Question

\[\int\frac{x \sin^{- 1} x^2}{\sqrt{1 - x^4}} dx\]

Answer 1

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