If $I_n=\int {\cot }^{n}x$ then $I_0+I_1+2(I_2+I_3)+I_4+I_5=$

$$\begin{gathered} I_n=\int {\cot }^{\mathrm{n}}\mathrm{x} \mathrm{dx} \\ \Rightarrow {\mathrm{I}}_{\mathrm{n}}+{\mathrm{I}}_{\mathrm{n}-2}=\frac{-{\cot }^{\mathrm{n}-1}\mathrm{x}}{\mathrm{n}-1} \\ {\mathrm{I}}_2+{\mathrm{I}}_0=\frac{-\cot \mathrm{x}}{1} \\ {\mathrm{I}}_3+{\mathrm{I}}_1=\frac{-{\cot }^2\mathrm{x}}{2} \\ {\mathrm{I}}_4+{\mathrm{I}}_2=\frac{-{\cot }^3\mathrm{x}}{3} \\ {\mathrm{I}}_5+{\mathrm{I}}_3=\frac{-{\cot }^4\mathrm{x}}{4} \\ \mathrm{Add} \mathrm{above} \mathrm{equations} \\ \Rightarrow {\mathrm{I}}_0+{\mathrm{I}}_1+2({\mathrm{I}}_2+{\mathrm{I}}_3)+{\mathrm{I}}_4+{\mathrm{I}}_5 \\ =-(\cot \mathrm{x}+\frac{co{\mathrm{t}}^2\mathrm{x}}{2}+\frac{{\cot }^3\mathrm{x}}{3}+\frac{{\cot }^4\mathrm{x}}{4})+\mathrm{C} \end{gathered}$$