Question

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

Answer 1

\[\int {x^2}_{II} . \sin^{- 1}_I x \text{ dx }\]
\[ = \sin^{- 1}_{} x\int x^2 dx - \int\left\{ \frac{d}{dx}\left( \sin^{- 1}_{} x \right)\int x^2 dx \right\}dx\]
\[ = \sin^{- 1} x . \frac{x^3}{3} - \int\frac{1}{\sqrt{1 - x^2}} \frac{x^3}{3}dx\]
\[\text{  Let 1} - x^2 = t\]
\[ \Rightarrow x^2 = 1 - t\]
\[ \Rightarrow -\text{  2x dx } = dt\]
\[ \Rightarrow\text{  x dx } = - \frac{dt}{2}\]

\[ \therefore \int {x^2}_{} . \sin^{- 1}_{} \text{ x dx } = \sin^{- 1} x . \frac{x^3}{3} - \frac{1}{3}\int \frac{x^2 . x}{\sqrt{1 - x^2}}dx\]

\[ = \sin^{- 1} x . \frac{x^3}{3} - \frac{1}{6}\int \frac{\left( 1 - t \right)}{\sqrt{t}}dt\]
\[ = \sin^{- 1} x . \frac{x^3}{3} + \frac{1}{6}\int t^{- \frac{1}{2}} dt - \frac{1}{6}\int t^\frac{1}{2} dt\]
\[ = \sin^{- 1} x . \frac{x^3}{3} + \frac{1}{6} \times 2\sqrt{t} - \frac{1}{6} \times \frac{2}{3} t^\frac{3}{2} + C\]
\[ = \sin^{- 1} x . \frac{x^3}{3} + \frac{\sqrt{1 - x^2}}{3} - \frac{1}{9} \left( 1 - x^2 \right)^\frac{3}{2} + C \left( \because 1 - x^2 = t \right)\]