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