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
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