Question

Let \(\int \frac{1 - 5 \cos^2 x}{\sin^5 x \cos^2 x} dx = f(x) + c\) then find \(f\left(\frac{\pi}{4}\right) - f\left(\frac{\pi}{6}\right)\) :

• A. \(4\sqrt{2} - \frac{32}{\sqrt{3}}\)
• B. \(4\sqrt{3} - \frac{32}{\sqrt{3}}\)
• C. \(4\sqrt{3} - \frac{32}{\sqrt{2}}\)
• D. \(4\sqrt{2} - \frac{32}{\sqrt{2}}\)

Hint:

When you see a combination of powers of \(\sin x\) and \(\cos x\) in the denominator with a subtraction in the numerator, consider integration by parts on one part to cancel out the difficult remainder of the other part.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is an indefinite integral problem involving trigonometric functions. We need to simplify the integrand into a form that is easily integrable or recognizable as a derivative of a quotient.
Step 2: Key Formula or Approach:
Split the numerator and simplify the fractions:
\[ \int \frac{1 - 5 \cos^2 x}{\sin^5 x \cos^2 x} dx = \int \frac{\sec^2 x}{\sin^5 x} dx - 5 \int \csc^5 x dx \] Step 3: Detailed Explanation:
Let \(I = \int \frac{\sec^2 x}{\sin^5 x} dx\). We can use integration by parts by taking \(u = \csc^5 x\) and \(dv = \sec^2 x dx\).
Then \(du = -5 \csc^5 x \cot x dx\) and \(v = \tan x\).
Using \(\int u dv = uv - \int v du\):
\[ I = \csc^5 x \tan x - \int \tan x (-5 \csc^5 x \cot x) dx \] Since \(\tan x \cot x = 1\):
\[ I = \frac{\tan x}{\sin^5 x} + 5 \int \csc^5 x dx \] Now substitute this back into the original integral:
\[ \int \frac{1 - 5 \cos^2 x}{\sin^5 x \cos^2 x} dx = \left( \frac{\tan x}{\sin^5 x} + 5 \int \csc^5 x dx \right) - 5 \int \csc^5 x dx \] \[ = \frac{\tan x}{\sin^5 x} = \frac{\sin x / \cos x}{\sin^5 x} = \frac{1}{\sin^4 x \cos x} \] Thus, \(f(x) = \frac{1}{\sin^4 x \cos x}\).
Evaluating the difference:
\(f(\pi/4) = \frac{1}{(1/\sqrt{2})^4 \cdot (1/\sqrt{2})} = \frac{1}{(1/4) \cdot (1/\sqrt{2})} = 4\sqrt{2}\).
\(f(\pi/6) = \frac{1}{(1/2)^4 \cdot (\sqrt{3}/2)} = \frac{1}{(1/16) \cdot (\sqrt{3}/2)} = \frac{32}{\sqrt{3}}\).
\[ f(\pi/4) - f(\pi/6) = 4\sqrt{2} - \frac{32}{\sqrt{3}} \] Step 4: Final Answer:
The result is \(4\sqrt{2} - \frac{32}{\sqrt{3}}\).