The value of the integral $$\displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left(\frac{4 - \operatorname{cosec}^2 x}{\cos^4 x} \right) dx$$ is equal to :

To solve this integral quickly, we simplify the integrand into terms that are easy to integrate using the substitution $$u = \tan x$$.

The integral is $$I = \int_{\pi/6}^{\pi/3} \frac{4 - \csc^2 x}{\cos^4 x} dx$$.

Rewrite the terms:

$$\frac{4}{\cos^4 x} - \frac{\csc^2 x}{\cos^4 x} = 4\sec^4 x - \frac{1}{\sin^2 x \cos^4 x}$$

Using the identity $$1 = \sin^2 x + \cos^2 x$$ for the second term:

$$\frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^4 x} = \frac{1}{\cos^4 x} + \frac{1}{\sin^2 x \cos^2 x} = \sec^4 x + \sec^2 x \csc^2 x$$

Now, substitute back into $$I$$:

$$I = \int_{\pi/6}^{\pi/3} (4\sec^4 x - \sec^4 x - \sec^2 x \csc^2 x) dx$$

$$I = \int_{\pi/6}^{\pi/3} (3\sec^4 x - \sec^2 x \csc^2 x) dx$$

Recall that $$\sec^2 x = 1 + \tan^2 x$$ and $$\csc^2 x = 1 + \cot^2 x$$:

$$I = \int_{\pi/6}^{\pi/3} [3\sec^2 x(1 + \tan^2 x) - \sec^2 x(1 + \cot^2 x)] dx$$

$$I = \int_{\pi/6}^{\pi/3} [3\sec^2 x + 3\tan^2 x \sec^2 x - \sec^2 x - \frac{\sec^2 x}{\tan^2 x}] dx$$

$$I = \int_{\pi/6}^{\pi/3} (2\sec^2 x + 3\tan^2 x \sec^2 x - \cot^2 x \sec^2 x) dx$$

Since $$\cot^2 x \sec^2 x = \frac{1}{\tan^2 x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin^2 x} = \csc^2 x$$:

$$I = \int_{\pi/6}^{\pi/3} (2\sec^2 x + 3\tan^2 x \sec^2 x - \csc^2 x) dx$$

The antiderivative is:

$$2\tan x + \tan^3 x + \cot x$$

Applying the limits from $$\pi/6$$ to $$\pi/3$$:

• At $$x = \pi/3$$: $$2(\sqrt{3}) + (\sqrt{3})^3 + \frac{1}{\sqrt{3}} = 2\sqrt{3} + 3\sqrt{3} + \frac{1}{\sqrt{3}} = 5\sqrt{3} + \frac{1}{\sqrt{3}} = \frac{16}{\sqrt{3}}$$
• At $$x = \pi/6$$: $$2(\frac{1}{\sqrt{3}}) + (\frac{1}{\sqrt{3}})^3 + \sqrt{3} = \frac{2}{\sqrt{3}} + \frac{1}{3\sqrt{3}} + \sqrt{3} = \frac{6 + 1 + 9}{3\sqrt{3}} = \frac{16}{3\sqrt{3}}$$

$$I = \frac{16}{\sqrt{3}} - \frac{16}{3\sqrt{3}} = \frac{48 - 16}{3\sqrt{3}} = \frac{32}{3\sqrt{3}}$$

Correct Option: C