If $$\int (\sin x) ^{\frac{-11}{2}}(\cos x)^{\frac{-5}{2}}dx= -\frac{p_{1}}{q_{1}}(\cot x)^{\frac{9}{2}}-\frac{p_{2}}{q_{2}}(\cot x)^{\frac{5}{2}}-\frac{p_{3}}{q_{3}}(\cot x)^{\frac{1}{2}}+ \frac{p_{4}}{q_{4}}(\cot x)^{\frac{-3}{2}}+C,\text{ where }p_{i} \text{ and } q_{i} $$ are positive integers with $$gcd(p_{i}, q_{i}) = l$$ for i = l, 2, 3, 4 and C is the constant of integration, then $$\frac{15p_{1}p_{2}p_{3}p_{4}}{q_{1}q_{2}q_{3}q_{4}} $$ is equal to ______

To solve $$I = \int (\sin x)^{-\frac{11}{2}} (\cos x)^{-\frac{5}{2}} dx$$, we transform the integrand into terms of $$\cot x$$ and $$\csc x$$.

1. Transform the Integrand

Multiply and divide by $$(\sin x)^{-\frac{5}{2}}$$ to create a $$\cot x$$ term:

$$I = \int \frac{1}{(\sin x)^{11/2} (\cos x)^{5/2}} dx = \int \frac{1}{(\sin x)^8 (\frac{\cos x}{\sin x})^{5/2}} dx = \int \frac{\csc^8 x}{(\cot x)^{5/2}} dx$$

2. Substitution

Let $$t = \cot x \implies dt = -\csc^2 x dx$$.

Using $$\csc^2 x = 1 + \cot^2 x = 1 + t^2$$, we rewrite $$\csc^8 x$$ as $$(\csc^2 x)^3 \cdot \csc^2 x = (1 + t^2)^3 \csc^2 x$$.

$$I = -\int \frac{(1 + t^2)^3}{t^{5/2}} dt = -\int \frac{1 + 3t^2 + 3t^4 + t^6}{t^{5/2}} dt$$

$$I = -\int (t^{-5/2} + 3t^{-1/2} + 3t^{3/2} + t^{7/2}) dt$$

3. Integrate

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

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

4. Compare with Given Form

Rearranging to match $$-\frac{p_1}{q_1}(\cot x)^{\frac{9}{2}} - \frac{p_2}{q_2}(\cot x)^{\frac{5}{2}} - \frac{p_3}{q_3}(\cot x)^{\frac{1}{2}} + \frac{p_4}{q_4}(\cot x)^{-\frac{3}{2}}$$:

• $$\frac{p_1}{q_1} = \frac{2}{9}$$
• $$\frac{p_2}{q_2} = \frac{6}{5}$$
• $$\frac{p_3}{q_3} = 6 = \frac{6}{1}$$
• $$\frac{p_4}{q_4} = \frac{2}{3}$$

5. Calculate Final Value

$$\frac{15 p_1 p_2 p_3 p_4}{q_1 q_2 q_3 q_4} = 15 \left( \frac{2}{9} \right) \left( \frac{6}{5} \right) \left( \frac{6}{1} \right) \left( \frac{2}{3} \right)$$

$$= 15 \cdot \frac{2 \cdot 6 \cdot 6 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3} = 15 \cdot \frac{144}{135} = 15 \cdot \frac{16}{15} = \mathbf{16}$$