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How can I solve:

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

without changing variables, by parts?

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  • $\begingroup$ Have you tried rationalizing the denominator? $\endgroup$ – daOnlyBG Oct 15 '14 at 14:57
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    $\begingroup$ Why is it important to not change variables? $\endgroup$ – djhaskin987 Oct 15 '14 at 20:02
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Hint: Integrate by parts 2 times and remember that $$(\arcsin(x))' = \frac{1}{\sqrt{1-x^2}}$$

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$$\int\frac{x^2}{\sqrt{1-x^2}}\,dx=-\frac{1}{2}\int\frac{x}{\sqrt{1-x^2}}\,d(1-x^2)=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\int\sqrt{1-x^2}\,dx\Big)=..$$\begin{align}&...=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}\int(\sqrt{1-x^2}+\frac{1-x^2}{\sqrt{1-x^2}})\,dx\Big)\\&=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}\int(\sqrt{1-x^2}+\frac{1}{\sqrt{1-x^2}}+\frac{-x^2}{\sqrt{1-x^2}})\,dx\Big)\\&=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}\int(\sqrt{1-x^2}+\frac{-x^2}{\sqrt{1-x^2}})\,dx-\frac{1}{2}\int\frac{1}{\sqrt{1-x^2}}\,dx\Big)\\&= -\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}\int\,d(x\sqrt{1-x^2})-\frac{1}{2}\int\frac{1}{\sqrt{1-x^2}}\,dx\Big)\\&=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\frac{1}{2}x\sqrt{1-x^2}-\frac{1}{2}\arcsin{(x)}+c\Big)\\&= -\frac{3}{4}x\sqrt{1-x^2}+\frac{1}{4}\arcsin{(x)}+c \end{align}

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  • $\begingroup$ That's what I did, yet how to integrate the last right integral without substitution? $\endgroup$ – Timbuc Oct 15 '14 at 15:05

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