1. **State the problem:** 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. **Analyze the integral:** The integral is over $\theta$ from $0$ to $\frac{\pi}{2}$ and $x$ from $0$ to $\infty$. The integrand involves $x$, $\sin \theta$, $\cos \theta$, a square root, and a logarithm of a ratio. 3. **Consider symmetry and substitution:** Note that the logarithm argument is a ratio of terms symmetric in $\sin^2 \theta$ and $\cos^2 \theta$. This suggests exploring the substitution $\theta \to \frac{\pi}{2} - \theta$ to check if the integral over $\theta$ simplifies. 4. **Check the integrand under $\theta \to \frac{\pi}{2} - \theta$:** Replace $\theta$ by $\frac{\pi}{2} - \theta$: - $\sin \theta \to \cos \theta$ - $\cos \theta \to \sin \theta$ The integrand becomes $$\frac{x \cos \theta}{\sqrt{1 + x^2 \cos^2 \theta}} \ln \left(\frac{1 + x^2 \sin^2 \theta}{1 + x^2 \cos^2 \theta}\right)$$ which is the negative of the original integrand because the logarithm argument is inverted. 5. **Add the integral and its transformed version:** Let $$I = \int_0^{\frac{\pi}{2}} \frac{x \sin \theta}{\sqrt{1 + x^2 \sin^2 \theta}} \ln \left(\frac{1 + x^2 \cos^2 \theta}{1 + x^2 \sin^2 \theta}\right) d\theta$$ and $$I' = \int_0^{\frac{\pi}{2}} \frac{x \cos \theta}{\sqrt{1 + x^2 \cos^2 \theta}} \ln \left(\frac{1 + x^2 \sin^2 \theta}{1 + x^2 \cos^2 \theta}\right) d\theta$$ Since $I' = -I$, adding gives $$I + I' = 0$$ But $I + I'$ is the integral of the sum of the two integrands, which is zero. 6. **Therefore, the integral over $\theta$ is zero for each fixed $x$.** 7. **Conclude the value of the double integral:** Since the inner integral over $\theta$ is zero for all $x$, the entire double integral is $$\int_0^\infty 0 \, dx = 0$$ **Final answer:** $$\boxed{0}$$