Question

If \[ \int (\cos x)^{-5/2}(\sin x)^{-11/2}\,dx = \frac{p_1}{q_1}(\cot x)^{9/2} + \frac{p_2}{q_2}(\cot x)^{5/2} + \frac{p_3}{q_3}(\cot x)^{1/2} - \frac{p_4}{q_4}(\cot x)^{-3/2} + C, \] where \(C\) is the constant of integration, then find the value of \[ \frac{15\,p_1p_2p_3p_4}{q_1q_2q_3q_4}. \]

• A. \(14\)
• B. \(16\)
• C. \(10\)
• D. \(9\)

Hint:

For integrals involving powers of \(\sin x\) and \(\cos x\):
Use \(\cot x\) substitution when both powers are negative
Convert everything into powers of \(\cot x\)
Expand using binomial theorem and integrate termwise

Answer 1

The Correct Option is B

Concept: Integrals of the form \[ \int (\sin x)^m(\cos x)^n\,dx \] with negative fractional powers can often be simplified using the substitution: \[ t=\cot x \quad \Rightarrow \quad dt = -\csc^2 x\,dx \] and rewriting all trigonometric functions in terms of \(\cot x\).
Step 1: Rewrite the integrand. \[ (\cos x)^{-5/2}(\sin x)^{-11/2} = (\cot x)^{-5/2}(\sin x)^{-8} \] Using: \[ \sin^2 x = \frac{1}{1+\cot^2 x} \] \[ (\sin x)^{-8} = (1+\cot^2 x)^4 \] Thus, \[ \int (\cot x)^{-5/2}(1+\cot^2 x)^4 \csc^2 x\,(-d(\cot x)) \]
Step 2: Substitute \(t=\cot x\). \[ I = -\int t^{-5/2}(1+t^2)^4\,dt \] Expand: \[ (1+t^2)^4 = 1+4t^2+6t^4+4t^6+t^8 \] So, \[ I = -\int \left( t^{-5/2} +4t^{-1/2} +6t^{3/2} +4t^{7/2} +t^{11/2} \right)dt \]
Step 3: Integrate term by term. \[ \int t^{-5/2}dt = -\frac{2}{3}t^{-3/2} \] \[ \int t^{-1/2}dt = 2t^{1/2} \] \[ \int t^{3/2}dt = \frac{2}{5}t^{5/2} \] \[ \int t^{7/2}dt = \frac{2}{9}t^{9/2} \] \[ \int t^{11/2}dt = \frac{2}{13}t^{13/2} \] After applying the negative sign and arranging in descending powers: \[ I = \frac{2}{9}(\cot x)^{9/2} + \frac{12}{5}(\cot x)^{5/2} + 8(\cot x)^{1/2} - \frac{2}{3}(\cot x)^{-3/2} + C \]
Step 4: Identify coefficients. \[ \frac{p_1}{q_1}=\frac{2}{9},\quad \frac{p_2}{q_2}=\frac{12}{5},\quad \frac{p_3}{q_3}=8,\quad \frac{p_4}{q_4}=\frac{2}{3} \] Thus: \[ p_1=2,\ q_1=9;\quad p_2=12,\ q_2=5;\quad p_3=8,\ q_3=1;\quad p_4=2,\ q_4=3 \]
Step 5: Compute the required value. \[ \frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4} = \frac{15\times2\times12\times8\times2}{9\times5\times1\times3} \] \[ = \frac{5760}{360}=16 \] \[ \boxed{16} \]