1. **Stating 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. **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 a logarithm. 3. **Interchange the order of integration:** We consider integrating first with respect to $x$: $$I = \int_0^{\frac{\pi}{2}} \sin\theta \int_0^\infty \frac{x \ln(1 + x^2 \cos^2\theta)}{(1 + x^2 \sin^2\theta)^{3/2}} \, dx \, d\theta.$$ 4. **Substitution for the inner integral:** Let $a = \sin\theta$ and $b = \cos\theta$. Then the inner integral is $$J(\theta) = \int_0^\infty \frac{x \ln(1 + b^2 x^2)}{(1 + a^2 x^2)^{3/2}} \, dx.$$ 5. **Change variable:** Set $x = \frac{t}{a}$ so that $dx = \frac{dt}{a}$ and $$J(\theta) = \int_0^\infty \frac{\frac{t}{a} \ln\left(1 + b^2 \frac{t^2}{a^2}\right)}{(1 + a^2 \frac{t^2}{a^2})^{3/2}} \frac{dt}{a} = \int_0^\infty \frac{t \ln\left(1 + \frac{b^2}{a^2} t^2\right)}{a^2 (1 + t^2)^{3/2}} \, dt.$$ 6. **Simplify:** $$J(\theta) = \frac{1}{a^2} \int_0^\infty \frac{t \ln\left(1 + \frac{b^2}{a^2} t^2\right)}{(1 + t^2)^{3/2}} \, dt.$$ 7. **Rewrite the full integral:** $$I = \int_0^{\frac{\pi}{2}} \sin\theta \cdot J(\theta) \, d\theta = \int_0^{\frac{\pi}{2}} \sin\theta \cdot \frac{1}{\sin^2\theta} \int_0^\infty \frac{t \ln\left(1 + \frac{\cos^2\theta}{\sin^2\theta} t^2\right)}{(1 + t^2)^{3/2}} \, dt \, d\theta.$$ This simplifies to $$I = \int_0^{\frac{\pi}{2}} \frac{\sin\theta}{\sin^2\theta} \int_0^\infty \frac{t \ln\left(1 + t^2 \cot^2\theta\right)}{(1 + t^2)^{3/2}} \, dt \, d\theta = \int_0^{\frac{\pi}{2}} \frac{1}{\sin\theta} \int_0^\infty \frac{t \ln\left(1 + t^2 \cot^2\theta\right)}{(1 + t^2)^{3/2}} \, dt \, d\theta.$$ 8. **Change variable in $\theta$ integral:** Let $u = \cot\theta = \frac{\cos\theta}{\sin\theta}$, then $d\theta = -\frac{1}{1+u^2} du$ and when $\theta=0$, $u=\infty$, when $\theta=\frac{\pi}{2}$, $u=0$. Also, $\sin\theta = \frac{1}{\sqrt{1+u^2}}$, so $$\frac{1}{\sin\theta} = \sqrt{1+u^2}.$$ Thus, $$I = \int_\infty^0 \sqrt{1+u^2} \int_0^\infty \frac{t \ln(1 + t^2 u^2)}{(1 + t^2)^{3/2}} \, dt \cdot \left(-\frac{1}{1+u^2}\right) du = \int_0^\infty \frac{1}{\sqrt{1+u^2}} \int_0^\infty \frac{t \ln(1 + t^2 u^2)}{(1 + t^2)^{3/2}} \, dt \, du.$$ 9. **Interchange the order of integration:** $$I = \int_0^\infty t \int_0^\infty \frac{\ln(1 + t^2 u^2)}{\sqrt{1+u^2} (1 + t^2)^{3/2}} \, du \, dt.$$ 10. **Focus on the inner integral over $u$:** $$K(t) = \int_0^\infty \frac{\ln(1 + t^2 u^2)}{\sqrt{1+u^2}} \, du.$$ 11. **Substitute $v = u t$:** $$K(t) = \int_0^\infty \frac{\ln(1 + v^2)}{\sqrt{1 + \frac{v^2}{t^2}}} \frac{dv}{t} = \int_0^\infty \frac{\ln(1 + v^2)}{\sqrt{1 + \frac{v^2}{t^2}}} \frac{dv}{t}.$$ Rewrite the denominator: $$\sqrt{1 + \frac{v^2}{t^2}} = \frac{\sqrt{t^2 + v^2}}{t}.$$ So $$K(t) = \int_0^\infty \ln(1 + v^2) \frac{t}{\sqrt{t^2 + v^2}} \frac{dv}{t} = \int_0^\infty \frac{\ln(1 + v^2)}{\sqrt{t^2 + v^2}} \, dv.$$ 12. **Therefore, the integral becomes:** $$I = \int_0^\infty \frac{t}{(1 + t^2)^{3/2}} \int_0^\infty \frac{\ln(1 + v^2)}{\sqrt{t^2 + v^2}} \, dv \, dt.$$ 13. **Symmetry in $t$ and $v$:** Interchange the order of integration again: $$I = \int_0^\infty \ln(1 + v^2) \int_0^\infty \frac{t}{(1 + t^2)^{3/2} \sqrt{t^2 + v^2}} \, dt \, dv.$$ 14. **Evaluate the inner integral over $t$:** Set $$L(v) = \int_0^\infty \frac{t}{(1 + t^2)^{3/2} \sqrt{t^2 + v^2}} \, dt.$$ 15. **Substitute $t = v w$:** $$L(v) = \int_0^\infty \frac{v w}{(1 + v^2 w^2)^{3/2} \sqrt{v^2 w^2 + v^2}} v dw = \int_0^\infty \frac{v w}{(1 + v^2 w^2)^{3/2} v \sqrt{w^2 + 1}} v dw.$$ Simplify: $$L(v) = v \int_0^\infty \frac{w}{(1 + v^2 w^2)^{3/2} \sqrt{w^2 + 1}} \, dw.$$ 16. **Rewrite:** $$L(v) = v \int_0^\infty \frac{w}{(1 + v^2 w^2)^{3/2} \sqrt{w^2 + 1}} \, dw.$$ 17. **Change variable $z = v w$:** $$L(v) = v \int_0^\infty \frac{w}{(1 + v^2 w^2)^{3/2} \sqrt{w^2 + 1}} \, dw = v \int_0^\infty \frac{\frac{z}{v}}{(1 + z^2)^{3/2} \sqrt{\frac{z^2}{v^2} + 1}} \frac{dz}{v} = \int_0^\infty \frac{z}{(1 + z^2)^{3/2} \sqrt{z^2 + v^2}} \, dz.$$ 18. **Notice $L(v)$ is symmetric in $v$ and $z$:** This is the same form as before, so by symmetry, $$L(v) = L(z)$$ for all $v,z$. 19. **This implies $L(v)$ is constant in $v$.** Evaluate at $v=0$: $$L(0) = \int_0^\infty \frac{t}{(1 + t^2)^{3/2} \sqrt{t^2 + 0}} dt = \int_0^\infty \frac{t}{(1 + t^2)^{3/2} t} dt = \int_0^\infty \frac{1}{(1 + t^2)^{3/2}} dt.$$ 20. **Evaluate:** $$\int_0^\infty \frac{1}{(1 + t^2)^{3/2}} dt = 1$$ (standard integral). 21. **Therefore, $L(v) = 1$ for all $v$.** 22. **Return to the integral:** $$I = \int_0^\infty \ln(1 + v^2) \cdot 1 \, dv = \int_0^\infty \ln(1 + v^2) \, dv,$$ which diverges. 23. **Re-examining the problem:** The integral over $v$ diverges, so the original integral diverges. **Final answer:** The given integral diverges (does not converge to a finite value).