On Apr 30, 6:26*am, Dono wrote:
On Apr 30, 7:05 am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-



SperM.hotmail.com wrote:
"Dono" wrote in ...
On Apr 30, 6:12 am, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
"David W. Cantrell" wrote in .. .

"Dirk Van de moortel" wrote:
"David W. Cantrell" wrote in message
...
Eric Gisse wrote:
On Apr 29, 8:42=A0pm, David W. Cantrell
wrote:
Dono wrote:
On Apr 29, 6:21 pm, Eric Gisse wrote:
On Apr 29, 4:02 pm, Dono wrote:

Sorry, wrong integrand, here is the correct one:

sin[x]*sqrt(1-a*(cos[x])^2((1+sin[x])^2/(1+a*(sin[x])^2)+(1-
a)*(cos[x])^2))

http://img291.imageshack.us/img291/8664/integralvi2.jpg

Excellent! Thank you , Eric!

Your jubilation is, I think, premature. If you differentiate the
supposed antiderivative shown there, you get just

sin(x) sqrt(1 - a cos(x)^2)

which is not equal to the given integrand.

Yes, it is. If you have Maple, Mathematica, or MATLAB, have one of
them [preferably Maple] apply the relevant simplify command to the
expression.

You seem to be saying that the given integrand simplifies to

sin(x) sqrt(1 - a cos(x)^2)

but that is easily shown to be false. For example, if we take a = 1 and
x = pi/4, the expression *sin(x) sqrt(1 - a cos(x)^2) *has the value
1/2, while the original integrand has a value which is approximately
0.119573 instead.

Concerning Dono's question "So, what is the correct answer?":
Perhaps it is not possible to express an antiderivative in closed form
using standard functions.

David W. Cantrell

Indeed, it is wrong.
Maple makes an error interpreting
* sin(x)*sqrt(1-a*(cos(x))^2((1+sin(x))^2/(1+a*sin(x)^2)+(1-a)*cos(x)^2))

There is a multiplication operator missing after the first "^2"

It should be
*sin(x)*sqrt(1-a*(cos(x))^2*((1+sin(x))^2/(1+a*sin(x)^2)+(1-a)*cos(x)^2))

Try entering the bare expression without the "*" and you see
that Maple drops everything beyond "^2".

Dirk Vdm

Wow! I'm startled. I have never encountered a CAS which requires that
multiplication be _explicitly_ indicated in such a case. I would have
supposed that multiplication could be implied by juxtaposition.

Surely Maple's dropping everything after the first "^2" is a bug,
not a "feature".

Vladimir will be thrilled :-)

Dirk Vdm

Great,

You found a bug in Maple. Question is, if you put in the "*" by hand ,
what is the answer? Does Maple return the same (incorrect) result?

Nope... no result.

Dirk Vdm


Drat, I simply assumed that Maple made a simplification that I didn't
see. It does that quite a lot.

Taking the new expression, Maple 10 times out as well.

This is why Mahematica timed out on me.
OK, there is some hope: the integral is really a definite integral ,
from 0 to 2pi.

Oooh. There might be a way to analytically do this. Have you tried
contour integration?


I transformed the domain into -pi/2 to +3pi/2
Then I separate it into two integrals, one from -pi/2 to +pi/2 and the
second one from pi/2 to 3pi/2.
In the second one, I made the variable change y=x-pi, thus shifting
the integration interval from -p/2 to pi/2. All terms in cos^2(x)
become cos^2(y+pi), so they do not change sign or value. The lone term
in sin(x) becomes sin(y+pi), so it changes sign. The two integrals
cancel each other, so , the result for the definite integral is ....0.
Unless I made a mistake, I am done.