Mathematics

$I = \int {{{{x^2}} \over {\sqrt {1 - {x^6}} }}dx}$

SOLUTION
$I= \displaystyle\int \dfrac{x^{2}}{\sqrt{1-x^{6}}}dx$
Let $x^{3}=4\Rightarrow 3x^{2}=dx= du$
$\therefore I=\dfrac{1}{3} \displaystyle\int \dfrac{du}{\sqrt{1-u^{2}}}$
$=\dfrac{1}{3}\sin^{-1}(u)+c$
$=\dfrac{1}{3}\sin^{-1}(x^{3})+c$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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