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