
Re: first anniversary of the IITS
Posted:
Mar 4, 2014 4:39 AM


On Sunday, March 2, 2014 12:29:28 AM UTC10, 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(1SQRT(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[(1Tan[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

