1. **Stating the problem:** Evaluate the double integral $$\int_0^\infty \int_0^{\frac{\pi}{2}} \frac{x \sin(\theta) \ln\left(1 + x^2 \cos^2(\theta)\right)}{\left(1 + x^2 \sin^2(\theta)\right)^{3/2}} \, d\theta \, dx$$ 2. **Understanding 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. **Strategy:** We will first integrate with respect to $\theta$, then with respect to $x$. To simplify, consider the substitution or symmetry in $\theta$. 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. **Change order of integration or use substitution:** Define $$J(x) = \int_0^{\frac{\pi}{2}} \frac{\sin(\theta) \ln\left(1 + x^2 \cos^2(\theta)\right)}{\left(1 + x^2 \sin^2(\theta)\right)^{3/2}} \, d\theta$$ so that $$I = \int_0^\infty x J(x) \, dx$$ 6. **Substitute $t = \sin(\theta)$:** Then $d\theta = \frac{dt}{\sqrt{1 - t^2}}$, $\cos^2(\theta) = 1 - t^2$, and $\sin(\theta) = t$. Rewrite $J(x)$: $$J(x) = \int_0^1 \frac{t \ln\left(1 + x^2 (1 - t^2)\right)}{(1 + x^2 t^2)^{3/2}} \cdot \frac{1}{\sqrt{1 - t^2}} \, dt = \int_0^1 \frac{t \ln\left(1 + x^2 - x^2 t^2\right)}{(1 + x^2 t^2)^{3/2} \sqrt{1 - t^2}} \, dt$$ 7. **Simplify the logarithm:** $$\ln\left(1 + x^2 - x^2 t^2\right) = \ln\left(1 + x^2 (1 - t^2)\right)$$ 8. **Interchange integration order:** Rewrite $I$ as $$I = \int_0^1 \frac{t}{\sqrt{1 - t^2}} \int_0^\infty \frac{x \ln\left(1 + x^2 - x^2 t^2\right)}{(1 + x^2 t^2)^{3/2}} \, dx \, dt$$ 9. **Simplify inside the logarithm:** $$1 + x^2 - x^2 t^2 = 1 + x^2 (1 - t^2)$$ 10. **Substitute $a = t^2$, $b = 1 - t^2$:** Then $$I = \int_0^1 \frac{t}{\sqrt{1 - t^2}} \int_0^\infty \frac{x \ln(1 + b x^2)}{(1 + a x^2)^{3/2}} \, dx \, dt$$ 11. **Evaluate inner integral:** Use substitution $u = x^2$, $du = 2x dx$, so $$\int_0^\infty \frac{x \ln(1 + b x^2)}{(1 + a x^2)^{3/2}} \, dx = \frac{1}{2} \int_0^\infty \frac{\ln(1 + b u)}{(1 + a u)^{3/2}} \, du$$ 12. **Use integration by parts:** Let $$f(u) = \ln(1 + b u), \quad dg = (1 + a u)^{-3/2} du$$ Then $$df = \frac{b}{1 + b u} du, \quad g = -\frac{2}{\sqrt{1 + a u}}$$ Integration by parts gives $$\int_0^\infty \frac{\ln(1 + b u)}{(1 + a u)^{3/2}} du = \left. -2 \frac{\ln(1 + b u)}{\sqrt{1 + a u}} \right|_0^\infty + 2 b \int_0^\infty \frac{1}{(1 + b u) \sqrt{1 + a u}} du$$ 13. **Evaluate boundary terms:** As $u \to \infty$, $\ln(1 + b u) \sim \ln u$, denominator grows as $\sqrt{a u}$, so the term goes to zero. At $u=0$, term is $$-2 \frac{\ln(1)}{\sqrt{1}} = 0$$ 14. **So integral reduces to:** $$2 b \int_0^\infty \frac{1}{(1 + b u) \sqrt{1 + a u}} du$$ 15. **Evaluate integral:** Use substitution or known integral formula: $$\int_0^\infty \frac{du}{(1 + b u) \sqrt{1 + a u}} = \frac{\pi}{\sqrt{a} + \sqrt{b}}$$ 16. **Therefore:** $$\int_0^\infty \frac{\ln(1 + b u)}{(1 + a u)^{3/2}} du = 2 b \frac{\pi}{\sqrt{a} + \sqrt{b}}$$ 17. **Recall $a = t^2$, $b = 1 - t^2$:** $$\int_0^\infty \frac{x \ln(1 + b x^2)}{(1 + a x^2)^{3/2}} dx = \frac{1}{2} \times 2 b \frac{\pi}{\sqrt{a} + \sqrt{b}} = b \frac{\pi}{\sqrt{a} + \sqrt{b}} = (1 - t^2) \frac{\pi}{t + \sqrt{1 - t^2}}$$ 18. **Substitute back into $I$:** $$I = \int_0^1 \frac{t}{\sqrt{1 - t^2}} (1 - t^2) \frac{\pi}{t + \sqrt{1 - t^2}} dt = \pi \int_0^1 \frac{t (1 - t^2)}{\sqrt{1 - t^2} (t + \sqrt{1 - t^2})} dt$$ 19. **Simplify integrand:** $$\frac{t (1 - t^2)}{\sqrt{1 - t^2} (t + \sqrt{1 - t^2})} = \frac{t \sqrt{1 - t^2} (1 - t^2)}{(t + \sqrt{1 - t^2}) (1 - t^2)} = \frac{t \sqrt{1 - t^2}}{t + \sqrt{1 - t^2}}$$ 20. **Rewrite integral:** $$I = \pi \int_0^1 \frac{t \sqrt{1 - t^2}}{t + \sqrt{1 - t^2}} dt$$ 21. **Substitute $t = \sin \phi$, $dt = \cos \phi d\phi$:** $$I = \pi \int_0^{\pi/2} \frac{\sin \phi \cos \phi}{\sin \phi + \cos \phi} \cos \phi d\phi = \pi \int_0^{\pi/2} \frac{\sin \phi \cos^2 \phi}{\sin \phi + \cos \phi} d\phi$$ 22. **Use symmetry:** Let $$K = \int_0^{\pi/2} \frac{\sin \phi \cos^2 \phi}{\sin \phi + \cos \phi} d\phi$$ Interchange $\sin$ and $\cos$: $$K = \int_0^{\pi/2} \frac{\cos \phi \sin^2 \phi}{\sin \phi + \cos \phi} d\phi$$ Add both expressions: $$2K = \int_0^{\pi/2} \frac{\sin \phi \cos^2 \phi + \cos \phi \sin^2 \phi}{\sin \phi + \cos \phi} d\phi = \int_0^{\pi/2} \frac{\sin \phi \cos \phi (\cos \phi + \sin \phi)}{\sin \phi + \cos \phi} d\phi = \int_0^{\pi/2} \sin \phi \cos \phi d\phi$$ 23. **Evaluate:** $$\int_0^{\pi/2} \sin \phi \cos \phi d\phi = \frac{1}{2} \int_0^{\pi/2} \sin(2 \phi) d\phi = \frac{1}{2} \left[-\frac{\cos(2 \phi)}{2}\right]_0^{\pi/2} = \frac{1}{4} (1 - (-1)) = \frac{1}{2}$$ 24. **Therefore:** $$2K = \frac{1}{2} \implies K = \frac{1}{4}$$ 25. **Final answer:** $$I = \pi K = \pi \times \frac{1}{4} = \frac{\pi}{4}$$ **Answer:** $$\boxed{\frac{\pi}{4}}$$