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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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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