Question

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

Answer 1

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

$=\frac{1}{2}\int 2m{\sin }^{-1}m.\left(dm\right)$

$\left[\begin{array}{c}Let{x}^2=m\\ \begin{array}{c}\therefore dm=2xdx\\ \begin{array}{c}and\frac{d}{dx}\left({\sin }^{-1}m\right)\\ dm=\frac{1}{\sqrt{1-x^2}}\end{array}\end{array}\end{array}\right]$

$=\frac{1}{2}[\frac{1}{2}{m}^2{\sin }^{-1}m-\int (\frac{m^2}{2\sqrt{1-{\sin }^2}}\left)dx\right]$ By integration by parts

$\frac{1}{2}\int {\sin }^{2}udx$ Let $m=\sin \left(u\right)$

$\int \frac{1-\cos 2u}{2}dv$

$\frac{1}{4}[\int 1.dv-\int \cos 2udv]$

$\frac{1}{4}(u-\frac{1}{2}\sin 2u)$

On back Substituting :

$\frac{1}{4}\left[{\sin }^{-1}\right(m)-\frac{1}{2}\sin (2{\sin }^{-1}m\left)\right]$

$2\left[\frac{1}{2}{m}^2{\sin }^{-1}\right(m)-\frac{1}{4}({\sin }^{-1}m-\frac{1}{2}\sin \left(2{\sin }^{-1}\right(m\left)\right)\left)\right]$

$m^2{\sin }^{-1}\left(m\right)-\frac{1}{2}{\sin }^{-1}\left(m\right)+\frac{1}{2}\sin \left(2{\sin }^{-1}\right(m)+c)$

as $(m=x^2)$

So value of integer is :-

$\left(x^2{)}^2{\sin }^{-1}\right(x^2)-\frac{1}{2}{\sin }^{-1}(x^2)+\frac{1}{2}\sin (2{\sin }^{-1}\left(x^2\right))+c$

$\{x^4.{\sin }^{-1}(x^2)-\frac{1}{2}{\sin }^{-1}(x^2\left)\right(2{\sin }^{-1}\left(x^2\right))+e\}$