"David W. Cantrell" wrote in message ...
Dono wrote:
[snip]
OK, there is some hope: the integral is really a definite integral ,
from 0 to 2pi.
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.

I suspect you made a mistake. Consider the following numerical integration.

In[7]:= a = 1/2; NIntegrate[Sin[x]*
Sqrt[1-a*(Cos[x])^2((1+Sin[x])^2/(1+a*(Sin[x])^2)+(1- a)*(Cos[x])^2)],
{x, 0, 2 Pi}]

Out[7]= -0.444846

David

Yep, same with Maple.
When I plot the function, it is clearly not symmetric.

Dirk Vdm

On Apr 30, 7:55 am, David W. Cantrell wrote:
Dono wrote:

[snip]

OK, there is some hope: the integral is really a definite integral ,
from 0 to 2pi.
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.

I suspect you made a mistake. Consider the following numerical integration.

In[7]:= a = 1/2; NIntegrate[Sin[x]*
Sqrt[1-a*(Cos[x])^2((1+Sin[x])^2/(1+a*(Sin[x])^2)+(1- a)*(Cos[x])^2)],
{x, 0, 2 Pi}]

Out[7]= -0.444846

David



Arrrg, there is a term in (1+sinx)/(1+a*sinx) that changes to (1-siny)/
(1-a*siny) :-)
I missed it, dang. Thank you.

"Dono" wrote in message ...
On Apr 30, 7:55 am, David W. Cantrell wrote:
Dono wrote:

[snip]

OK, there is some hope: the integral is really a definite integral ,
from 0 to 2pi.
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.

I suspect you made a mistake. Consider the following numerical integration.

In[7]:= a = 1/2; NIntegrate[Sin[x]*
Sqrt[1-a*(Cos[x])^2((1+Sin[x])^2/(1+a*(Sin[x])^2)+(1- a)*(Cos[x])^2)],
{x, 0, 2 Pi}]

Out[7]= -0.444846

David



Arrrg, there is a term in (1+sinx)/(1+a*sinx) that changes to (1-siny)/
(1-a*siny) :-)
I missed it, dang. Thank you.

for a from -0.5 by 0.05 to 0.5 do
print( a, evalf( Int( sin(x)*sqrt(1-a*(cos(x))^2*((1+sin(x))^2/(1+a*sin(x)^2)+(1-a)*cos(x)^2)), x=0.0..2*Pi ) ) );
end do;

-.50, 0.4313135266
-.45, 0.3807384152
-.40, 0.3328483124
-.35, 0.2871377155
-.30, 0.2431931891
-.25, 0.2006682901
-.20, 0.1592657918
-.15, 0.1187245853
-.10, 0.07880959525
-.05, 0.03930358362
0., -1.702775486 10^(-16)
0.05, -0.03930385591
0.10, -0.07881414095
0.15, -.1187492837
0.20, -.1593519644
0.25, -.2009072950
0.30, -.2437734550
0.35, -.2884393677
0.40, -.3356491928
0.45, -.3867305258
0.50, -.4448463419

On Apr 30, 5:55 am, David W. Cantrell wrote:
"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".

Maple 11 (in the full, Java-interface version) interprets the
expression without the '*', at least if one leaves a space between the
"^2" and what follows. Maple 9.5 does not; I don't know about Maple
10.x because I lack access to it.

R.G. Vickson



David

"Ray Vickson" wrote in message ...
On Apr 30, 5:55 am, David W. Cantrell wrote:
"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".

Maple 11 (in the full, Java-interface version) interprets the
expression without the '*', at least if one leaves a space between the
"^2" and what follows. Maple 9.5 does not; I don't know about Maple
10.x because I lack access to it.

R.G. Vickson

Ah, I should have mentioned I made them check on 9.03.
It looks like they have to buy an upgrade :-)

Dirk Vdm

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.

In article ,
"Dirk Van de moortel"
wrote:

"David W. Cantrell" wrote in message ...
"Dirk Van de moortel" wrote:

[...]

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 :-)

Vlad will be cheesed he did not find it.

--
Michael Press