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∫ Sin X √ 1 − Cos 2 X D X - Mathematics

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प्रश्न

` ∫   sin x  \sqrt (1-cos 2x)    dx `

 

योग
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उत्तर

` ∫   sin x  .  \sqrt (1-cos 2x)    dx `

` ∫   sin x  \sqrt (2 sin^2 x )    dx `   `[∴ 1 - cos 2A = 2 sin^2 A]`

` =  \sqrt2    ∫   sin^2 x    dx `
\[ = \sqrt{2}\int\left( \frac{1 - \cos 2x}{2} \right)dx\]
\[ = \frac{1}{\sqrt{2}}\int\left( 1 - \cos 2x \right)dx\]
\[ = \frac{1}{\sqrt{2}}\left[ x - \frac{\sin 2x}{2} \right] + C\]

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.06 [पृष्ठ ३६]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.06 | Q 8 | पृष्ठ ३६

वीडियो ट्यूटोरियलVIEW ALL [1]

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