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