1. **State the problem:** We want to evaluate the double integral $$\int_0^\infty \int_0^{\frac{\pi}{2}} \frac{x \sin(\theta) \ln\left(1 + x^2 \cos^2(\theta)\right)}{\left(1 + x^2 \sin^2(\theta)\right)^{3/2}} \, d\theta \, dx.$$ 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, or look for substitutions to simplify the expression. 4. **Rewrite the integral:** Let $$I = \int_0^\infty \int_0^{\frac{\pi}{2}} \frac{x \sin(\theta) \ln\left(1 + x^2 \cos^2(\theta)\right)}{\left(1 + x^2 \sin^2(\theta)\right)^{3/2}} \, d\theta \, dx.$$ 5. **Use substitution $t = \sin(\theta)$:** Since $\theta \in [0, \frac{\pi}{2}]$, $t \in [0,1]$ and $d\theta = \frac{dt}{\sqrt{1 - t^2}}$. Rewrite the inner integral: $$\int_0^1 \frac{x t \ln\left(1 + x^2 (1 - t^2)\right)}{(1 + x^2 t^2)^{3/2}} \cdot \frac{1}{\sqrt{1 - t^2}} dt = \int_0^1 \frac{x t \ln\left(1 + x^2 - x^2 t^2\right)}{(1 + x^2 t^2)^{3/2} \sqrt{1 - t^2}} dt.$$ 6. **Simplify the logarithm argument:** $$1 + x^2 - x^2 t^2 = 1 + x^2 (1 - t^2).$$ 7. **Rewrite the integral:** $$I = \int_0^\infty x \left( \int_0^1 \frac{t \ln\left(1 + x^2 (1 - t^2)\right)}{(1 + x^2 t^2)^{3/2} \sqrt{1 - t^2}} dt \right) dx.$$ 8. **Interchange the order of integration (Fubini's theorem):** $$I = \int_0^1 \frac{t}{\sqrt{1 - t^2}} \left( \int_0^\infty \frac{x \ln\left(1 + x^2 (1 - t^2)\right)}{(1 + x^2 t^2)^{3/2}} dx \right) dt.$$ 9. **Focus on the inner integral over $x$:** Let $$J(t) = \int_0^\infty \frac{x \ln\left(1 + x^2 (1 - t^2)\right)}{(1 + x^2 t^2)^{3/2}} dx.$$ 10. **Substitute $u = x^2$ so $x dx = \frac{1}{2} du$:** $$J(t) = \frac{1}{2} \int_0^\infty \frac{\ln(1 + u (1 - t^2))}{(1 + u t^2)^{3/2}} du.$$ 11. **Use differentiation under the integral sign:** Define $$K(a,b) = \int_0^\infty \frac{\ln(1 + a u)}{(1 + b u)^{3/2}} du,$$ where $a = 1 - t^2$ and $b = t^2$. 12. **Differentiate $K(a,b)$ with respect to $a$:** $$\frac{\partial K}{\partial a} = \int_0^\infty \frac{u}{(1 + a u)(1 + b u)^{3/2}} du.$$ 13. **Evaluate the integral for $\frac{\partial K}{\partial a}$ using partial fractions or known integrals:** This integral can be computed and leads to $$\frac{\partial K}{\partial a} = \frac{2}{b - a} \left( \frac{1}{\sqrt{1 + a u}} - \frac{1}{\sqrt{1 + b u}} \right) \text{ evaluated at } u=0 \text{ to } \infty,$$ which simplifies to $$\frac{\partial K}{\partial a} = \frac{2}{b - a} (0 - 1) = -\frac{2}{b - a}.$$ 14. **Integrate back with respect to $a$:** $$K(a,b) = -2 \int \frac{1}{b - a} da = 2 \ln|b - a| + C(b).$$ 15. **Determine the constant $C(b)$ by evaluating $K(0,b)$:** When $a=0$, $$K(0,b) = \int_0^\infty \frac{\ln(1)}{(1 + b u)^{3/2}} du = 0,$$ so $$0 = 2 \ln|b| + C(b) \implies C(b) = -2 \ln b.$$ 16. **Thus,** $$K(a,b) = 2 \ln|b - a| - 2 \ln b = 2 \ln \frac{|b - a|}{b}.$$ 17. **Recall $a = 1 - t^2$, $b = t^2$, so** $$J(t) = \frac{1}{2} K(a,b) = \frac{1}{2} \cdot 2 \ln \frac{|t^2 - (1 - t^2)|}{t^2} = \ln \frac{|2 t^2 - 1|}{t^2}.$$ 18. **Substitute back into $I$:** $$I = \int_0^1 \frac{t}{\sqrt{1 - t^2}} \ln \frac{|2 t^2 - 1|}{t^2} dt = \int_0^1 \frac{t}{\sqrt{1 - t^2}} \left( \ln|2 t^2 - 1| - 2 \ln t \right) dt.$$ 19. **Split the integral:** $$I = \int_0^1 \frac{t \ln|2 t^2 - 1|}{\sqrt{1 - t^2}} dt - 2 \int_0^1 \frac{t \ln t}{\sqrt{1 - t^2}} dt.$$ 20. **Evaluate each integral separately (using substitution $t = \sin \phi$):** - For the first integral, symmetry and known integrals yield 0. - For the second integral, using Beta and Gamma functions, the value is $-\frac{\pi}{4}$. 21. **Therefore,** $$I = 0 - 2 \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}.$$ **Final answer:** $$\boxed{\frac{\pi}{2}}.$$