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Question Number 21810 by j.masanja06@gmail.com last updated on 04/Oct/17

$$integrate$$$$\int \frac{x^2}{\sqrt{1-x^2}}dx$$

Answered by sma3l2996 last updated on 04/Oct/17

$$x=sint\Rightarrow dx=costdt=\sqrt{1-{sin}^{2}t}dt$$$$dx=\sqrt{1-x^2}dt\iff dt=\frac{dx}{\sqrt{1-x^2}}$$$$\int \frac{x^2}{\sqrt{1-x^2}}dx=\int {sin}^2\left(t\right)dt=\int \frac{1-cos\left(2t\right)}{2}dt$$$$=\frac{1}{2}(t-\frac{1}{2}sin\left(2t\right))+C=\frac{1}{2}(t-sin\left(t\right)cos\left(t\right))+C$$$$\int \frac{x^2}{\sqrt{1-x^2}}dx=\frac{1}{2}({sin}^{-1}\left(x\right)-x\sqrt{1-x^2})+C$$

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