To evaluate the definite integral \( \int^{\pi / 3}_{-\pi / 3} \frac {x \sin x}{\cos^2 x}\ dx \), we first analyze the integrand and the limits of integration.

Analyzing the Integrand and Limits

Let the integrand be \( f(x) = \frac {x \sin x}{\cos^2 x} \). The limits of integration are from \( -\frac{\pi}{3} \) to \( \frac{\pi}{3} \), which are symmetric around 0. This suggests checking if the function is odd or even.

Determining if the Function is Odd or Even

We replace \( x \) with \( -x \) in the integrand:

\( f(-x) = \frac {(-x) \sin(-x)}{\cos^2(-x)} \)

Using the properties \( \sin(-x) = -\sin x \) and \( \cos(-x) = \cos x \), we get:

\( f(-x) = \frac {(-x) (-\sin x)}{(\cos x)^2} = \frac {x \sin x}{\cos^2 x} \)

Since \( f(-x) = f(x) \), the function \( f(x) = \frac {x \sin x}{\cos^2 x} \) is an even function.

Applying the Property for Even Functions

For a definite integral with symmetric limits \( -a \) to \( a \), if the integrand \( f(x) \) is even, then \( \int_{-a}^{a} f(x)\ dx = 2 \int_{0}^{a} f(x)\ dx \). Applying this property here:

\( \int^{\pi / 3}_{-\pi / 3} \frac {x \sin x}{\cos^2 x}\ dx = 2 \int_{0}^{\pi / 3} \frac {x \sin x}{\cos^2 x}\ dx \)

Evaluating the Definite Integral

We need to evaluate \( \int_{0}^{\pi / 3} \frac {x \sin x}{\cos^2 x}\ dx \). We can use integration by parts, where \( \int u\ dv = uv - \int v\ du \).

Let \( u = x \) and \( dv = \frac {\sin x}{\cos^2 x}\ dx \).

Then \( du = dx \).

To find \( v \), we integrate \( dv \):

\( v = \int \frac {\sin x}{\cos^2 x}\ dx \)

Use substitution: let \( t = \cos x \), then \( dt = -\sin x\ dx \), so \( \sin x\ dx = -dt \).

\( v = \int \frac {-dt}{t^2} = -\int t^{-2}\ dt = -(-t^{-1}) = t^{-1} = \frac {1}{\cos x} = \sec x \)

Now apply integration by parts formula:

\( \int \frac {x \sin x}{\cos^2 x}\ dx = x \sec x - \int \sec x\ dx \)

The integral of \( \sec x \) is \( \log|\sec x + \tan x| \).

So, \( \int \frac {x \sin x}{\cos^2 x}\ dx = x \sec x - \log|\sec x + \tan x| + C \)

Applying the Limits of Integration

Now we evaluate the definite integral from 0 to \( \pi/3 \):

\( 2 \int_{0}^{\pi / 3} \frac {x \sin x}{\cos^2 x}\ dx = 2 [x \sec x - \log|\sec x + \tan x|]_{0}^{\pi / 3} \)

Evaluate at the upper limit \( x = \frac{\pi}{3} \):

• \( \sec(\frac{\pi}{3}) = \frac{1}{\cos(\frac{\pi}{3})} = \frac{1}{1/2} = 2 \)
• \( \tan(\frac{\pi}{3}) = \sqrt{3} \)
• Value at \( \frac{\pi}{3} \): \( \frac{\pi}{3} \sec(\frac{\pi}{3}) - \log|\sec(\frac{\pi}{3}) + \tan(\frac{\pi}{3})| = \frac{\pi}{3} (2) - \log|2 + \sqrt{3}| = \frac{2\pi}{3} - \log(2 + \sqrt{3}) \)

Evaluate at the lower limit \( x = 0 \):

• \( \sec(0) = 1 \)
• \( \tan(0) = 0 \)
• Value at \( 0 \): \( 0 \sec(0) - \log|\sec(0) + \tan(0)| = 0(1) - \log|1 + 0| = 0 - \log(1) = 0 - 0 = 0 \)

Subtract the value at the lower limit from the value at the upper limit:

\( 2 \left[ (\frac{2\pi}{3} - \log(2 + \sqrt{3})) - 0 \right] = 2 (\frac{2\pi}{3} - \log(2 + \sqrt{3})) = \frac{4\pi}{3} - 2 \log(2 + \sqrt{3}) \)

Relating the Result to the Options

We need to check if \( \log(2 + \sqrt{3}) \) can be expressed in terms of \( \log \tan \frac{5\pi}{12} \). Let's calculate \( \tan \frac{5\pi}{12} \):

\( \frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6} \)

\( \tan(\frac{5\pi}{12}) = \tan(\frac{\pi}{4} + \frac{\pi}{6}) = \frac{\tan(\frac{\pi}{4}) + \tan(\frac{\pi}{6})}{1 - \tan(\frac{\pi}{4})\tan(\frac{\pi}{6})} \)

\( \tan(\frac{5\pi}{12}) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \)

Rationalize the denominator:

\( \tan(\frac{5\pi}{12}) = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{3 + 1 + 2\sqrt{3}}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \)

Therefore, \( \log(2 + \sqrt{3}) = \log \tan \frac{5\pi}{12} \).

Substituting this back into our result:

Value = \( \frac{4\pi}{3} - 2 \log \tan \frac{5\pi}{12} \)

This matches one of the given options.

Final Answer

The value of the integral \( \int^{\pi / 3}_{-\pi / 3} \frac {x \sin x}{\cos^2 x}\ dx \) is \( \frac {4\pi} {3} -2 \log \tan \frac {5\pi}{12} \).