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Topic: first anniversary of the IITS
Replies: 15   Last Post: Mar 26, 2014 4:42 PM

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Albert D. Rich

Posts: 214
From: Hawaii Island
Registered: 5/30/09
Re: first anniversary of the IITS
Posted: Mar 4, 2014 4:39 AM
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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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