Question
Evaluate the following integrals:
$\int\text{cosec}^43\text{x}\text{ dx}$
$\int\text{cosec}^43\text{x}\text{ dx}$
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Answer
Let $\text{I}=\int\text{cosec}^43\text{x}\text{ dx}$ Then
$\text{I}=\int\text{cosec}^23\text{x }\text{cosec}^23\text{x}\text{ dx}$
$=\int\big(1+\cot^23\text{x}\big)\text{cosec}^23\text{x}\text{ dx}$
$=\int\big(\text{cosec}^23\text{x}+\cot^23\text{x }\text{cosec}^23\text{x}\big)\text{dx}$
$\text{I}=\int\text{cosec}^23\text{x}\text{ dx}+\int\cot^23\text{x }\text{cosec}^23\text{x}\text{ dx}$
Sunbstituting $\cot3\text{x}=\text{t}$ and $\text{cosec}^23\text{x}\text{ dx}=-\text{dt}$ in $2^{nd}$ integral, we get
$\text{I}=\int\text{cosec}^23\text{x}-\int\text{t}^2\frac{\text{dt}}{3}$
$=\frac{-1}{3}\cot3\text{x}-\frac{\text{t}^3}{9}+\text{C}$
$=\frac{-1}{3}\cot3\text{x}-\frac{\cot^33\text{x}}{9}+\text{C}$
$\therefore\ \text{I}=\frac{-1}{3}\cot3\text{x}-\frac{1}{9}\infty\text{t}^3\text{3x}+\text{C}$
$\text{I}=\int\text{cosec}^23\text{x }\text{cosec}^23\text{x}\text{ dx}$
$=\int\big(1+\cot^23\text{x}\big)\text{cosec}^23\text{x}\text{ dx}$
$=\int\big(\text{cosec}^23\text{x}+\cot^23\text{x }\text{cosec}^23\text{x}\big)\text{dx}$
$\text{I}=\int\text{cosec}^23\text{x}\text{ dx}+\int\cot^23\text{x }\text{cosec}^23\text{x}\text{ dx}$
Sunbstituting $\cot3\text{x}=\text{t}$ and $\text{cosec}^23\text{x}\text{ dx}=-\text{dt}$ in $2^{nd}$ integral, we get
$\text{I}=\int\text{cosec}^23\text{x}-\int\text{t}^2\frac{\text{dt}}{3}$
$=\frac{-1}{3}\cot3\text{x}-\frac{\text{t}^3}{9}+\text{C}$
$=\frac{-1}{3}\cot3\text{x}-\frac{\cot^33\text{x}}{9}+\text{C}$
$\therefore\ \text{I}=\frac{-1}{3}\cot3\text{x}-\frac{1}{9}\infty\text{t}^3\text{3x}+\text{C}$
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