Answer
$$\int x\sqrt{1-x^2}dx=-\frac{1}{3}(1-x^2)^{3/2}+c$$
Work Step by Step
To solve the integral $\int x\sqrt{1-x^2}dx$ we will use substitution $u=1-x^2$ which gives us $du=-2xdx\Rightarrow-\frac{1}{2}du=xdx$. Putting this in the integral we get:
$$\int x\sqrt{1-x^2}dx=\int\sqrt u\cdot(-\frac{1}{2}du)=-\frac{1}{2}\int u^{1/2}du=-\frac{1}{2}\frac{u^{3/2}}{\frac{3}{2}}+c=-\frac{1}{2}\frac{2}{3}u^{3/2}+c=-\frac{1}{3}u^{3/2}+c,$$
where $c$ is arbitrary constant. Now we have to express solution in terms of $x$ by expressing $u$ in terms of $x$:
$$\int x\sqrt{1-x^2}dx=-\frac{1}{3}u^{3/2}+c=-\frac{1}{3}(1-x^2)^{3/2}+c$$
