1. **State the problem:** We want to evaluate the double integral $$\int_0^\infty \int_0^{\frac{\pi}{2}} \frac{x \sin \theta \ln(1 + x^2 \cos^2 \theta)}{(1 + x^2 \sin^2 \theta)^{3/2}} \, d\theta \, dx.$$ 2. **Analyze 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 a logarithm. 3. **Consider changing the order of integration or substitution:** We can try to evaluate the inner integral with respect to $\theta$ first. 4. **Inner integral:** Define $$I(x) = \int_0^{\frac{\pi}{2}} \frac{x \sin \theta \ln(1 + x^2 \cos^2 \theta)}{(1 + x^2 \sin^2 \theta)^{3/2}} \, d\theta.$$ 5. **Substitution for inner integral:** Let $t = \sin \theta$, so when $\theta=0$, $t=0$ and when $\theta=\frac{\pi}{2}$, $t=1$. Then $d\theta = \frac{dt}{\cos \theta} = \frac{dt}{\sqrt{1 - t^2}}$. Rewrite the integral: $$I(x) = \int_0^1 \frac{x t \ln(1 + x^2 (1 - t^2))}{(1 + x^2 t^2)^{3/2}} \cdot \frac{1}{\sqrt{1 - t^2}} \, dt = x \int_0^1 \frac{t \ln(1 + x^2 - x^2 t^2)}{(1 + x^2 t^2)^{3/2} \sqrt{1 - t^2}} \, dt.$$ 6. **Simplify the logarithm:** $$\ln(1 + x^2 - x^2 t^2) = \ln(1 + x^2 (1 - t^2)) = \ln(1 + x^2 - x^2 t^2).$$ 7. **This integral is complicated, but we can try to interchange the order of integration:** Original integral: $$\int_0^\infty I(x) \, dx = \int_0^\infty \int_0^{\frac{\pi}{2}} \frac{x \sin \theta \ln(1 + x^2 \cos^2 \theta)}{(1 + x^2 \sin^2 \theta)^{3/2}} \, d\theta \, dx.$$ Interchange integrals: $$\int_0^{\frac{\pi}{2}} \sin \theta \int_0^\infty \frac{x \ln(1 + x^2 \cos^2 \theta)}{(1 + x^2 \sin^2 \theta)^{3/2}} \, dx \, d\theta.$$ 8. **Focus on the inner integral over $x$:** $$J(\theta) = \int_0^\infty \frac{x \ln(1 + x^2 \cos^2 \theta)}{(1 + x^2 \sin^2 \theta)^{3/2}} \, dx.$$ 9. **Substitute $u = x \sin \theta$, so $x = \frac{u}{\sin \theta}$ and $dx = \frac{du}{\sin \theta}$:** Rewrite $J(\theta)$: $$J(\theta) = \int_0^\infty \frac{\frac{u}{\sin \theta} \ln\left(1 + \frac{u^2 \cos^2 \theta}{\sin^2 \theta}\right)}{(1 + u^2)^{3/2}} \cdot \frac{du}{\sin \theta} = \frac{1}{\sin^2 \theta} \int_0^\infty \frac{u \ln\left(1 + u^2 \cot^2 \theta\right)}{(1 + u^2)^{3/2}} \, du.$$ 10. **Rewrite the original integral:** $$\int_0^{\frac{\pi}{2}} \sin \theta J(\theta) \, d\theta = \int_0^{\frac{\pi}{2}} \sin \theta \cdot \frac{1}{\sin^2 \theta} \int_0^\infty \frac{u \ln(1 + u^2 \cot^2 \theta)}{(1 + u^2)^{3/2}} \, du \, d\theta = \int_0^{\frac{\pi}{2}} \frac{1}{\sin \theta} \int_0^\infty \frac{u \ln(1 + u^2 \cot^2 \theta)}{(1 + u^2)^{3/2}} \, du \, d\theta.$$ 11. **Interchange the order of integration again:** $$= \int_0^\infty \frac{u}{(1 + u^2)^{3/2}} \int_0^{\frac{\pi}{2}} \frac{\ln(1 + u^2 \cot^2 \theta)}{\sin \theta} \, d\theta \, du.$$ 12. **Substitute $v = \cot \theta$, so $\theta = \cot^{-1} v$, $d\theta = -\frac{1}{1 + v^2} dv$, and when $\theta=0$, $v=\infty$, when $\theta=\frac{\pi}{2}$, $v=0$:** Inner integral becomes $$\int_\infty^0 \frac{\ln(1 + u^2 v^2)}{\sin(\cot^{-1} v)} \cdot \left(-\frac{1}{1 + v^2}\right) dv = \int_0^\infty \frac{\ln(1 + u^2 v^2)}{\sin(\cot^{-1} v)} \cdot \frac{1}{1 + v^2} dv.$$ 13. **Note that $\sin(\cot^{-1} v) = \frac{1}{\sqrt{1 + v^2}}$:** So inner integral is $$\int_0^\infty \ln(1 + u^2 v^2) \cdot \frac{\sqrt{1 + v^2}}{1 + v^2} dv = \int_0^\infty \frac{\ln(1 + u^2 v^2)}{\sqrt{1 + v^2}} dv.$$ 14. **Therefore, the original integral is** $$\int_0^\infty \frac{u}{(1 + u^2)^{3/2}} \int_0^\infty \frac{\ln(1 + u^2 v^2)}{\sqrt{1 + v^2}} dv \, du.$$ 15. **By symmetry, the integral is symmetric in $u$ and $v$ after change of variables, so the value is zero.** **Final answer:** $$\boxed{0}.$$