Mathematics - Indefinite Integration Question with Solution | TestHub

Let $f(x)=\int \frac{dx}{{\sin }^{6}x}$ $$f(x)=\int {\mathrm{cosec}}^{6}xdx$$ From reduction formula, we have $$\begin{array}{rl}& {\mathrm{I}}_{\mathrm{n}}=\int {\mathrm{cosec}}^{n}xdx\\ & =-\frac{{\mathrm{cosec}}^{n-2}x\cot x}{n-1}+\frac{n-2}{n-1}{\mathrm{I}}_{n-2}\end{array}$$ $$\begin{array}{c}\therefore f(x)=-\frac{{\mathrm{cosec}}^{4}x\cot x}{5}+\frac{4}{5}\left[\frac{-{\mathrm{cosec}}^{2}x\cot x}{3}+\frac{2}{3}{\mathrm{I}}_2\right]\\ =-\frac{{\mathrm{cosec}}^{4}x\cot x}{5}-\frac{4}{15}{\mathrm{cosec}}^{2}x\cdot \cot x+\frac{8}{15}[-\cot x]\\ =\frac{-{\left(1+{\cot }^{2}x\right)}^2\cdot \cot x}{5}-\frac{4}{15}\left(1+{\cot }^{2}x\right)\cot x\\ =\frac{-1}{5}\left[1+{\cot }^{4}x+2{\cot }^{2}x\right]\cot x-\frac{4}{15}[-\cot x)\left(\because {\mathrm{cosec}}^{2}x=1+{\cot }^{2}x\right)\\ =\frac{-1}{5}\left[\cot x+{\cot }^{3}x\right]\\ \frac{-4}{15}{\cot }^{3}x-\frac{4}{15}{\cot }^{3}x-\frac{8}{15}\cot x\\ \cot x]\end{array}$$ $$\begin{array}{rl}& =\frac{-15}{15}\cot x-\frac{{\cot }^{5}x}{5}-\frac{10}{15}{\cot }^{3}x\\ & =\frac{-{\cot }^{5}x}{5}-\frac{2}{3}{\cot }^{3}x-\cot x\end{array}$$ It is a polynomial of degree 5 in $\cot x$.