Question

If \[ \int \frac{1-5\cos^2 x}{\sin^5 x\cos^2 x}\,dx=f(x)+C, \] where \( C \) is the constant of integration, then \[ f\!\left(\frac{\pi}{6}\right)-f\!\left(\frac{\pi}{4}\right) \] is equal to:

• A. \( \dfrac{1}{\sqrt{3}}(26-\sqrt{3}) \)
• B. \( \dfrac{1}{\sqrt{3}}(26+\sqrt{3}) \)
• C. \( \dfrac{4}{\sqrt{3}}(8-\sqrt{6}) \)
• D. \( \dfrac{2}{\sqrt{3}}(4+\sqrt{6}) \)

Hint:

Always try to reduce trigonometric integrals to powers of \( \tan x \) or \( \sec x \); substitutions then become straightforward.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to integrate the given expression. A good strategy is to split the integrand or find a function whose derivative matches it.
Step 2: Detailed Explanation:
Let \(f(x) = \frac{1}{\sin^4 x \cos x}\). Let's check its derivative:
\[ \frac{d}{dx} (\sin^{-4} x \cos^{-1} x) = -4 \sin^{-5} x \cos x \cdot \cos^{-1} x + \sin^{-4} x \cdot (-1) \cos^{-2} x (-\sin x) \] \[ = \frac{-4}{\sin^5 x} + \frac{\sin x}{\sin^4 x \cos^2 x} = \frac{-4 \cos^2 x + \sin^2 x}{\sin^5 x \cos^2 x} = \frac{-4 \cos^2 x + 1 - \cos^2 x}{\sin^5 x \cos^2 x} \] \[ = \frac{1 - 5 \cos^2 x}{\sin^5 x \cos^2 x} \] This matches our integrand. Thus, \(f(x) = \frac{1}{\sin^4 x \cos x}\).
Now, calculate \(f(\pi/6) - f(\pi/4)\):
At \(x = \pi/6\): \(\sin(\pi/6) = 1/2, \cos(\pi/6) = \sqrt{3}/2\).
\[ f(\pi/6) = \frac{1}{(1/16) \cdot (\sqrt{3}/2)} = \frac{32}{\sqrt{3}} \] At \(x = \pi/4\): \(\sin(\pi/4) = 1/\sqrt{2}, \cos(\pi/4) = 1/\sqrt{2}\).
\[ f(\pi/4) = \frac{1}{(1/4) \cdot (1/\sqrt{2})} = 4\sqrt{2} \] Difference:
\[ \frac{32}{\sqrt{3}} - 4\sqrt{2} = \frac{32 - 4\sqrt{6}}{\sqrt{3}} = \frac{4}{\sqrt{3}} (8 - \sqrt{6}) \] Step 3: Final Answer:
The result is \(\frac{4}{\sqrt{3}}(8 - \sqrt{6})\).