If $\int \sqrt{\frac{{\cos }^3\mathrm{x}}{{\sin }^{11}\mathrm{x}}}\mathrm{dx}=-2\left\{{\mathrm{Atan}}^{-9/2}\mathrm{x}+{\mathrm{Btan}}^{-5/2}\mathrm{x}\right\}+\mathrm{C}$  then $\frac{1}{\mathrm{A}}+\frac{1}{\mathrm{B}}$ is equal to

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

So, we divide numerator and denominator by cos⁴ x

$\begin{array}{l}\therefore \mathrm{I}=\int \sqrt{\frac{{\cos }^3\mathrm{x}}{{\sin }^{11}\mathrm{x}}}\mathrm{dx}=\int \frac{{\cos }^{3/2}\mathrm{x}}{{\sin }^{11/2}\mathrm{x}}\mathrm{dx}\\ \Rightarrow \mathrm{I}=\int \frac{{\cos }^{3/2}\mathrm{x}\times {\cos }^4\mathrm{x}}{{\sin }^{11/2}\mathrm{x}}\times \frac{1}{{\cos }^4\mathrm{x}}\mathrm{dx}\\ \Rightarrow \mathrm{I}=\int \frac{1}{{\tan }^{11/2}\mathrm{x}}\left(1+{\tan }^2\mathrm{x}\right)\mathrm{d}(\tan \mathrm{x})\\ \Rightarrow \mathrm{I}=\int \left({\tan }^{-11/2}\mathrm{x}+{\tan }^{-7/2}\mathrm{x}\right)\mathrm{d}(\tan \mathrm{x})\\ \Rightarrow \mathrm{I}=-\frac{2}{9}{\tan }^{-9/2}\mathrm{x}-\frac{2}{5}{\tan }^{-5/2}\mathrm{x}+\mathrm{C}\\ \Rightarrow \mathrm{I}=-2\left\{\frac{1}{9}{\tan }^{-9/2}\mathrm{x}+\frac{1}{5}{\tan }^{-5/2}\mathrm{x}\right\}+\mathrm{C}\end{array}$

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

$\therefore \frac{1}{\mathrm{A}}+\frac{1}{\mathrm{B}}=14$