Mathematics

Evaluate the following integrals:
$$\displaystyle \int { \cot ^{ 6 }{ x }  } dx$$


SOLUTION
$$\displaystyle\int{{\cot}^{6}{x}\,dx}$$

$$=\displaystyle\int{{\cot}^{4}{x}\,{\cot}^{2}{x}\,dx}$$

$$=\displaystyle\int{{\cot}^{4}{x}\left({cosec}^{2}{x}-1\right)\,dx}$$

$$=\displaystyle\int{{\cot}^{4}{x}\,{cosec}^{2}{x}\,dx}-\displaystyle\int{{\cot}^{4}{x}\,dx}$$

$$=\displaystyle\int{{\cot}^{4}{x}\,{cosec}^{2}{x}\,dx}-\displaystyle\int{{\cot}^{2}{x}{\cot}^{2}{x}\,dx}$$

$$=\displaystyle\int{{\cot}^{4}{x}\,{cosec}^{2}{x}\,dx}-\displaystyle\int{{\cot}^{2}{x}\left({cosec}^{2}{x}-1\right)\,dx}$$

$$=\displaystyle\int{{\cot}^{4}{x}\,{cosec}^{2}{x}\,dx}-\displaystyle\int{{\cot}^{2}{x}\,{cosec}^{2}{x}\,dx}+\displaystyle\int{{\cot}^{2}{x}\,dx}$$

$$=\displaystyle\int{{\cot}^{4}{x}\,{cosec}^{2}{x}\,dx}-\displaystyle\int{{\cot}^{2}{x}{cosec}^{2}{x}\,dx}+\displaystyle\int{\left({cosec}^{2}{x}-1\right)\,dx}$$

$$=\displaystyle\int{{\cot}^{4}{x}{cosec}^{2}{x}\,dx}-\displaystyle\int{{\cot}^{2}{x}{cosec}^{2}{x}\,dx}+\displaystyle\int{{cosec}^{2}{x}\,dx}-\displaystyle\int{dx}$$
Let $$t=\cot{x}\Rightarrow\,dt=-{cosec}^{2}{x}\,dx$$

$$=-\displaystyle\int{{t}^{4}\,dx}+\displaystyle\int{{t}^{2}\,dx}-\cot{x}-x+c$$

$$=-\dfrac{1}{5}{t}^{5}+\dfrac{1}{3}{t}^{3}-\cot{x}-x+c$$

$$=-\dfrac{1}{5}{\cot}^{5}{x}+\dfrac{1}{3}{\cot}^{3}{x}-\cot{x}-x+c$$ where 

$$t=\cot{x}$$
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