan[x]])]/Sqrt[2]   
      
   Did you not do this because of continuity problems?   
      
   Albert   
      
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