hi im solving some electrodynamics problem but im troubled with integration.
i don`t know how to integrate this integrand analytically.
what i want to integrate is this(posted picture)
picture
this equation is originated from definition of solid angle; d(omega)=da/r^2 i chose primed notation for the source point(on the circular loop which has radius a) and not primed notation for the observation point(any point except on the circular loop). radius of loop is a so i integrate r' from 0 to a and because of circular loop, azimutahl angle is o to 2pi
thank you for reading my question
The integral in the picture was $$\begin{align}I&=\int_0^{2\pi}\int_0^a\frac{r^{\prime}}{\left(r^{\prime}\cos\phi^{\prime}-r\sin\theta\cos\phi\right)^2+\left(r^{\prime}\sin\phi^{\prime}-r\sin\theta\sin\phi\right)^2+r^2\cos^2\theta}dr^{\prime}\,d\phi^{\prime}\\ &=\int_0^{2\pi}\int_0^a\frac{r^{\prime}}{{r^{\prime}}^2-2rr^{\prime}\sin\theta\cos\left(\phi^{\prime}-\phi\right)+r^2}dr^{\prime}\,d\phi^{\prime}\\ &=\int_0^a\frac{r^{\prime}}{{r^{\prime}}^2+r^2}\int_0^{2\pi}\frac{d\phi^{\prime}}{1-\frac{2rr^{\prime}\sin\theta}{{r^{\prime}}^2+r^2}\cos\left(\phi^{\prime}-\phi\right)}dr^{\prime}\end{align}$$ To simplify the azimuthal integral we let $e=\frac{2rr^{\prime}\sin\theta}{{r^{\prime}}^2+r^2}$ and $\phi^{\prime}-\phi=\nu+\pi$ and then we can use the eccentric anomaly $$\sin E=\frac{\sqrt{1-e^2}\sin\nu}{1+e\cos\nu},\,\cos E=\frac{\cos\nu+e}{1+e\cos\nu},\,dE=\frac{\sqrt{1-e^2}d\nu}{1+e\cos\nu}$$ So we get $$\begin{align}\int_0^{2\pi}\frac{d\phi^{\prime}}{1-\frac{2rr^{\prime}\sin\theta}{{r^{\prime}}^2+r^2}\cos\left(\phi^{\prime}-\phi\right)}&=\int_{-\pi-\phi}^{\pi-\phi}\frac{d\nu}{1+e\cos\nu}=\int_{\nu=-\pi-\phi}^{\nu=\pi-\phi}\frac{dE}{\sqrt{1-e^2}}\\ &=\frac{2\pi}{\sqrt{1-e^2}}=\frac{2\pi}{\sqrt{1-\frac{4r^2{r^{\prime}}^2\sin^2\theta}{\left({r^{\prime}}^2+r^2\right)^2}}}\end{align}$$ And now we are back to $$\begin{align}I&=2\pi\int_0^a\frac{r^{\prime}dr^{\prime}}{\sqrt{{r^{\prime}}^4+2r^2{r^{\prime}}^2+r^4-4r^2{r^{\prime}}^2\sin^2\theta}}\\ &=2\pi\int_0^a\frac{r^{\prime}dr^{\prime}}{\sqrt{\left({r^{\prime}}^2+r^2-2r^2\sin^2\theta\right)^2+4r^4\sin^2\theta\cos^2\theta}}\\ &=2\pi\int_0^a\frac{r^{\prime}dr^{\prime}}{\sqrt{\left({r^{\prime}}^2+r^2\cos2\theta\right)^2+r^4\sin^22\theta}}\\ &=\pi\int_{\sinh^{-1}\frac{r^2\cos2\theta}{r^2\sin2\theta}}^{\sinh^{-1}\frac{a^2+r^2\cos2\theta}{r^2\sin2\theta}}\frac{r^2\sin2\theta\cosh u\,du}{r^2\sin2\theta\cosh u}\\ &=\pi\sinh^{-1}\frac{a^2+r^2\cos2\theta}{r^2\sin2\theta}-\pi\sinh^{-1}\frac{r^2\cos2\theta}{r^2\sin2\theta}\\ &=\pi\ln\left(\frac{a^2+r^2\cos2\theta+\sqrt{(a^2+r^2\cos2\theta)^2+r^4\sin^22\theta}}{r^2\cos2\theta+\sqrt{r^4\cos^22\theta+r^4\sin^22\theta}}\right)\\ &=\pi\ln\left(\frac{a^2+r^2\cos2\theta+\sqrt{a^4+2a^2r^2\cos2\theta+r^4}}{r^2+r^2\cos2\theta}\right)\end{align}$$ Where we have made the substitution $${r^{\prime}}^2+r^2\cos2\theta=r^2\sin2\theta\sinh u$$ EDIT: Had some serious errors in my first version; hope this one is better.