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Re: Integrate 5.0

The desired result, though, is not valid for all x.  When x=Pi,   2*Sqrt[1 + Cos[x]]*Tan[x/2]  equals 0.  However, the integral shouldn't give 0 since the original function is always positive.  In the integral command, try the assumtions:

Assumptions -> 0<x< Pi     or
Assumptions -> Pi<x<2 Pi

These may actually give more useful answers.

>>> Selwyn Hollis <sh2.7183 at> 10/29 12:34 AM >>>
I've come to the conclusion that Integrate has become nearly worthless 
for computing definite integrals with symbolic limits. To cite a simple 

	Integrate[Sqrt[Cos[t] + 1], {t, 0, x}]

returns an awful mess inside of an If statement (very mild in this 
case) that no one should have to deal with if they're only concerned 
with real numbers (specifically calculus students and a great many 
applied mathematicians).

On the other hand, DSolve gives the simple, clean answer that Integrate 
used to give:

    y[t]/. DSolve[{y'[t] == Sqrt[Cos[t] + 1], y[0] == 0}, y[t], t]

	   2*Sqrt[1 + Cos[t]]*Tan[t/2]

Could it be that we need a new function such as this:

		(y[x]/. First@DSolve[{y'[x] ==expr, y[a] == 0}, y[t], t])/.x->b

that would be associated with \[Integral] ? ... leaving the current 
Integrate to be associated with \[ContourIntegral]??

Or perhaps a simple option for Integrate like RealLimits->True?

Selwyn Hollis 

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