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
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