Evaluate the following integrals: ^6x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\cot^6\text{x}\text{ dx}$

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Answer

$\int\cot^6\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\big(\text{cosec}^2-1\big)\text{dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^4\text{x}\text{dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}-\text{ cosec}^2\text{x}\text{ dx}-\int\big(\text{cosec}^2\text{x}-1\big)\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}+\int\cot^2\text{x}\text{ dx}$
$=\int\cot^4\text{x}\cdot\text{ cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}+\int(\text{cosec}^2\text{x}-1)\text{ dx}$
Now, let $\text{I}_1=\int\cot^4\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}$
And $\text{I}_2=\int(\text{cosec}^2\text{x}-1)\text{dx}$
First we integrate $I_1$
$\text{I}_1=\int\cot^4\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}-\int\cot^2\text{x}\cdot\text{cosec}^2\text{x}\text{ dx}$
Let $\cot\text{x}=\text{t}$
$-\text{cosec}^2\text{x}\text{ dx}=\text{dt}$
$\text{cosec}^2\text{x}\text{ dx}=-\text{dt}$
$\text{I}_1=-\int\text{t}^4\text{dt}+\int\text{t}^2\text{dt}$
$=\frac{-\text{t}^5}{5}+\frac{\text{t}^3}{3}+\text{C}_1$
$=-\frac{\cot^5\text{x}}{5}+\frac{\cot^3\text{x}}{3}+\text{C}_1$
Now we integrate $I_2$
$\text{I}_2=\int(\text{cosec}^2\text{x}-1)\text{dx}$
$=-\cot\text{x}-\text{x}+\text{C}_1$
Now, $\int\cot^6\text{x}\text{ dx}=\text{I}_1+\text{I}_2$
$=-\frac{1}{5}\cot^5\text{x}+\frac{1}{3}\cot^3\text{x}-\cot\text{x}-\text{x}+\text{x}+\text{C}_1+\text{C}_2$
$=-\frac{1}{5}\cot^5\text{x}+\frac{1}{3}\cot^3\text{x}-\cot\text{x}-\text{x}+\text{C}$ $\big[\therefore\text{ C}=\text{C}_1+\text{C}_2\big]$

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