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How can I evaluate this? Is there a trick or simplification that will make it nicer?

$$ \frac{1}{4\pi} \frac{\partial}{\partial t}\left(2t\int_0^{2\pi}\int_0^1{(x+tr\cos\theta)^2 (y+tr\sin\theta)\over\sqrt{1-r^2}}\,r dr\,d\theta\right)$$

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Integrate first with respect to $\theta$. We then get $$\int_0^{2 \pi }(x+tr \cos\theta)^2 (y+tr \sin\theta) d \theta = \pi y(2x^2+t^2r^2)$$ making use of the fact that $\displaystyle \int_0^{2 \pi }(x+tr \cos\theta)^2 \sin\theta d \theta = 0$, $\displaystyle \int_0^{2 \pi} \cos(\theta) d \theta = 0$ and $\displaystyle \int_0^{2 \pi} \cos^2(\theta) d \theta = \pi$. Now the rest should be easy.

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