Evaluate the following integrals: ^5x dx

Question

Evaluate the following integrals:
$\int\cot^5\text{x}\text{ dx}$

✓

Answer

Let $\text{I}=\int\cot^5\text{x}\text{ dx}$ Then
$\text{I}=\int\cot^3\text{x}\times\big(\text{cosec}^2-1\big)\text{dx}$
$=\int\cot^3\text{x}\times\big(\text{cosec}^2\text{x}-1\big)\text{dx}$
$=\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}-\int\cot^3\text{x}\text{dx}$
$=\int\cot^3\text{x}\text{ cosec}^2\text{x}\text{ dx}-\int\big(\text{cosec}^2\text{x}-1\big)\cot\text{dx}$
$=\int\cot^3\text{x}\text{cosec}^2\text{x}\text{ dx}-\int\text{cosec}^2\text{x}\cot\text{x}\text{ dx}+\int\cot\text{x}\text{ dx}$
$\text{I}=\int\cot^3\text{x}\text{cosec}^2\text{x}\text{ dx}-\int\text{cosec}^2\text{x}\cot\text{x}\text{ dx}+\int\cot\text{x}\text{ dx}$
Substituting $\cot\text{x}=\text{t}$ and $-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$ in first two integral, we get
$\text{I}=\int\text{t}^3(-\text{dt})-\int\text{t}\times(-\text{dt})+\int\cot\text{x}\text{ dx}$
$=-\frac{\text{t}^4}{4}+\frac{\text{t}^2}{2}+\log|\sin\text{x}|+\text{C}$
$=-\frac{1}{4}\cot^4\text{x}+\frac{1}{2}\cot^2\text{x}+\log|\sin\text{x}|+\text{C}$
$\therefore\ \text{I}=-\frac{1}{4}\cot^4\text{x}+\frac{1}{2}\cot^2\text{x}+\log|\sin\text{x}|+\text{C}$

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