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प्रश्न
\[\int\frac{1 + \cos x}{1 - \cos x} dx\]
बेरीज
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उत्तर
\[\int\left( \frac{1 + \cos x}{1 - \cos x} \right) dx\]
\[ = \int\left( \frac{2 \cos^2 \frac{x}{2}}{2 \sin^2 \frac{x}{2}} \right) dx \left[ \therefore 1 + \cos x = 2 \cos^2 \frac{x}{2} \text{and} 1 - \cos x = 2 \sin^2 \frac{x}{2} \right]\]
\[ = \int \cot^2 \frac{x}{2} dx\]
` = ∫ ( "cosec"^2 x/2 -1)` dx
\[ = \frac{- \cot \left( \frac{x}{2} \right)}{\frac{1}{2}} - x + C\]
\[ = - 2 \cot \left( \frac{x}{2} \right) - x + C\]
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