Question

\[\int\frac{x^3 \sin^{- 1} x^2}{\sqrt{1 - x^4}} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int \frac{x^3 \times \sin^{- 1} x^2}{\sqrt{1 - x^4}}dx\]
\[\text{ Putting } \sin^{- 1} x^2 = t \]
\[ \Rightarrow x^2 = \sin t\]
\[ \Rightarrow \frac{1 \times 2x  \text{ dx }}{\sqrt{1 - \left( x^2 \right)^2}} = dt\]
\[ \Rightarrow \frac{x    \text{ dx }}{\sqrt{1 - x^4}} = \frac{dt}{2}\]
\[ \therefore I = \int x^2 . \frac{\sin^{- 1} x^2}{\sqrt{1 - x^4}} . \text{ x   dx }\]
\[ = \int \left( \sin t \right) . t . \frac{dt}{2}\]
\[ = \frac{1}{2}\int t_I . \sin_{II} t    \text{ dt }\]
\[ = \frac{1}{2}\left[ t\int\text{ sin  t  dt} - \int\left\{ \frac{d}{dt}\left( t \right)\int\text{ sin  t  dt } \right\}dt \right]\]
\[ = \frac{1}{2} \left[ t . \left( - \cos t \right) - \int 1 . \left( - \cos t \right) dt \right]\]
\[ = \frac{1}{2}\left[ - t \cos t + \sin t \right] + C\]
\[ = \frac{1}{2} \left[ - t\sqrt{1 - \sin^2 t} + \sin t \right] + C\]
\[ = \frac{1}{2} \left[ - \sin^{- 1} \left( x^2 \right) \sqrt{1 - x^4} + x^2 \right] + C\]