1. **Stating the problem:** We want to evaluate the double integral $$\int_0^\infty \int_0^{\frac{\pi}{2}} \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln \left( \frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta} \right) d\theta \, dx$$ 2. **Observing the integral:** The integral is over $x$ from 0 to $\infty$ and $\theta$ from 0 to $\frac{\pi}{2}$. The integrand involves $x$, $\sin \theta$, $\cos \theta$, and logarithms of rational expressions in $x^2$. 3. **Key insight:** The integrand is symmetric in $\sin^2 \theta$ and $\cos^2 \theta$ inside the logarithm, but with a minus sign due to the division inside the log. 4. **Interchange the order of integration or consider substitution:** We can attempt to evaluate the inner integral with respect to $\theta$ first. 5. **Simplify the logarithm:** $$\ln \left( \frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta} \right) = \ln(1 + x^2 \cos^2 \theta) - \ln(1 + x^2 \sin^2 \theta)$$ 6. **Split the integral accordingly:** $$\int_0^{\frac{\pi}{2}} \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln(1 + x^2 \cos^2 \theta) d\theta - \int_0^{\frac{\pi}{2}} \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln(1 + x^2 \sin^2 \theta) d\theta$$ 7. **Use substitution $\theta \to \frac{\pi}{2} - \theta$ in the first integral:** Since $\sin(\frac{\pi}{2} - \theta) = \cos \theta$ and $\cos(\frac{\pi}{2} - \theta) = \sin \theta$, the first integral becomes $$\int_0^{\frac{\pi}{2}} \frac{x \cos \theta}{\sqrt{1 + x^2 \cos^2 \theta}} \ln(1 + x^2 \sin^2 \theta) d\theta$$ 8. **Rewrite the original integral as:** $$\int_0^{\frac{\pi}{2}} \ln(1 + x^2 \sin^2 \theta) \left( \frac{x \cos \theta}{\sqrt{1 + x^2 \cos^2 \theta}} - \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \right) d\theta$$ 9. **Note the symmetry and the integrand structure:** The expression inside the parentheses is antisymmetric with respect to swapping $\sin \theta$ and $\cos \theta$, and the limits are symmetric. 10. **Integral over $\theta$ evaluates to zero for each fixed $x$:** Because the integrand is antisymmetric over $[0, \frac{\pi}{2}]$ with respect to $\theta \to \frac{\pi}{2} - \theta$, the integral over $\theta$ is zero. 11. **Therefore, the entire double integral is zero:** $$\boxed{0}$$