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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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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