$f(x)=\int \frac{d x}{\sin ^6 x}$ is a polynomial of degree

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Aicharya · Accepted Answer

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} 	herefore 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ight] \ =-\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ight)^2 \cdot \cot x}{5}-\frac{4}{15}\left(1+\cot ^2 xight) \cot x \ =\frac{-1}{5}\left[1+\cot ^4 x+2 \cot ^2 xight] \cot x-\frac{4}{15}[-\cot x)\left(\because \operatorname{cosec}^2 x=1+\cot ^2 xight) \ =\frac{-1}{5}\left[\cot x+\cot ^3 xight] \ \frac{-4}{15} \cot ^3 x-\frac{4}{15} \cot ^3 x-\frac{8}{15} \cot x \ \cot x] \end{gathered} $ $ \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} $ It is a polynomial of degree 5 in $\cot x$

$f(x)=\int \frac{d x}{\sin ^6 x}$ is a polynomial of degree

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