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प्रश्न
Evaluate:
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उत्तर
\[\int\sqrt{\frac{1 - \cos 2x}{2}}dx\]
`=∫ \sqrt {{2 sin_2 x }/{2}} dx` `[∴ 1 - cos 2x = 2 sin ^2 x]`
\[ = \int\text{sin x dx}\]
\[ = - \cos x + C\]
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