$\int \sqrt{\frac{\cos ^3 x}{\sin ^{11} x}} d x=-2\left\{A \tan ^{-9 / 2} x+B \tan ^{-5 / 2} x\right\}+C$, then

$A=\frac{1}{9}, B=\frac{1}{5}$

We observe that the sum of the exponents of $\cos x$ and $\sin x$ is $\left(\frac{3}{4}-\frac{11}{2}\right)=-4=$ a negative even integer.

So, we divide numerator and denominator by $\cos ^4 x$

∴  $I=\int \sqrt{\frac{\cos ^3 x}{\sin ^{11} x}} d x=\int \frac{\cos ^{3 / 2} x}{\sin ^{11 / 2} x} d x$

$\Rightarrow I=\int \frac{\cos ^{3 / 2} x \times \cos ^4 x}{\sin ^{11 / 2} x} \times \frac{1}{\cos ^4 x} d x$

$\Rightarrow I=\int \frac{1}{\tan ^{11 / 2} x}\left(1+\tan ^2 x\right) d(\tan x)$

$\Rightarrow I=\int\left(\tan ^{-11 / 2} x+\tan ^{-7 / 2} x\right) d(\tan x)$

$\Rightarrow I=-\frac{2}{9} \tan ^{-9 / 2} x-\frac{2}{5} \tan ^{-5 / 2} x+C$

$\Rightarrow I=-2\left\{\frac{1}{9} \tan ^{-9 / 2} x+\frac{1}{5} \tan ^{-5 / 2} x\right\}+C$

Hence, $A=\frac{1}{9}$ and $B=\frac{1}{5}$