Answer
$$\int x\sqrt{1-x^2}dx=-\frac{1}{3}\sqrt{(1-x^2)^3}+C$$
Work Step by Step
$$A=\int x\sqrt{1-x^2}dx$$
Let $u=1-x^2$
Then $du=(1-x^2)'dx=-2xdx$. So $xdx=-\frac{1}{2}du$
Substitute into $A$, we have $$A=\int \sqrt u(-\frac{1}{2})du$$ $$A=-\frac{1}{2}\int u^{1/2}du$$ $$A=-\frac{1}{2}\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=-\frac{\sqrt{u^3}}{3}+C$$ $$A=-\frac{1}{3}\sqrt{(1-x^2)^3}+C$$
