Step 1: Simplify the Integral

Let $I = \int (\cos x)^{-5/2} (\sin x)^{-11/2} dx = \int \frac{dx}{\cos^{5/2}x \sin^{11/2}x}$

To integrate, we rearrange terms to use $\sec x$ and $\tan x$. Multiply numerator and denominator by $\sec^{11/2} x$ (effectively dividing denominator by $\cos^{11/2} x$):

$I = \int \frac{\sec^8 x}{\tan^{11/2} x} dx$

Step 2: Substitution

Let $t = \tan x$. Then $dt = \sec^2 x dx$.

We can rewrite $\sec^8 x$ as $\sec^6 x \cdot \sec^2 x = (\sec^2 x)^3 \cdot \sec^2 x = (1 + t^2)^3 \sec^2 x$.

Substituting into the integral:

$I = \int \frac{(1 + t^2)^3}{t^{11/2}} dt$

Step 3: Expansion and Integration

Expand $(1 + t^2)^3$ using the binomial theorem:

$I = \int \frac{1 + 3t^2 + 3t^4 + t^6}{t^{11/2}} dt = \int (t^{-11/2} + 3t^{-7/2} + 3t^{-3/2} + t^{1/2}) dt$

Integrate term by term:

$I = \frac{t^{-9/2}}{-9/2} + 3\frac{t^{-5/2}}{-5/2} + 3\frac{t^{-1/2}}{-1/2} + \frac{t^{3/2}}{3/2} + C$

$I = -\frac{2}{9}t^{-9/2} - \frac{6}{5}t^{-5/2} - 6t^{-1/2} + \frac{2}{3}t^{3/2} + C$

Step 4: Back-Substitution

Convert back to $x$ using $t = \tan x = \frac{1}{\cot x}$:

$I = -\frac{2}{9}(\cot x)^{9/2} - \frac{6}{5}(\cot x)^{5/2} - 6(\cot x)^{1/2} + \frac{2}{3}(\cot x)^{-3/2} + C$

Step 5: Coefficient Comparison

Given the form: $\frac{P_{1}}{q_{1}}(\cot x)^{9/2}+\frac{P_{2}}{q_{2}}(\cot x)^{5/2}+\frac{P_{3}}{q_{3}}(\cot x)^{1/2}-\frac{P_{4}}{q_{4}}(\cot x)^{-3/2}+C$

Comparing coefficients:

$\frac{P_1}{q_1} = -\frac{2}{9}$

$\frac{P_2}{q_2} = -\frac{6}{5}$

$\frac{P_3}{q_3} = -6$

$-\frac{P_4}{q_4} = \frac{2}{3} \implies \frac{P_4}{q_4} = -\frac{2}{3}$

Step 6: Final Calculation

Value = $15 \cdot \frac{P_1}{q_1} \cdot \frac{P_2}{q_2} \cdot \frac{P_3}{q_3} \cdot \frac{P_4}{q_4}$

$= 15 \cdot (-\frac{2}{9}) \cdot (-\frac{6}{5}) \cdot (-6) \cdot (-\frac{2}{3})$

Since there are four negatives, the result is positive:

$= 15 \cdot \frac{2}{9} \cdot \frac{6}{5} \cdot 6 \cdot \frac{2}{3} = \frac{15 \cdot 2 \cdot 6 \cdot 6 \cdot 2}{9 \cdot 5 \cdot 3} = \frac{2160}{135} = 16$