# How to integrate by parts, without changing variable

How can I solve:

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

without changing variables, by parts?

• Have you tried rationalizing the denominator? – daOnlyBG Oct 15 '14 at 14:57
• Why is it important to not change variables? – djhaskin987 Oct 15 '14 at 20:02

Hint: Integrate by parts 2 times and remember that $$(\arcsin(x))' = \frac{1}{\sqrt{1-x^2}}$$
$$\int\frac{x^2}{\sqrt{1-x^2}}\,dx=-\frac{1}{2}\int\frac{x}{\sqrt{1-x^2}}\,d(1-x^2)=-\frac{1}{2}\Big(2x\sqrt{1-x^2}-\int\sqrt{1-x^2}\,dx\Big)=..$$\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}