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Re: ArcCos[x] with x > 1


jaegerm at ibmt.fhg.de wrote:
> Hi,
> 
> as the result of a definite integral (Integrate[r ArcCos[1/(2 r)] (1 +
> 4 r^2)^(1/2), {r, 1/2, 1/2^(1/2)}]) I received the expression
> i*ArcCos[2]. How does Mathematica calculate ArcCos[x] with x > 1 which
> is outside the function's domain [-1,1]?

By using Euler's Exp[I x] == Cos[x] + I Sin[x]

 > In Mathematica's
> opinion, ArcCos[2] is a purely imaginary number, so why isn't
> i*ArcCos[2] reduced to a real expession?

Because (I ArcCos[2]) is a perfectly good analytic expression, and 
Mathematica has no built-in preference for the representation you want.

> How do I get an analytical
> expression for the imaginary part of ArcCos[2]?

ComplexExpand[I ArcCos[2]] will convert your result to the 
representation you prefer.

-jpd



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