$\int {\sec }^2{\mathrm{xcosec}}^4\mathrm{xdx}=-\frac{1}{3}{\cot }^3\mathrm{x}+\mathrm{ktanx}-2\mathrm{cotx}+\mathrm{c}\Rightarrow \mathrm{k}=$

$$\begin{gathered} \int {\sec }^2{\mathrm{xcosec}}^4\mathrm{xdx}=\int \frac{{\sin }^2\mathrm{x}+{\cos }^2\mathrm{x}}{{\cos }^2\mathrm{x}\cdot {\sin }^4\mathrm{x}}\mathrm{dx} \\ G.I=\int \frac{{\sin }^2\mathrm{x}}{{\cos }^2\mathrm{x}\cdot {\sin }^4\mathrm{x}}\mathrm{dx}+\int \frac{{\cos }^2\mathrm{x}}{{\cos }^2\mathrm{x}\cdot {\sin }^4\mathrm{x}}\mathrm{dx} \\ G.I=\int \frac{1}{{\cos }^2\mathrm{x}\cdot {\sin }^2\mathrm{x}}dx+\int \frac{1}{{\sin }^4\mathrm{x}}dx \\ G.I=\int \left({\sec }^{2}x+{\mathrm{cosec}}^{2}x\right)dx+\int \cos e{c}^{2}x(1+co{t}^{2}x)dx \\ G.I=\int se{c}^{2}xdx+2\int \cos e{c}^{2}xdx+\int co{t}^{2}x\cos e{c}^{2}dx \\ G.I=\tan x-2cotx-\frac{co{t}^{3}x}{3} \\ \therefore k=1 \end{gathered}$$