1. **State the problem:** Evaluate the double integral $$\int_0^\infty \int_0^{\pi/2} \frac{x \sin(\theta) \ln(1 + x^2 \cos^2(\theta))}{(1 + x^2 \sin^2(\theta))^{3/2}} \, d\theta \, dx$$ 2. **Analyze the integral:** The integral is over $\theta$ from $0$ to $\pi/2$ and $x$ from $0$ to $\infty$. 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. **Substitution for inner integral:** Let $t = \sin(\theta)$, so $d\theta = \frac{dt}{\cos(\theta)} = \frac{dt}{\sqrt{1 - t^2}}$. When $\theta = 0$, $t=0$; when $\theta = \pi/2$, $t=1$. Rewrite the inner integral: $$\int_0^{\pi/2} \frac{x \sin(\theta) \ln(1 + x^2 \cos^2(\theta))}{(1 + x^2 \sin^2(\theta))^{3/2}} d\theta = \int_0^1 \frac{x t \ln(1 + x^2 (1 - t^2))}{(1 + x^2 t^2)^{3/2}} \cdot \frac{1}{\sqrt{1 - t^2}} dt$$ 5. **Simplify the logarithm argument:** $$1 + x^2 (1 - t^2) = 1 + x^2 - x^2 t^2$$ 6. **Rewrite the integral:** $$\int_0^1 \frac{x t \ln(1 + x^2 - x^2 t^2)}{(1 + x^2 t^2)^{3/2} \sqrt{1 - t^2}} dt$$ 7. **Change the order of integration:** The original integral is $$\int_0^\infty \left( \int_0^{\pi/2} \cdots d\theta \right) dx = \int_0^{\pi/2} \left( \int_0^\infty \cdots dx \right) d\theta$$ Try integrating with respect to $x$ first. 8. **Inner integral with respect to $x$:** Fix $\theta$, consider $$I(\theta) = \int_0^\infty \frac{x \sin(\theta) \ln(1 + x^2 \cos^2(\theta))}{(1 + x^2 \sin^2(\theta))^{3/2}} dx$$ 9. **Substitute $u = x \sin(\theta)$:** Then $x = \frac{u}{\sin(\theta)}$, $dx = \frac{du}{\sin(\theta)}$. Rewrite the integral: $$I(\theta) = \int_0^\infty \frac{\frac{u}{\sin(\theta)} \sin(\theta) \ln\left(1 + \frac{u^2}{\sin^2(\theta)} \cos^2(\theta)\right)}{(1 + u^2)^{3/2}} \cdot \frac{du}{\sin(\theta)}$$ Simplify numerator and denominator: $$I(\theta) = \int_0^\infty \frac{u \ln\left(1 + u^2 \frac{\cos^2(\theta)}{\sin^2(\theta)}\right)}{(1 + u^2)^{3/2} \sin(\theta)} du$$ 10. **Rewrite logarithm argument:** $$1 + u^2 \frac{\cos^2(\theta)}{\sin^2(\theta)} = 1 + u^2 \cot^2(\theta)$$ So $$I(\theta) = \frac{1}{\sin(\theta)} \int_0^\infty \frac{u \ln(1 + u^2 \cot^2(\theta))}{(1 + u^2)^{3/2}} du$$ 11. **Define $a = \cot(\theta)$:** Then $$I(\theta) = \frac{1}{\sin(\theta)} \int_0^\infty \frac{u \ln(1 + a^2 u^2)}{(1 + u^2)^{3/2}} du$$ 12. **Evaluate the integral with respect to $u$:** Use integration by parts: Let $$f(u) = \ln(1 + a^2 u^2), \quad dg = \frac{u}{(1 + u^2)^{3/2}} du$$ Then $$df = \frac{2 a^2 u}{1 + a^2 u^2} du, \quad g = -\frac{1}{\sqrt{1 + u^2}}$$ Integration by parts: $$\int f dg = fg - \int g df$$ Calculate: $$\int_0^\infty \frac{u \ln(1 + a^2 u^2)}{(1 + u^2)^{3/2}} du = \left[-\frac{\ln(1 + a^2 u^2)}{\sqrt{1 + u^2}}\right]_0^\infty + \int_0^\infty \frac{2 a^2 u}{1 + a^2 u^2} \cdot \frac{1}{\sqrt{1 + u^2}} du$$ 13. **Evaluate boundary term:** As $u \to \infty$, $$\ln(1 + a^2 u^2) \sim 2 \ln u + \ln a^2$$ and $$\frac{1}{\sqrt{1 + u^2}} \sim \frac{1}{u}$$ So $$\frac{\ln(1 + a^2 u^2)}{\sqrt{1 + u^2}} \sim \frac{2 \ln u + \ln a^2}{u} \to 0$$ At $u=0$, term is zero. So boundary term is zero. 14. **Integral reduces to:** $$2 a^2 \int_0^\infty \frac{u}{(1 + a^2 u^2) \sqrt{1 + u^2}} du$$ 15. **Substitute $u = \frac{1}{a} \tan \phi$:** Then $$du = \frac{1}{a} \sec^2 \phi d\phi$$ Rewrite integral: $$2 a^2 \int_0^{\pi/2} \frac{\frac{1}{a} \tan \phi}{(1 + a^2 \frac{1}{a^2} \tan^2 \phi) \sqrt{1 + \frac{1}{a^2} \tan^2 \phi}} \cdot \frac{1}{a} \sec^2 \phi d\phi$$ Simplify denominator: $$1 + \tan^2 \phi = \sec^2 \phi$$ So denominator is $$(1 + \tan^2 \phi) \sqrt{1 + \frac{1}{a^2} \tan^2 \phi} = \sec^2 \phi \sqrt{1 + \frac{1}{a^2} \tan^2 \phi}$$ Integral becomes $$2 a^2 \int_0^{\pi/2} \frac{\frac{1}{a} \tan \phi \cdot \frac{1}{a} \sec^2 \phi}{\sec^2 \phi \sqrt{1 + \frac{1}{a^2} \tan^2 \phi}} d\phi = 2 \int_0^{\pi/2} \frac{\tan \phi}{\sqrt{1 + \frac{1}{a^2} \tan^2 \phi}} d\phi$$ 16. **Simplify inside the square root:** $$1 + \frac{1}{a^2} \tan^2 \phi = \frac{a^2 + \tan^2 \phi}{a^2}$$ So $$\sqrt{1 + \frac{1}{a^2} \tan^2 \phi} = \frac{\sqrt{a^2 + \tan^2 \phi}}{a}$$ Integral is $$2 \int_0^{\pi/2} \frac{\tan \phi}{\frac{\sqrt{a^2 + \tan^2 \phi}}{a}} d\phi = 2 a \int_0^{\pi/2} \frac{\tan \phi}{\sqrt{a^2 + \tan^2 \phi}} d\phi$$ 17. **Rewrite $\tan \phi = t$:** When $\phi=0$, $t=0$; when $\phi=\pi/2$, $t \to \infty$. Then $$d\phi = \frac{dt}{1 + t^2}$$ Integral becomes $$2 a \int_0^\infty \frac{t}{\sqrt{a^2 + t^2}} \cdot \frac{1}{1 + t^2} dt$$ 18. **Rewrite integral:** $$2 a \int_0^\infty \frac{t}{(1 + t^2) \sqrt{a^2 + t^2}} dt$$ 19. **Substitute $t = a u$:** Then $$dt = a du$$ Integral becomes $$2 a \int_0^\infty \frac{a u}{(1 + a^2 u^2) \sqrt{a^2 + a^2 u^2}} a du = 2 a^3 \int_0^\infty \frac{u}{(1 + a^2 u^2) a \sqrt{1 + u^2}} du$$ Simplify: $$2 a^2 \int_0^\infty \frac{u}{(1 + a^2 u^2) \sqrt{1 + u^2}} du$$ 20. **Notice this is the same integral as before:** This implies the integral satisfies $$I = 2 a^2 I$$ which is only possible if $I=0$ or $2 a^2 = 1$. Since $a = \cot(\theta)$ varies, the integral must be zero. 21. **Conclusion:** The inner integral with respect to $x$ is zero for all $\theta$. 22. **Therefore, the entire double integral is zero:** $$\int_0^{\pi/2} I(\theta) d\theta = 0$$ **Final answer:** $$\boxed{0}$$