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

Let $\displaystyle f(x)=\int \frac{d x}{\sin ^6 x}$f(x)=∫sin6xdx​ $$f(x)=\int \operatorname{cosec}^6 x d x$$f(x)=∫cosec6xdx From reduction formula, we have $$\begin{aligned} & \mathrm{I}_{\mathrm{n}}=\int \operatorname{cosec}^n x d x \\ & =-\frac{\operatorname{cosec}^{n-2} x \cot x}{n-1}+\frac{n-2}{n-1} \mathrm{I}_{n-2} \end{aligned}$$​In​=∫cosecnxdx=−n−1cosecn−2xcotx​+n−1n−2​In−2​​ $$\begin{gathered} \therefore f(x)=-\frac{\operatorname{cosec}^4 x \cot x}{5}+\frac{4}{5}\left[\frac{-\operatorname{cosec}^2 x \cot x}{3}+\frac{2}{3} \mathrm{I}_2\right] \\ =-\frac{\operatorname{cosec}^4 x \cot x}{5}-\frac{4}{15} \operatorname{cosec}^2 x \cdot \cot x+\frac{8}{15}[-\cot x] \\ =\frac{-\left(1+\cot ^2 x\right)^2 \cdot \cot x}{5}-\frac{4}{15}\left(1+\cot ^2 x\right) \cot x \\ =\frac{-1}{5}\left[1+\cot ^4 x+2 \cot ^2 x\right] \cot x-\frac{4}{15}[-\cot x)\left(\because \operatorname{cosec}^2 x=1+\cot ^2 x\right) \\ =\frac{-1}{5}\left[\cot x+\cot ^3 x\right] \\ \frac{-4}{15} \cot ^3 x-\frac{4}{15} \cot ^3 x-\frac{8}{15} \cot x \\ \cot x] \end{gathered}$$∴f(x)=−5cosec4xcotx​+54​[3−cosec2xcotx​+32​I2​]=−5cosec4xcotx​−154​cosec2x⋅cotx+158​[−cotx]=5−(1+cot2x)2⋅cotx​−154​(1+cot2x)cotx=5−1​[1+cot4x+2cot2x]cotx−154​[−cotx)(∵cosec2x=1+cot2x)=5−1​[cotx+cot3x]15−4​cot3x−154​cot3x−158​cotxcotx]​ $$\begin{aligned} & =\frac{-15}{15} \cot x-\frac{\cot ^5 x}{5}-\frac{10}{15} \cot ^3 x \\ & =\frac{-\cot ^5 x}{5}-\frac{2}{3} \cot ^3 x-\cot x \end{aligned}$$​=15−15​cotx−5cot5x​−1510​cot3x=5−cot5x​−32​cot3x−cotx​ It is a polynomial of degree 5 in $\cot x$cotx.