I input the following into the web-based Mathematica and it timed out
before giving an answer. Can anyone help finding out if there is a
symbolic answer:

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

Please copy and paste the formula in order to avoid introducing
errors. Thank you.

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))

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

Mathematica uses Sin[] and Cos[].

Tom Roberts


Dono wrote:
I input the following into the web-based Mathematica and it timed out
before giving an answer. Can anyone help finding out if there is a
symbolic answer:

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

Please copy and paste the formula in order to avoid introducing
errors. Thank you.

On Apr 29, 6:22 pm, Tom Roberts wrote:
Mathematica uses Sin[] and Cos[].

Tom Roberts

Dono wrote:
I input the following into the web-based Mathematica and it timed out
before giving an answer. Can anyone help finding out if there is a
symbolic answer:

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

Please copy and paste the formula in order to avoid introducing
errors. Thank you.

It recognized the lower case, this is not the problem.

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!

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.

David

On Apr 29, 9:42 pm, 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.

David



Hmm,

So, what is the correct answer?

On Apr 29, 8:42*pm, 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.


David

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