f(x)=∫ (d x)/( ^6 x) is a polynomial of degree

Let $f(x)=\int \frac{d x}{\sin ^6 x}$ f(x)=\int \operatorname{cosec}^6 x d x From reduction formula, we have $$\begin{aligned} \mathrm{I}_{\mathrm{n}}=\int \operatorname{cosec}^ n x d x \\ & =-\frac{\operatorname{cosec}^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} \mathrm{I}_{n-2} \end{aligned}$$\begin{gathered} \therefore f(x)=-\frac{\operatorname{cosec}^4 x \cot x}{5}+\frac{4}{5}\left[\frac{-\operatorname{cosec}^2 x \cot x}{3}+\frac{2}{3} \mathrm{I}_{2}\right] \\ =-\frac{\operatorname{cosec}^4 x \cot x}{5}-\frac{4}{15} \operatorname{cosec}^2 x \cdot \cotx+\frac{8}{15}[-\cotx] \\ =\frac{-\left(1+\cot ^2 x\right)^2 \cdot \cot x}{5}-\frac{4}{15}\left(1+\cot ^2 x\right) \cot x \\ =\frac{-1}{5}\left[1+\cot ^4 x+2 \cot ^2 x\right] \cot x-\frac{4}{15}[-\cotx)\left(\because \operatorname{cosec}^2 x=1+\cot ^2 x\right) \\ =\frac{-1}{5}\left[\cotx+\cot ^3 x\right] \\ \frac{-4}{15} \cot ^3 x-\frac{4}{15} \cot ^3 x-\frac{8}{15} \cotx \\ \cot x] \end{gathered} $$\begin{aligned} & =\frac{-15}{15} \cotx-\frac{\cot ^5 x}{5}-\frac{10}{15} \cot ^3 x \\ & =\frac{-\cot ^5 x}{5}-\frac{2}{3} \cot ^3 x-\cot x \end{aligned}$$ It is a polynomial$of$degree 5 in \cot x$.