Question

$\int \frac{{\left({\sin }^{-1}x\right)}^3}{\sqrt{1-x^2}}dx$

Answer 1

$$\begin{gathered} \int \frac{{\left({\sin }^{-1}x\right)}^3}{\sqrt{1-x^2}}dx \\ \mathrm{Let}{\sin }^{-1}x=t \\ \Rightarrow \frac{1}{\sqrt{1-x^2}}dx=dt \\ \mathrm{Now},\int \frac{{\left({\sin }^{-1}x\right)}^3}{\sqrt{1-x^2}}dx \\ =\int t^{3}dt \\ =\frac{t^4}{4}+\mathrm{C} \\ =\frac{{\left({\sin }^{-1}x\right)}^4}{4}+\mathrm{C} \end{gathered}$$