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Any suggestions on how to integrate this beast?:

$$\int_0^{\omega_t}\int_{\omega_t}^f\sin^2(\omega_{12}/2)\sin^2(\omega_{23}/2)d\omega_{23}d\omega_{12}$$

where:

$f{} = 2\pi+2\tan^{-1}(y,x)$

$y = -A^2\sin^2(\omega_{12}/2)\cos(\omega_t/2)-r\cos(\omega_{12}/2)$

$x = A\sin(\omega_{12}/2)[\cos(\omega_t/2)\cos(\omega_{12}/2)-r]$

$r = \sqrt{A^2\sin^2(\omega_{12}/2)[A^2\sin^2(\omega_{12}/2)+\cos^2(\omega_{12}/2)-\cos^2(\omega_t/2)]}$

I can perform the first integration fine, but when you evaluate it at f you get something nasty that I can't seem to integrate. Here is the result after the first integration:

$$\int_0^{\omega_t}\sin^2(\omega_{12}/2)\left[\left(\pi+\tan^{-1}(y,x)-\frac{xy}{x^2+y^2}\right)-\left(\frac{\omega_t}{2}-\sin(\omega_t/2)\cos(\omega_t/2)\right)\right]d\omega_{12}$$

Note: $\tan^{-1}(y,x)$ is the two argument inverse tangent function, a.k.a. atan2.

  • 1
    A standard technique for evaluating double integrals is to switch the order of integration. Does that get you anything nicer?2011-05-05
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    You may have use for the usual conversion $\arctan(y,x)=2\arctan\left(\frac{\sqrt{x^2+y^2}-x}{y}\right)$.2011-05-05
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    @okj, as a side remark, have you tried using mathematica? It is pretty good at solving these types of questions where you are seeking exact and analytic solutions.2011-05-06
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    @picakhu: not always. My experience with elliptic integrals has shown me that just taking what *Mathematica* spits out isn't enough...2011-05-06
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    The second portion at least, you can integrate nicely. Note that $\sin^2\frac{\omega_{12}}{2}=\frac12\left(1-\cos\omega_{12}\right)$ and $\sin\frac{\omega_t}{2}\cos\frac{\omega_t}{2}=\frac12\sin\omega_t$.2011-05-06
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    @J.M.: yes, but I can't seem to integrate $\sin^2(\omega_{12}/2)\tan^{-1}(y,x)$ and I despair of any hope of integrating $\sin^2(\omega_{12}/2)\frac{xy}{x^2+y^2}$. I feel like a solution is possible though using something related to elliptic integrals, or the like, however, I'm very unfamiliar with these concepts. Any other thoughts?2011-05-06
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    Well, the way your $x$ and $y$ are formed makes it *nasty*... for now I've no insight. I'll get back to you when I think of something.2011-05-06
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    @okj where did you get this integral?2013-03-27
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    Did this integral come from a physics problem? What's the current status? Is it solved?2018-03-15

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