Mathematics

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


SOLUTION
So, $$\displaystyle \int \dfrac{x^2.x\sin^{-1}(x^2)}{\sqrt{1-(x^2)^2}}dx$$
$$\displaystyle =\dfrac{1}{2}\int 2m\, \sin^{-1}m.(dm)$$
$$\left[ \begin{matrix}  Let\,x^2=m\\ \begin{matrix} \therefore dm=2xdx \\ \begin{matrix} and\, \dfrac{d}{dx}(\sin^{-1}m) \\dm=\dfrac{1}{\sqrt{1-x^2}}  \end{matrix} \end{matrix} \end{matrix} \right] $$
$$\displaystyle =\dfrac{1}{2}\left[\dfrac{1}{2}m^2\sin^{-1}m-\int(\dfrac{m^2}{2\sqrt{1-\sin^2}})dx\right]$$          By integration by parts
$$\displaystyle \dfrac{1}{2}\int \sin^2 u dx$$         Let   $$m=\sin (u)$$
$$\displaystyle \int \dfrac{1-\cos 2u}{2}dv$$
$$\displaystyle \dfrac{1}{4}\left[\int 1.dv-\int \cos 2u dv\right]$$
$$\dfrac{1}{4}(u-\dfrac{1}{2}\sin 2u)$$

On back Substituting :
$$\dfrac{1}{4}\left[\sin^{-1}(m)-\dfrac{1}{2}\sin (2\sin^{-1}m)\right]$$
$$2\left[\dfrac{1}{2}m^2 \sin^{-1}(m)-\dfrac{1}{4}(\sin^{-1}m-\dfrac{1}{2}\sin (2\sin^{-1}(m)))\right]$$
$$m^2\sin^{-1}(m)-\dfrac{1}{2}\sin^{-1}(m)+\dfrac{1}{2}\sin(2\sin^{-1}(m)+c)$$
as $$(m=x^2)$$
So value of integer is :-
$$(x^2)^2\sin^{-1}(x^2)-\dfrac{1}{2}\sin^{-1}(x^2)+\dfrac{1}{2}\sin (2\sin^{-1}(x^2))+c$$
$$\{x^4.\sin^{-1}(x^2)-\dfrac{1}{2}\sin^{-1}(x^2)(2\sin^{-1}(x^2))+e\}$$
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