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प्रश्न
\[\int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}} \text{ dx }\]
योग
उत्तर
\[\text{ Let I } = \int\frac{\left( \sin^{- 1} x \right)^3}{\sqrt{1 - x^2}}dx\]
\[\text{ Putting sin}^{- 1} x = t\]
\[ \Rightarrow \frac{dx}{\sqrt{1 - x^2}} = dt\]
\[ \therefore I = \int t^3 \cdot dt\]
\[ = \frac{t^4}{4} + C\]
\[ = \frac{\left( \sin^{- 1} x \right)^4}{4} + C......... \left( \because t = \sin^{- 1} x \right)\]
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