$\displaystyle \int (\tan x)^{-\frac{11}{2}} \sec^8 x , dx$ $= \int (\tan x)^{-\frac{11}{2}} (1+\tan^2 x)^3 \sec^2 x , dx$ Put $\tan x = t$ $\Rightarrow \int t^{-\frac{11}{2}} (1+t^2)^3 , dt$ $= \int \left(t^{-\frac{11}{2}} + 3t^{-\frac{7}{2}} + 3t^{-\frac{3}{2}} + t^{\frac{1}{2}}\right) dt$ $= -\frac{2}{9}t^{-\frac{9}{2}} - \frac{6}{5}t^{-\frac{5}{2}} - 2t^{-\frac{1}{2}} + \frac{2}{3}t^{\frac{3}{2}} + C$ Back substitute $t = \cot x$ $\Rightarrow p_1 = 2,; p_2 = 6,; p_3 = 2,; p_4 = 2$ $q_1 = 9,; q_2 = 5,; q_3 = 1,; q_4 = 3$ $\displaystyle \frac{15p_1p_2p_3p_4}{q_1q_2q_3q_4} = \frac{15 \cdot 2 \cdot 6 \cdot 2 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3} = 16$