Evaluate: \[ \int_{-\pi/3}^{\pi/3} \frac{x \sin x}{\cos^2 x} \, dx \] Step 1: Identify the type of function
We define: \[ f(x) = \frac{x \sin x}{\cos^2 x} \] Let's test for evenness: \[ f(-x) = \frac{-x \sin(-x)}{\cos^2(-x)} = \frac{-x (-\sin x)}{\cos^2 x} = \frac{x \sin x}{\cos^2 x} = f(x) \] So it's an even function.
Step 2: Use property of even function \[ \int_{-a}^{a} f(x) dx = 2 \int_0^a f(x) dx \] So, \[ \int_{-\pi/3}^{\pi/3} \frac{x \sin x}{\cos^2 x} dx = 2 \int_0^{\pi/3} \frac{x \sin x}{\cos^2 x} dx \] Step 3: Integration by parts
Let
\( u = x \Rightarrow du = dx \)
\( dv = \frac{\sin x}{\cos^2 x} dx = \tan x \sec x dx \Rightarrow v = \sec x \)
So, \[ \int x \cdot \frac{\sin x}{\cos^2 x} dx = x \sec x - \int \sec x dx \] And we know: \[ \int \sec x dx = \ln |\sec x + \tan x| \] So, putting it all together: \[ \int_0^{\pi/3} \frac{x \sin x}{\cos^2 x} dx = \left[ x \sec x - \ln |\sec x + \tan x| \right]_0^{\pi/3} \] Step 4: Evaluate the expression
At \( x = \frac{\pi}{3} \):
\( \sec(\pi/3) = 2 \)
\( \tan(\pi/3) = \sqrt{3} \)
\[ x \sec x = \frac{\pi}{3} \cdot 2 = \frac{2\pi}{3} \] \[ \ln |\sec x + \tan x| = \ln (2 + \sqrt{3}) \] At \( x = 0 \): \( x \sec x = 0 \), \( \ln(\sec x + \tan x) = \ln(1) = 0 \) So the value becomes: \[ 2 \left[ \frac{2\pi}{3} - \ln(2 + \sqrt{3}) \right] = \frac{4\pi}{3} - 2 \ln(2 + \sqrt{3}) \] Match with options:
Only option B matches: \[ \boxed{\frac{4\pi}{3} - 2 \log \tan \frac{5\pi}{12}} \] And we know: \[ \tan \frac{5\pi}{12} = 2 + \sqrt{3} \quad \text{(Identity)} \] So: \[ \log \tan \frac{5\pi}{12} = \log(2 + \sqrt{3}) \] Thus, the correct answer is: B