Question

$\int \frac{x^3{\sin }^{-1}{x}^2}{\sqrt{1-x^4}}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{x^3\times {\sin }^{-1}{x}^2}{\sqrt{1-x^4}}dx \\ \mathrm{Putting}{\sin }^{-1}{x}^2=t \\ \Rightarrow x^2=\sin t \\ \Rightarrow \frac{1\times 2xdx}{\sqrt{1-{\left(x^2\right)}^2}}=dt \\ \Rightarrow \frac{xdx}{\sqrt{1-x^4}}=\frac{dt}{2} \\ \therefore I=\int x^2.\frac{{\sin }^{-1}{x}^2}{\sqrt{1-x^4}}.xdx \\ =\int \left(\sin t\right).t.\frac{dt}{2} \\ =\frac{1}{2}\int \underset{\mathrm{I}}{t}.\underset{\mathrm{II}}{\sin t}dt \\ =\frac{1}{2}\left[t\int \sin tdt-\int \left\{\frac{d}{dt}\left(t\right)\int \sin tdt\right\}dt\right] \\ =\frac{1}{2}\left[t.\left(-\cos t\right)-\int 1.\left(-\cos t\right)dt\right] \\ =\frac{1}{2}\left[-t\cos t+\sin t\right]+C \\ =\frac{1}{2}\left[-t\sqrt{1-{\sin }^{2}t}+\sin t\right]+C \\ =\frac{1}{2}\left[-{\sin }^{-1}\left(x^2\right)\sqrt{1-x^4}+x^2\right]+C \end{gathered}$$