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प्रश्न
उत्तर
\[\int\sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}}dx\]
\[ = \int\sqrt{\frac{\cos^2 x + \sin^2 x - 2 \sin x \cos x}{\cos^2 x + \sin^2 x + 2 \sin x \cos x}} dx\]
\[ = \sqrt{\frac{\left( \cos x - \sin x \right)^2}{\left( \cos x + \sin x \right)^2}}dx\]
\[ = \int\frac{\cos x - \sin x}{\cos x + \sin x}dx\]
\[ = \int\frac{1 - \tan x}{1 + \tan x}dx\]
\[ = \int\tan \left( \frac{\pi}{4} - x \right)dx\]
\[ = \frac{1}{- 1}\text{ln}\left| \sec \left( \frac{\pi}{4} - x \right) \right| \left[ \because \int\tan \left( ax + b \right)dx = \frac{1}{a}\text{ln }\left| \sec \left( ax + b \right) \right| + C \right]\]
\[ = \frac{- \text{In} \left| \text{cos} \left( \frac{\pi}{4} - x \right) \right|}{- 1} + C\]
\[ = \text{ln }\left| \text{cos} \left( \frac{\pi}{4} - x \right) \right| + C\]
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