Integrate Trigonometric Function

The problem asks for the value of $f\left(\frac{\pi}{6}\right) - f\left(\frac{\pi}{4}\right)$, where $f(x)$ is the indefinite integral of the given function:

$ \frac{1-5\cos^2x}{\sin^5x \cos^2x} $

We can rewrite the integrand by splitting the fraction:

$ \frac{1-5\cos^2x}{\sin^5x \cos^2x} = \frac{1}{\sin^5x \cos^2x} - \frac{5\cos^2x}{\sin^5x \cos^2x} = \frac{1}{\sin^5x \cos^2x} - \frac{5}{\sin^5x} $

Alternatively, we can express it using $\sec x$ and $\csc x$:

$ \frac{1-5\cos^2x}{\sin^5x \cos^2x} = \frac{\sec^2x - 5}{\sin^5x} = \sec^2x \csc^5x - 5\csc^5x $

Let's find the integral $f(x)$ by checking the derivative of a potential function. Consider the function $g(x) = \tan x \csc^5 x$. Its derivative is:

$ \frac{d}{dx}(\tan x \csc^5 x) = (\sec^2 x) \csc^5 x + \tan x (5 \csc^4 x (-\csc x \cot x)) $ $ = \sec^2 x \csc^5 x - 5 \tan x \csc^5 x \cot x $ $ = \frac{1}{\cos^2 x \sin^5 x} - 5 \frac{\sin x}{\cos x} \frac{1}{\sin^5 x} \frac{\cos x}{\sin x} $ $ = \frac{1}{\cos^2 x \sin^5 x} - \frac{5}{\sin^5 x} $ $ = \frac{1 - 5\cos^2 x}{\cos^2 x \sin^5 x} $

This matches the integrand. Thus, the indefinite integral is:

$ f(x) = \tan x \csc^5 x = \frac{\sin x}{\cos x} \frac{1}{\sin^5 x} = \frac{1}{\cos x \sin^4 x} $

Evaluate Function at Specific Points

Now, we evaluate $f(x)$ at the given points $\frac{\pi}{6}$ and $\frac{\pi}{4}$.

• For $x = \frac{\pi}{6}$:
  • $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
  • $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$
  • $f\left(\frac{\pi}{6}\right) = \frac{1}{\cos\left(\frac{\pi}{6}\right) \sin^4\left(\frac{\pi}{6}\right)} = \frac{1}{\left(\frac{\sqrt{3}}{2}\right) \left(\frac{1}{2}\right)^4} = \frac{1}{\frac{\sqrt{3}}{2} \cdot \frac{1}{16}} = \frac{1}{\frac{\sqrt{3}}{32}} = \frac{32}{\sqrt{3}}\)
• For $x = \frac{\pi}{4}$:
  • $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\)
  • $\sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\)
  • $f\left(\frac{\pi}{4}\right) = \frac{1}{\cos\left(\frac{\pi}{4}\right) \sin^4\left(\frac{\pi}{4}\right)} = \frac{1}{\left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right)^4} = \frac{1}{\frac{1}{\sqrt{2}} \cdot \frac{1}{4}} = \frac{1}{\frac{1}{4\sqrt{2}}} = 4\sqrt{2}\)

Calculate the Difference

Finally, calculate the difference $f\left(\frac{\pi}{6}\right) - f\left(\frac{\pi}{4}\right)$:

$ f\left(\frac{\pi}{6}\right) - f\left(\frac{\pi}{4}\right) = \frac{32}{\sqrt{3}} - 4\sqrt{2} $

To simplify and match the format of the options, we can write:

$ \frac{32}{\sqrt{3}} - 4\sqrt{2} = \frac{32}{\sqrt{3}} - \frac{4\sqrt{2} \cdot \sqrt{3}}{\sqrt{3}} = \frac{32 - 4\sqrt{6}}{\sqrt{3}} = \frac{4(8 - \sqrt{6})}{\sqrt{3}} $

This result corresponds to Option C.