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