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प्रश्न
\[\int\frac{1 - \cos x}{1 + \cos x} dx\]
बेरीज
उत्तर
\[\int\left( \frac{1 - \cos x}{1 + \cos x} \right)dx\]
\[ = \int\frac{\left( 1 - \cos x \right)^2}{1 - \cos^2 x}dx\]
\[ = \int\frac{1 + \cos^2 x - 2\cos x}{\sin^2 x}dx\]
\[ = \int \left( \frac{1}{\sin^2 x} + \frac{\cos^2 x}{\sin^2 x} - \frac{2\cos x}{\sin^2 x} \right)dx\]
\[ = \int \left( {cosec}^2 x + \cot^2 x - 2\cot x . \text{cosec x} \right)dx\]
\[ = \int \left( {cosec}^2 x + {cosec}^2 x - 1 - 2\cot x . cosec x \right)dx\]
\[ = \int \left( 2 {cosec}^2 x - 1 - 2\cot x . \text{cosec x} \right)dx\]
\[ = \int2 {cosec}^2 x dx - \int1 dx - \int2\cot x . \text{cosec x} dx\]
\[ = - 2\cot x - x + \text{2 cosec x} + C\]
\[ = 2\left( \text{cosec x }- \cot x \right) - x + C\]
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