Question

solve this integral

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Answer 1

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

To Integrate:

• $\dfrac{x^3}{\sqrt{1-x^2}}$

Solution:

$\displaystyle \implies \int \dfrac{x^3}{\sqrt{1-x^2}}dx \\\\\implies \int x^2\dfrac{d}{dx}\left(-\sqrt{1-x^2} \right)dx \\\\\implies -x^2\sqrt{1-x^2} + \int \dfrac{d}{dx}(x^2)\sqrt{1-x^2}dx \\\\\implies -x^2\sqrt{1-x^2} + \int 2x \sqrt{1-x^2}dx \\\\\implies -x^2\sqrt{1-x^2} + \int \dfrac{d}{dx}\left(-\dfrac{2}{3}(1-x^2)^{\frac{3}{2}}\right)dx \\\\\implies -x^2\sqrt{1-x^2} -\dfrac{2}{3}(1-x^2)^{\frac{3}{2}} + C \\\\\implies \sqrt{1-x^2} \left(-x^2 -\dfrac{2}{3}(1-x^2)\right) + C \\\\\implies \sqrt{1-x^2} \left(-\dfrac{3x^2}{3} -\dfrac{2}{3} + \dfrac{2x^2}{3}\right) + C \\\\\implies \sqrt{1-x^2} \left(-\dfrac{x^2}{3} -\dfrac{2}{3}\right) + C \\\\\bold{\implies \int \dfrac{x^3}{\sqrt{1-x^2}}dx \:=\: -\dfrac{1}{3}(x^2+2)\sqrt{1-x^2} + C}$

Hope it helps!!!

Answer 2

Answer

solved see the answer