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प्रश्न
\[\int\frac{1}{\left( \sin^{- 1} x \right) \sqrt{1 - x^2}} \text{ dx} \]
योग
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उत्तर
\[\text{ Let I} = \int\frac{1}{\sin^{- 1} x \cdot \sqrt{1 - x^2}}dx\]
\[\text{ Putting sin}^{- 1} x = t\]
\[ \Rightarrow \frac{dx}{\sqrt{1 - x^2}} = dt\]
\[ \therefore I = \int\frac{dt}{t}\]
\[ = \text{ ln }\left| t \right| + C\]
` = \text{ ln } | sin ^-1 x| + c ( ∵ t = sin ^-1 x ) `
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