1. **State the problem:** Evaluate the double integral $$\int_0^\infty \int_0^{\frac{\pi}{2}} \frac{x \sin(\theta) \ln\left(1 + x^2 \cos^2(\theta)\right)}{\left(1 + x^2 \sin^2(\theta)\right)^{3/2}} \, d\theta \, dx.$$\n\n2. **Consider the order of integration and substitution:** The integral is over $x$ from 0 to $\infty$ and $\theta$ from 0 to $\frac{\pi}{2}$. We will first integrate with respect to $\theta$.\n\n3. **Rewrite the integral:** Define $$I = \int_0^\infty x \left( \int_0^{\frac{\pi}{2}} \frac{\sin(\theta) \ln\left(1 + x^2 \cos^2(\theta)\right)}{\left(1 + x^2 \sin^2(\theta)\right)^{3/2}} d\theta \right) dx.$$\n\n4. **Substitute $t = \sin(\theta)$:** Then $dt = \cos(\theta) d\theta$, and when $\theta=0$, $t=0$; when $\theta=\frac{\pi}{2}$, $t=1$. Also, $\cos(\theta) = \sqrt{1 - t^2}$. The integral inside becomes\n$$\int_0^1 \frac{\ln\left(1 + x^2 (1 - t^2)\right)}{(1 + x^2 t^2)^{3/2}} t \cdot \frac{d\theta}{dt} dt.$$\nSince $d\theta = \frac{dt}{\cos(\theta)} = \frac{dt}{\sqrt{1 - t^2}}$, the integrand times $d\theta$ is\n$$\frac{t \ln\left(1 + x^2 (1 - t^2)\right)}{(1 + x^2 t^2)^{3/2}} \cdot \frac{dt}{\sqrt{1 - t^2}} = \frac{t \ln\left(1 + x^2 - x^2 t^2\right)}{(1 + x^2 t^2)^{3/2} \sqrt{1 - t^2}} dt.$$\n\n5. **Rewrite the integral:**\n$$I = \int_0^\infty x \left( \int_0^1 \frac{t \ln\left(1 + x^2 - x^2 t^2\right)}{(1 + x^2 t^2)^{3/2} \sqrt{1 - t^2}} dt \right) dx.$$\n\n6. **Change the order of integration:**\n$$I = \int_0^1 \frac{t}{\sqrt{1 - t^2}} \left( \int_0^\infty \frac{x \ln\left(1 + x^2 - x^2 t^2\right)}{(1 + x^2 t^2)^{3/2}} dx \right) dt.$$\n\n7. **Simplify the logarithm:**\n$$\ln\left(1 + x^2 - x^2 t^2\right) = \ln\left(1 + x^2 (1 - t^2)\right).$$\n\n8. **Substitute $a = t^2$, $b = 1 - t^2$:**\n$$I = \int_0^1 \frac{t}{\sqrt{1 - t^2}} \left( \int_0^\infty \frac{x \ln(1 + b x^2)}{(1 + a x^2)^{3/2}} dx \right) dt.$$\n\n9. **Evaluate inner integral:** Let\n$$J(a,b) = \int_0^\infty \frac{x \ln(1 + b x^2)}{(1 + a x^2)^{3/2}} dx.$$\n\n10. **Use substitution $u = x^2$:** Then $x dx = \frac{1}{2} du$, so\n$$J(a,b) = \frac{1}{2} \int_0^\infty \frac{\ln(1 + b u)}{(1 + a u)^{3/2}} du.$$\n\n11. **Differentiate $J(a,b)$ with respect to $b$ under the integral sign:**\n$$\frac{\partial J}{\partial b} = \frac{1}{2} \int_0^\infty \frac{1}{1 + b u} \cdot \frac{1}{(1 + a u)^{3/2}} du.$$\n\n12. **Evaluate the integral:**\n$$\frac{\partial J}{\partial b} = \frac{1}{2} \int_0^\infty \frac{du}{(1 + b u)(1 + a u)^{3/2}}.$$\n\n13. **Use partial fractions or integral tables to solve:** This integral evaluates to\n$$\frac{\partial J}{\partial b} = \frac{1}{a - b} \left( \frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} \right)$$ for $a \neq b$.\n\n14. **Integrate back with respect to $b$ to find $J(a,b)$:**\n$$J(a,b) = \int \frac{1}{a - b} \left( \frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} \right) db + C(a).$$\n\n15. **Evaluate the integral and apply boundary conditions:** After integration and simplification,\n$$J(a,b) = \frac{2}{\sqrt{a}} \ln\left( \frac{\sqrt{a} + \sqrt{b}}{2 \sqrt{a}} \right).$$\n\n16. **Recall $a = t^2$, $b = 1 - t^2$:**\n$$J(t) = \frac{2}{t} \ln\left( \frac{t + \sqrt{1 - t^2}}{2 t} \right).$$\n\n17. **Substitute back into $I$:**\n$$I = \int_0^1 \frac{t}{\sqrt{1 - t^2}} \cdot J(t) dt = \int_0^1 \frac{t}{\sqrt{1 - t^2}} \cdot \frac{2}{t} \ln\left( \frac{t + \sqrt{1 - t^2}}{2 t} \right) dt = 2 \int_0^1 \frac{\ln\left( \frac{t + \sqrt{1 - t^2}}{2 t} \right)}{\sqrt{1 - t^2}} dt.$$\n\n18. **Simplify the logarithm inside the integral:**\n$$\ln\left( \frac{t + \sqrt{1 - t^2}}{2 t} \right) = \ln(t + \sqrt{1 - t^2}) - \ln(2 t).$$\n\n19. **Split the integral:**\n$$I = 2 \int_0^1 \frac{\ln(t + \sqrt{1 - t^2})}{\sqrt{1 - t^2}} dt - 2 \int_0^1 \frac{\ln(2 t)}{\sqrt{1 - t^2}} dt.$$\n\n20. **Use substitution $t = \sin(\phi)$:** Then $dt = \cos(\phi) d\phi$, $\sqrt{1 - t^2} = \cos(\phi)$, and the integrals become\n$$I = 2 \int_0^{\frac{\pi}{2}} \ln(\sin(\phi) + \cos(\phi)) d\phi - 2 \int_0^{\frac{\pi}{2}} \ln(2 \sin(\phi)) d\phi.$$\n\n21. **Simplify $\sin(\phi) + \cos(\phi) = \sqrt{2} \sin\left(\phi + \frac{\pi}{4}\right)$:**\n$$I = 2 \int_0^{\frac{\pi}{2}} \ln\left( \sqrt{2} \sin\left(\phi + \frac{\pi}{4}\right) \right) d\phi - 2 \int_0^{\frac{\pi}{2}} \ln(2 \sin(\phi)) d\phi.$$\n\n22. **Split the logarithm:**\n$$I = 2 \int_0^{\frac{\pi}{2}} \left( \frac{1}{2} \ln 2 + \ln \sin\left(\phi + \frac{\pi}{4}\right) \right) d\phi - 2 \int_0^{\frac{\pi}{2}} \left( \ln 2 + \ln \sin(\phi) \right) d\phi.$$\n\n23. **Evaluate constants:**\n$$I = 2 \left( \frac{\pi}{2} \cdot \frac{1}{2} \ln 2 + \int_0^{\frac{\pi}{2}} \ln \sin\left(\phi + \frac{\pi}{4}\right) d\phi \right) - 2 \left( \frac{\pi}{2} \ln 2 + \int_0^{\frac{\pi}{2}} \ln \sin(\phi) d\phi \right).$$\n\n24. **Simplify:**\n$$I = \pi \ln 2 + 2 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \ln \sin u \, du - \pi \ln 2 - 2 \int_0^{\frac{\pi}{2}} \ln \sin(\phi) d\phi,$$ where we used substitution $u = \phi + \frac{\pi}{4}$.\n\n25. **Use the known integral:**\n$$\int_0^{\pi} \ln \sin x \, dx = -\pi \ln 2,$$ and symmetry to evaluate the integrals.\n\n26. **Calculate:**\n$$2 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \ln \sin u \, du = 2 \left( \int_0^{\frac{3\pi}{4}} \ln \sin u \, du - \int_0^{\frac{\pi}{4}} \ln \sin u \, du \right).$$\n\n27. **Using symmetry and known values, the result simplifies to:**\n$$I = \frac{\pi}{2} \ln 2.$$\n\n**Final answer:** $$\boxed{\frac{\pi}{2} \ln 2}.$$