Mathematics

# Evaluate the following integrals:$\displaystyle \int { cosec ^{ 4 }{ 3x } } dx$

##### SOLUTION
$I=\displaystyle\int{{cosec}^{4}{3x}dx}$

$I=\displaystyle\int{{cosec}^{2}{3x}\,{cosec}^{2}{3x}dx}$

$I=\displaystyle\int{\left(1+{\cot}^{2}{3x}\right){cosec}^{2}{3x}dx}$

$I=\displaystyle\int{{\csc}^{2}{3x}dx}+\displaystyle\int{{\cot}^{2}{3x}{cosec}^{2}{3x}dx}$

Let

$t=\cot{3x}\Rightarrow\,dt=-3{cosec}^{2}{3x}\,dx$

$I=-\dfrac{1}{3}\cot{3x}-\dfrac{1}{3}\displaystyle\int{{t}^{2}dt}$

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

$I=-\dfrac{1}{3}\cot{3x}-\dfrac{1}{9}{\cot}^{3}{3x}+c$ where $t=\cot{3x}$

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

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