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प्रश्न
\[\int \cos^{- 1} \left( \sin x \right) dx\]
बेरीज
उत्तर
\[\int \cos^{- 1} \left( \sin x \right)dx\]
\[ = \int \cos^{- 1} \left( \cos\left( \frac{\pi}{2} - x \right) \right)dx \left[ \therefore \sin x = \cos\left( \frac{\pi}{2} - x \right) \right]\]
\[ = \int\left( \frac{\pi}{2} - x \right)dx\]
` = (π x)/ 2 - x^2 / 2 + c `
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