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Evaluate: ∫ √ 1 − Cos 2 X 2 D X

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प्रश्न

Evaluate:

\[\int\sqrt{\frac{1 - \cos 2x}{2}}dx\]
योग
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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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अध्याय 18: Indefinite Integrals - Exercise 19.01 [पृष्ठ ४]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 18 Indefinite Integrals
Exercise 19.01 | Q 2.2 | पृष्ठ ४

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