Maple answer:
integrand := sin(x)*%e^(1/2*log(sin(x)^2-1))/cos(x);
integral := integrate(integrand, x);
integral := -exp(1/2*ln(4*tan(1/2*x)^2/(1+tan(1/2*x)^2)^2-1)*ln(%e))/ln(%e)
R.Rogers
> -----Original Message-----
> From: Raymond Toy [mailto:raymond.toy at ericsson.com]
> Sent: Friday, May 19, 2006 1:37 PM
> To: Billinghurst, David (CALCRTS)
> Cc: Maxima at math.utexas.edu
> Subject: Re: [Maxima] Another integration issue
>
>
> >>>>> "David" == David Billinghurst <Billinghurst> writes:
>
> David> Another integration issue from the ODE testsuite.
> David> With 5.9.3 (and before) I have
>
> Aargh!
>
> David> (%i1) display2d:false;
> David> (%o1) false
> David> (%i2) integrand:sin(x)*%e^(log(sin(x)^2-1)/2)/cos(x);
> David> (%o2) sin(x)*%e^(log(sin(x)^2-1)/2)/cos(x)
> David> (%i3) integral:integrate(integrand,x);
> David> (%o3) -sqrt(sin(x)^2-1)
>
> I can't look deeply into this right now, but this integral seems not
> right.
>
> If I take the rectform of the integrand and assume x is real so
> sin(x)^2-1 is negative, the result is
>
> %i*sin(x)/cos(x)*exp(log(1-sin(x)^2)/2)
>
> For small x, x > 0, the imaginary part is positive, so the integral
> should have a positive imaginary part too.
>
> The integral is -sqrt(sin(x)^2-1), which for small x > 0, has an
> imaginary part that is negative.
>
> At least the integral from maxima CVS has a positive imaginary part
> for small x > 0.
>
> Ray
>
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