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Any suggestions on how to integrate this beast?:

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

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
    Did this integral come from a physics problem? What's the current status? Is it solved?2018-03-15

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