Need help evaluating flux integral!

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  • #1
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I have constructed a formula to represent the neutron flux from a disc emitter through an aperture. I have got it down to a double integral of a form that I can't see how to evaluate. Mathematica crashes on this and I am not ready yet to give up and go to numerical solution. Does anyone know a good technique for evaluating integrals of this type?

Int dTheta*Sin(2*Theta)*Sqrt[A^2-(r-D*Tan(Theta))^2]

It is a product of a trig function and then a sqrt with a trig function inside. Integration by parts doesn't improve the radical and I'm looking for analytic solution if possible...any ideas?
 

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  • #2
fzero
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Trig identities and operations like completing the square inside the square root will put that in the form

$$\int d\theta \sin\theta \sqrt{\beta \cos^2\theta -1},$$

which can be done in closed form by a simple substitution.
 
  • #3
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Trig identities and operations like completing the square inside the square root will put that in the form

$$\int d\theta \sin\theta \sqrt{\beta \cos^2\theta -1},$$

which can be done in closed form by a simple substitution.

The integrand is of the form

$$\int d\theta \sin2\theta \sqrt{\alpha^2 - (r-\beta\tan\theta)^2}$$

Completing the square will simplify the radical, but the radicand becomes negative in this case:

$$\int d\theta \sin2\theta \sqrt{-\gamma-u^2}$$ where $$u=\tan\theta-r/\beta$$ and $$\gamma = (\alpha^2-2r^2)/\beta^2$$

Our integrating variable $$du/d\theta$$ is then equal to $$1+\tan\theta^2 = 1/\cos^2\theta$$

After rewriting the differential coefficient in terms of u, the integral can then be rewritten

$$2 \int \frac{du (u+\frac{r}{\beta})}{(1+(u+\frac{r}{\beta})^2)^2} \sqrt{-\gamma-u^2}$$

Mathematica returns a huge mess to this. I think the fact that the u^2 is negative inside the radical makes this a different form from what you might have been thinking. I am not sure this form is easier to evaluate...
 
  • #4
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I think I may be able to construct a different integrand if I parameterize the problem differently...will check back later.
 

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