$\int {\sin }^{-7/5}{\mathrm{xcos}}^{-3/5}\mathrm{dx}=-k(\cot \mathrm{x}{)}^{2/5}+\mathrm{C} \mathrm{then} \mathrm{k}=$ 

Here, 
$\mathrm{p}=-\frac{7}{5},\mathrm{q}=-\frac{3}{5}$
$\begin{array}{l}\mathrm{p}+\mathrm{q}=-2\\ \mathrm{I}=\int {\sin }^{-7/5}{\mathrm{xcos}}^{-3/5}\mathrm{xdx}=\int \frac{{\cos }^{-3/5}\mathrm{x}}{{\sin }^{-3/5}{\mathrm{xsin}}^2\mathrm{x}}\mathrm{dx}\\ =\int (\cot \mathrm{x}{)}^{-3/5}{\mathrm{cosec}}^2\mathrm{xdx}\end{array}$
Put, $\cot \mathrm{x}=\mathrm{t}\Rightarrow {\mathrm{cosec}}^2\mathrm{xdx}=-\mathrm{dt}$
So, $\mathrm{I}=-\int {\mathrm{t}}^{-3/5}\mathrm{dt}=-\frac{5}{2}(\cot \mathrm{x}{)}^{2/5}+\mathrm{C}$