Mathematics

# $\displaystyle \int {\dfrac{{x{{\sin }^{ - 1}}{x^2}dx}}{{\sqrt {1 - {x^4}} }}} = \,\ ?$

##### SOLUTION
Putting $\sin^{-1}x^2=t$

diff wrt $t$

$=\dfrac{1}{\sqrt{1-x^4}}(2x).dx=dt$

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

$\displaystyle\therefore I=\int\dfrac{t.dt}{2}$

$I=\dfrac{1}{2}\left(\dfrac{t^2}{2}\right)+C$

$=\dfrac{1}{4}(\sin^{-1}x^2)^2$

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

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