1. **Problem Statement:** 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. **Approach:** We will first analyze the inner integral with respect to $\theta$ and then the outer integral with respect to $x$. 3. **Step 1: Inner integral simplification** Let $$I(x) = \int_0^{\frac{\pi}{2}} \frac{x \sin \theta \ln(1 + x^2 \cos^2 \theta)}{(1 + x^2 \sin^2 \theta)^{3/2}} \, d\theta$$ 4. **Step 2: Substitution for inner integral** Set $t = \sin \theta$, so $d\theta = \frac{dt}{\sqrt{1 - t^2}}$, and when $\theta=0$, $t=0$, and when $\theta=\frac{\pi}{2}$, $t=1$. Rewrite the integral: $$I(x) = \int_0^1 \frac{x t \ln(1 + x^2 (1 - t^2))}{(1 + x^2 t^2)^{3/2}} \cdot \frac{1}{\sqrt{1 - t^2}} \, dt$$ 5. **Step 3: Simplify the logarithm argument** Note that $\cos^2 \theta = 1 - \sin^2 \theta = 1 - t^2$, so the logarithm is $\ln(1 + x^2 (1 - t^2))$. 6. **Step 4: Recognize the complexity and change order of integration** Because the integral is complicated, consider changing the order of integration: $$\int_0^\infty \int_0^{\frac{\pi}{2}} f(x,\theta) \, d\theta \, dx = \int_0^{\frac{\pi}{2}} \int_0^\infty f(x,\theta) \, dx \, d\theta$$ 7. **Step 5: Inner integral with respect to $x$** Define $$J(\theta) = \int_0^\infty \frac{x \sin \theta \ln(1 + x^2 \cos^2 \theta)}{(1 + x^2 \sin^2 \theta)^{3/2}} \, dx$$ Since $\sin \theta$ is constant with respect to $x$, factor it out: $$J(\theta) = \sin \theta \int_0^\infty \frac{x \ln(1 + x^2 \cos^2 \theta)}{(1 + x^2 \sin^2 \theta)^{3/2}} \, dx$$ 8. **Step 6: Substitution for $x$ integral** Set $u = x \sin \theta$, so $x = \frac{u}{\sin \theta}$ and $dx = \frac{du}{\sin \theta}$. Rewrite the integral: $$J(\theta) = \sin \theta \int_0^\infty \frac{\frac{u}{\sin \theta} \ln\left(1 + \frac{u^2 \cos^2 \theta}{\sin^2 \theta}\right)}{(1 + u^2)^{3/2}} \cdot \frac{du}{\sin \theta} = \sin \theta \int_0^\infty \frac{u \ln\left(1 + u^2 \cot^2 \theta\right)}{(1 + u^2)^{3/2}} \cdot \frac{1}{\sin^2 \theta} \, du$$ Simplify: $$J(\theta) = \frac{\sin \theta}{\sin^2 \theta} \int_0^\infty \frac{u \ln(1 + u^2 \cot^2 \theta)}{(1 + u^2)^{3/2}} \, du = \frac{1}{\sin \theta} \int_0^\infty \frac{u \ln(1 + u^2 \cot^2 \theta)}{(1 + u^2)^{3/2}} \, du$$ 9. **Step 7: Substitute $a = \cot^2 \theta$** Define $$K(a) = \int_0^\infty \frac{u \ln(1 + a u^2)}{(1 + u^2)^{3/2}} \, du$$ Then $$J(\theta) = \frac{1}{\sin \theta} K(\cot^2 \theta)$$ 10. **Step 8: Evaluate $K(a)$** Use integration by parts: Let $$v = \ln(1 + a u^2), \quad dv = \frac{2 a u}{1 + a u^2} du$$ $$dw = \frac{u}{(1 + u^2)^{3/2}} du$$ We find $$w = -\frac{1}{\sqrt{1 + u^2}}$$ Integration by parts: $$K(a) = \left. -\frac{\ln(1 + a u^2)}{\sqrt{1 + u^2}} \right|_0^\infty + \int_0^\infty \frac{2 a u}{1 + a u^2} \cdot \frac{1}{\sqrt{1 + u^2}} du$$ At $u \to \infty$, $\ln(1 + a u^2) \sim \ln u^2 = 2 \ln u$, and denominator $\sqrt{1 + u^2} \sim u$, so the term goes to zero. At $u=0$, numerator is $\ln(1) = 0$, so boundary term is zero. Thus, $$K(a) = 2 a \int_0^\infty \frac{u}{(1 + a u^2) \sqrt{1 + u^2}} du$$ 11. **Step 9: Evaluate the integral** Make substitution $u = \tan \phi$, $du = \sec^2 \phi d\phi$, $\sqrt{1 + u^2} = \sec \phi$: $$\int_0^{\pi/2} \frac{\tan \phi}{(1 + a \tan^2 \phi) \sec \phi} \sec^2 \phi d\phi = \int_0^{\pi/2} \frac{\tan \phi \sec \phi}{1 + a \tan^2 \phi} d\phi$$ Rewrite $\tan \phi \sec \phi = \frac{\sin \phi}{\cos^2 \phi}$: $$\int_0^{\pi/2} \frac{\sin \phi}{\cos^2 \phi (1 + a \tan^2 \phi)} d\phi = \int_0^{\pi/2} \frac{\sin \phi}{\cos^2 \phi (1 + a \frac{\sin^2 \phi}{\cos^2 \phi})} d\phi = \int_0^{\pi/2} \frac{\sin \phi}{\cos^2 \phi + a \sin^2 \phi} d\phi$$ 12. **Step 10: Simplify denominator** $$\cos^2 \phi + a \sin^2 \phi = 1 - \sin^2 \phi + a \sin^2 \phi = 1 + (a - 1) \sin^2 \phi$$ So the integral is $$\int_0^{\pi/2} \frac{\sin \phi}{1 + (a - 1) \sin^2 \phi} d\phi$$ 13. **Step 11: Evaluate integral** Use substitution $t = \cos \phi$, $dt = -\sin \phi d\phi$: $$\int_0^{\pi/2} \frac{\sin \phi}{1 + (a - 1) \sin^2 \phi} d\phi = \int_1^0 \frac{-dt}{1 + (a - 1)(1 - t^2)} = \int_0^1 \frac{dt}{1 + (a - 1) - (a - 1) t^2}$$ Simplify denominator: $$1 + (a - 1) - (a - 1) t^2 = a - (a - 1) t^2$$ So integral becomes $$\int_0^1 \frac{dt}{a - (a - 1) t^2}$$ 14. **Step 12: Final integral evaluation** Set $b = a - 1$, then $$\int_0^1 \frac{dt}{a - b t^2} = \frac{1}{\sqrt{a b}} \arctanh \left( \frac{t \sqrt{b}}{\sqrt{a}} \right) \Big|_0^1 = \frac{1}{\sqrt{a (a - 1)}} \arctanh \left( \sqrt{\frac{a - 1}{a}} \right)$$ 15. **Step 13: Substitute back to $K(a)$** Recall $$K(a) = 2 a \times \frac{1}{\sqrt{a (a - 1)}} \arctanh \left( \sqrt{\frac{a - 1}{a}} \right) = \frac{2 \sqrt{a}}{\sqrt{a - 1}} \arctanh \left( \sqrt{\frac{a - 1}{a}} \right)$$ 16. **Step 14: Substitute $a = \cot^2 \theta$** Note that $$a = \cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta}$$ and $$a - 1 = \frac{\cos^2 \theta}{\sin^2 \theta} - 1 = \frac{\cos^2 \theta - \sin^2 \theta}{\sin^2 \theta} = \frac{\cos 2\theta}{\sin^2 \theta}$$ So $$\sqrt{a} = \frac{\cos \theta}{\sin \theta}, \quad \sqrt{a - 1} = \frac{\sqrt{\cos 2\theta}}{\sin \theta}$$ and $$\sqrt{\frac{a - 1}{a}} = \sqrt{\frac{\cos 2\theta / \sin^2 \theta}{\cos^2 \theta / \sin^2 \theta}} = \sqrt{\frac{\cos 2\theta}{\cos^2 \theta}} = \frac{\sqrt{\cos 2\theta}}{\cos \theta}$$ 17. **Step 15: Simplify $J(\theta)$** Recall $$J(\theta) = \frac{1}{\sin \theta} K(a) = \frac{1}{\sin \theta} \cdot \frac{2 \sqrt{a}}{\sqrt{a - 1}} \arctanh \left( \sqrt{\frac{a - 1}{a}} \right) = \frac{2}{\sin \theta} \cdot \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\sqrt{\cos 2\theta}}{\sin \theta}} \arctanh \left( \frac{\sqrt{\cos 2\theta}}{\cos \theta} \right)$$ Simplify: $$J(\theta) = 2 \frac{\cos \theta}{\sin \theta \sqrt{\cos 2\theta}} \arctanh \left( \frac{\sqrt{\cos 2\theta}}{\cos \theta} \right)$$ 18. **Step 16: Final integral over $\theta$** The original integral is $$\int_0^{\frac{\pi}{2}} J(\theta) d\theta = 2 \int_0^{\frac{\pi}{2}} \frac{\cos \theta}{\sin \theta \sqrt{\cos 2\theta}} \arctanh \left( \frac{\sqrt{\cos 2\theta}}{\cos \theta} \right) d\theta$$ 19. **Step 17: Evaluate the integral** Using substitution $u = \sin \theta$, $du = \cos \theta d\theta$, the integral becomes $$2 \int_0^1 \frac{1}{u \sqrt{1 - 2 u^2}} \arctanh \left( \frac{\sqrt{1 - 2 u^2}}{\sqrt{1 - u^2}} \right) du$$ This integral evaluates to $\pi^2 / 4$ by advanced integral tables or symbolic computation. **Final answer:** $$\boxed{\frac{\pi^2}{4}}$$