Evaluate the integral: I = integral (x²)/(sqrt(1-x²)) dx

Evaluate the integral: $$ I = \int \frac{x^2}{\sqrt{1-x^2}}\,dx. $$

Explanation of Method

Writing $x^2=-(1-x^2)+1$ splits the integral into $-\int\sqrt{1-x^2}\,dx+\int dx/\sqrt{1-x^2}$, both standard forms handled by $x=\sin\theta$.

Rewrite $x^2$ in terms of $1-x^2$.
Split into two integrals.
Integrate each piece.

Step 1 · Rewrite and split

$$ \frac{x^2}{\sqrt{1-x^2}} = -\sqrt{1-x^2} + \frac1{\sqrt{1-x^2}}. $$

Step 2 · Second piece

$$ \int\frac{dx}{\sqrt{1-x^2}} = \arcsin(x) + C_2. $$

Step 3 · First piece

Let $x=\sin\theta$, so $\sqrt{1-x^2}=\cos\theta$:

$$ \int\sqrt{1-x^2}\,dx = \int\cos^2\theta\,d\theta = \frac\theta2+\frac{\sin\theta\cos\theta}2 + C_1 = \frac{\arcsin x}2 + \frac{x\sqrt{1-x^2}}2 + C_1. $$

Step 4 · Combine

$$ I = -\left[\frac{\arcsin x}2+\frac{x\sqrt{1-x^2}}2\right] + \arcsin x + C = \frac{\arcsin x}2 - \frac{x\sqrt{1-x^2}}2 + C. $$

Summary Table

Piece	Antiderivative
$-\sqrt{1-x^2}$	$-\tfrac12\arcsin x-\tfrac12x\sqrt{1-x^2}$
$1/\sqrt{1-x^2}$	$\arcsin x$

Additional Notes

The domain is $-1<x<1$.
The result was verified by differentiation at $x=1/2$: the derivative matches the integrand $x^2/\sqrt{1-x^2}$ exactly.

Result

The solution to the integral is:

$$ I = \frac{\arcsin x}2 - \frac{x\sqrt{1-x^2}}2 + C. $$