Evalute the following integrals: cosec^2x 1+ dx - Vidyadip

Question

Evalute the following integrals:
$\int\frac{\text{cosec}^2\text{x}}{1+\cot\text{x}}\text{dx}$

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Answer

Let $\text{I}=\int\frac{\text{cosec}^2\text{x}}{1+\cot\text{x}}\text{dx}$
Putting $\cot\text{x}=\text{t}$
$\Rightarrow-\text{cosec}^2\text{x}=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow\text{cosec}^2\text{x dx}=-\text{dt}$
$\therefore\text{I}=\int\frac{-\text{dt}}{1+\text{t}}$
$=-\text{ln}|1+\text{t}|+\text{C}$
$=-\text{ln}|1+\cot\text{x}|+\text{C}\ \big[\because\text{t}=\cot\text{x}\big]$

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