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प्रश्न
\[\int\frac{1}{\sqrt{2x - x^2}} dx\]
योग
उत्तर
\[\int\frac{dx}{\sqrt{2x - x^2}}\]
\[ = \int\frac{dx}{\sqrt{2x - x^2 - 1 + 1}}\]
\[ = \int\frac{dx}{\sqrt{1 - \left( x^2 - 2x + 1 \right)}}\]
\[ = \int\frac{dx}{\sqrt{1 - \left( x - 1 \right)^2}} \]
\[ = \sin^{- 1} \left( x - 1 \right) + C \left[ \because \int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{- 1} \left( \frac{x}{a} \right) + C \right]\]
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