Question

If \(\int(\sin x)^{-11/2} (\cos x)^{-5/2} dx = \frac{p_1}{q_1}(\cot x)^{9/2} - \frac{p_2}{q_2}(\cot x)^{5/2} - \frac{p_3}{q_3}(\cot x)^{1/2} + \frac{p_4}{q_4}(\cot x)^{-3/2} + C\) where H.C.F. \(\{p_i, q_i\} = 1\) \& \(i \in \{1, 2, 3, 4\}\). Then value of \(\sum_{i=1}^{4}(p_i +q_i)\) is:

• A. 30
• B. 25
• C. 24
• D. 34

Hint:

For integrals of the form \(\int\sin^m x \cos^n x \, dx\), if \(m+n\) is a negative even integer, the substitution \(t = \tan x\) or \(t = \cot x\) is a very effective strategy to simplify the integral.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are asked to evaluate a complex trigonometric integral and then compare the resulting coefficients with a given form to find the sum of certain values.

Step 2: Key Formula or Approach:
The integral is: \[ I = \int (\sin x)^{-11/2} (\cos x)^{-5/2} \, dx \] The sum of the powers is \(-11/2 - 5/2 = -16/2 = -8\), which is a negative even integer. In such cases, substitution with \( t = \tan x \) or \( t = \cot x \) is effective. Let’s use \( t = \cot x \).
If \( t = \cot x \), then \( dt = -\csc^2 x \, dx \), so \( dx = -\sin^2 x \, dt \).
We also need expressions for \( \sin x \) and \( \cos x \) in terms of \( t \): From \( 1 + \cot^2 x = \csc^2 x \), we get \( \sin^2 x = \frac{1}{1 + \cot^2 x} = \frac{1}{1 + t^2} \).
From \( \tan^2 x = \frac{1}{\cot^2 x} = \frac{1}{t^2} \), we get \( \sec^2 x = 1 + \tan^2 x = 1 + \frac{1}{t^2} = \frac{t^2 + 1}{t^2} \), so \( \cos^2 x = \frac{t^2}{t^2 + 1} \).

Step 3: Detailed Explanation:
The integral becomes: \[ I = \int (\sin x)^{-11/2} (\cos x)^{-5/2} \, dx \] Substitute \( dx = -\sin^2 x \, dt \): \[ I = \int (\sin x)^{-11/2} (\cos x)^{-5/2} (-\sin^2 x) \, dt = -\int (\sin x)^{-7/2} (\cos x)^{-5/2} \, dt \] Substitute \( \sin x = (1 + t^2)^{-1/2} \) and \( \cos x = t(1 + t^2)^{-1/2} \): \[ I = -\int (1 + t^2)^{7/4} t^{-5/2} (1 + t^2)^{5/4} \, dt = -\int t^{-5/2} (1 + t^2)^3 \, dt \] Expanding \( (1 + t^2)^3 = 1 + 3t^2 + 3t^4 + t^6 \): \[ I = -\int (t^{-5/2} + 3t^{-1/2} + 3t^{3/2} + t^{7/2}) \, dt \] Now, integrate term by term: \[ I = - \left[ \frac{t^{-3/2}}{-3/2} + \frac{3t^{1/2}}{1/2} + \frac{3t^{5/2}}{5/2} + \frac{t^{9/2}}{9/2} \right] + C \] \[ I = -\left[ -\frac{2}{3} t^{-3/2} + 6 t^{1/2} + \frac{6}{5} t^{5/2} + \frac{2}{9} t^{9/2} \right] + C \] \[ I = -\frac{2}{9} t^{9/2} - \frac{6}{5} t^{5/2} - 6 t^{1/2} + \frac{2}{3} t^{-3/2} + C \] Substitute back \( t = \cot x \): \[ I = -\frac{2}{9} (\cot x)^{9/2} - \frac{6}{5} (\cot x)^{5/2} - 6 (\cot x)^{1/2} + \frac{2}{3} (\cot x)^{-3/2} + C \]
Step 4: Final Answer:
The value of the sum is 34.