Question: A problem with integration

I do a calculation o a simple integral


      int(cosh(cos(2*t)), t = 0 .. 2*Pi)


and get the answer 0 (zero), which I know is incorrect. The cosh function of a real argument is always positive.


If I represent the cosh function as a sum of two exponentials

(1/2)*(int(cosh(cos(2*t)), t = 0 .. 2*Pi) + int(cosh(-cos(2*t)), t = 0 .. 2*Pi))


I get    

2*Pi*BesselI(0, 1)

which looks much better.


What's the matter? Why Maple yields two different results? Victor.



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