Question

\[\int \left( \sin^{- 1} x \right)^3 dx\]

Answer 1

\[\text{We have}, \]

\[I = \int \left( \sin^{- 1} x \right)^3 dx\]

\[\text{ Let,} \sin^{- 1} x = t\]

\[ \Rightarrow \sin t = x \Rightarrow \cos t = \sqrt{1 - x^2}\]

\[\text{Differentiating both sides we get}\]

\[\text{ cos t dt = dx}\]

\[\text{Now, integral becomes}\]

\[I = \int \left( \sin^{- 1} x \right)^3 dx\]

\[ = \int t^3 \text{ cos  t dt }\]

\[ = t^3 \sin t - \int3 t^2 \text{ sin t dt}.......... \left( \text{ Using by parts} \right)\]

\[ = t^3 \sin t - 3\int t^2 \text{ sin  t  dt } \]

\[ = t^3 \sin t - 3\left[ - t^2 \cos t - \int - 2t \text{ cos t dt} \right]\]

\[ = t^3 \sin t + 3 t^2 \cos t - 6\int t \text{ cos t dt }\]

\[ = t^3 \sin t + 3 t^2 \cos t - 6\left[ t\sin t - \int\text{ sin t dt} \right]\]

\[ = t^3 \sin t + 3 t^2 \cos t - 6\left[ t \sin t + \cos t \right] + C\]

\[ = \left( \sin^{- 1} x \right)^3 x + 3 \left( \sin^{- 1} x \right)^2 \sqrt{1 - x^2} - 6\left( \sin^{- 1} x \right) x - 6\sqrt{1 - x^2} + C\]

\[ = x\left( \sin^{- 1} x \right)\left[ \left( \sin^{- 1} x \right)^2 - 6 \right] + 3\left[ \left( \sin^{- 1} x \right)^2 - 2 \right]\sqrt{1 - x^2} + C\]