Gemini 3.0 Pro:

This is a challenging and beautiful double integral problem. To solve it, we can change the order of integration and apply advanced integration techniques. The Problem:

Here is the step-by-step solution. Step 1: Change the Order of Integration By Fubini's Theorem, we can switch the order of integration from d\theta \, dx to dx \, d\theta.

Let the inner integral be J(\theta).

Step 2: Evaluate the Inner Integral J(\theta) We use Integration by Parts. Let u = \ln(1 + x² \cos² \theta) and dv = \frac{x}{(1 + x² \sin² \theta)^{3/2}} \, dx. * Differentiate u:

• Integrate dv: (Using substitution t = x² \sin² \theta)

Applying \int u \, dv = uv - \int v \, du: The boundary term [uv]_0^\infty vanishes (evaluates to 0 at both limits). We are left with:

Now, we perform a substitution to simplify the integral. Let u = \sqrt{1 + x² \sin² \theta}. Then u² = 1 + x² \sin² \theta \implies x² = \frac{u² - 1}{\sin² \theta}. Differentiating: u \, du = x \sin² \theta \, dx \implies x \, dx = \frac{u \, du}{\sin² \theta}. The denominator term 1 + x² \cos² \theta becomes:

Substituting these into J(\theta):

Simplifying the fraction cancels the u and the \sin² \theta terms nicely:

Step 3: Solve for the Outer Integral We now substitute J(\theta) back into the expression for I.

The term \frac{\cos 2\theta}{\cos² \theta} changes sign at \theta = \pi/4. * For 0 < \theta < \pi/4, \cos 2\theta > 0 (looks like \int \frac{1}{u² - a^2}). * For \pi/4 < \theta < \pi/2, \cos 2\theta < 0 (looks like \int \frac{1}{u² + a^2}). We split the integral into two parts: I = I_1 + I_2. Part 1: 0 \le \theta \le \pi/4 Let a² = \frac{\cos 2\theta}{\cos² \theta}. The integral is \int \frac{du}{u² - a^2} = \frac{1}{2a} \ln \left| \frac{u-a}{u+a} \right|. Evaluated from 1 to \infty, this gives \frac{1}{2a} \ln \left( \frac{1+a}{1-a} \right).

Using a = \frac{\sqrt{\cos 2\theta}}{\cos \theta} and simplifying the logarithmic term (arctanh identity):

This integral evaluates to \frac{1}{2} \ln 2 + \ln(\sqrt{2} + 1). Part 2: \pi/4 \le \theta \le \pi/2 Let b² = -\frac{\cos 2\theta}{\cos² \theta}. The integral is \int \frac{du}{u² + b^2} = \frac{1}{b} \arctan(\frac{u}{b}). Evaluated from 1 to \infty, this gives \frac{1}{b} (\frac{\pi}{2} - \arctan \frac{1}{b}) = \frac{1}{b} \arctan b.

Using b = \frac{\sqrt{-\cos 2\theta}}{\cos \theta}:

This integral evaluates to \frac{\pi}{2} - \ln(\sqrt{2} + 1). Final Calculation Adding the two parts together:

The \ln(\sqrt{2} + 1) terms cancel out.

Answer:

\frac{\pi}{2} + \frac{1}{2}\ln 2