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प्रश्न
\[\int\frac{1}{1 - \sin x} dx\]
बेरीज
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उत्तर
\[\int\frac{dx}{1 - \sin x}\]
\[ = \int\frac{\left( 1 + \sin x \right)}{\left( 1 - \sin x \right) \times \left( 1 + \sin x \right)}dx\]
\[ = \int\left( \frac{1 + \sin x}{1 - \sin^2 x} \right)dx\]
\[ = \int\left( \frac{1 + \sin x}{\cos^2 x} \right)dx\]
\[ = \int\left( \frac{1}{\cos^2 x} + \frac{\sin x}{\cos x} \times \frac{1}{\cos x} \right)dx\]
\[ = \int\left( \sec^2 x + \sec x \tan x \right)dx\]
\[ = \tan x + \sec x + C\]
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