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
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