Step-by-step Solution:

\[ I = \int_{-\pi/3}^{\pi/3} \frac{x \sin x}{\cos^2 x} \, dx \] Step 1: Use Symmetry
Consider the transformation \( x \to -x \). Rewriting the function: \[ f(x) = \frac{x \sin x}{\cos^2 x} \] Checking its behavior under \( x \to -x \): \[ f(-x) = \frac{-x \sin(-x)}{\cos^2(-x)} \] Since \( \sin(-x) = -\sin x \) and \( \cos(-x) = \cos x \), we get: \[ f(-x) = \frac{-x (-\sin x)}{\cos^2 x} = \frac{x \sin x}{\cos^2 x} = f(x) \] This confirms that \( f(x) \) is an even function, meaning: \[ I = 2 \int_0^{\pi/3} \frac{x \sin x}{\cos^2 x} \, dx \] Step 2: Substituting \( u = \cos x \)
Let \( u = \cos x \), so that: \[ du = -\sin x \, dx \] Limits change as follows:
When \( x = 0 \), \( u = \cos 0 = 1 \).
When \( x = \pi/3 \), \( u = \cos(\pi/3) = 1/2 \).
Rewriting the integral: \[ I = -2 \int_1^{1/2} x \cdot \frac{-du}{u^2} \] \[ = 2 \int_{1/2}^{1} \frac{x \, du}{u^2} \] Using \( x = \cos^{-1} u \), we rewrite: \[ I = 2 \int_{1/2}^{1} \frac{\cos^{-1} u}{u^2} \, du \] Using standard integral results and evaluations, we get: \[ I = \frac{4\pi}{3} - 2 \log \tan \frac{5\pi}{12} \] Thus, the correct answer is: B