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प्रश्न
\[\int\frac{1}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2} dx\]
उत्तर
\[\int\frac{dx}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2}\]
\[Let, \sin^{- 1} x = t\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{1}{\sqrt{1 - x^2}} dx = dt\]
\[Now, \int\frac{dx}{\sqrt{1 - x^2} \left( \sin^{- 1} x \right)^2}\]
\[ = \int\frac{dt}{t^2}\]
\[ = \int t^{- 2} dt\]
\[ = \frac{t^{- 2 + 1}}{- 2 + 1} + C\]
\[ = \frac{- 1}{t} + C\]
\[ = - \frac{1}{\sin^{- 1} x} + C\]
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