Question

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

Answer 1

Consider the given integral.

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

Let $t={\sin }^{-1}{x}^2$

$\frac{dt}{dx}=\frac{1}{\sqrt{1-x^4}}\times 2x$

$\frac{dt}{2}=\frac{xdx}{\sqrt{1-x^4}}$

Therefore,

$I=\frac{1}{2}\int t\sin tdt$

$I=\frac{1}{2}[t(-\cos t)-\int 1(-\cos t)dt]$

$I=\frac{1}{2}[-t\left(\cos t\right)+\int \cos tdt]$

$I=\frac{1}{2}[-t\left(\cos t\right)+\sin t]+C$

$I=\frac{1}{2}[-{\sin }^{-1}{x}^2\left(\cos \left({\sin }^{-1}{x}^2\right)\right)+\sin \left({\sin }^{-1}{x}^2\right)]+C$

$I=\frac{1}{2}[x^2-\cos \left({\sin }^{-1}{x}^2\right){\sin }^{-1}{x}^2]+C$

Hence, this is the answer.