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प्रश्न
\[\int\frac{1 - \cot x}{1 + \cot x} dx\]
योग
उत्तर
\[\text{Let I} = \int\left( \frac{1 - \cot x}{1 + \cot x} \right)dx\]
\[ = \int\left( \frac{1 - \frac{\cos x}{\sin x}}{1 + \frac{\cos x}{\sin x}} \right)dx\]
\[ = \int\left( \frac{\sin x - \cos x}{\sin x + \cos x} \right)dx\]
\[\text{Putting }\sin x + \cos x = t\]
\[ \Rightarrow \left( \cos x - \sin x \right)dx = dt\]
\[ \Rightarrow \left( \sin x - \cos x \right)dx = - dt\]
\[ \therefore I = \int\frac{- dt}{t}\]
\[ = - \text{ln }\left| t \right| + C\]
\[ = - \text{ln} \left| \sin x + \cos x \right| + C\]
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